y = x h = x+ a 1 or x = y a 1

Lecture 6 Solution Of Bi-quadratic Equations By Using Descartes Method Let the given Bi-quadratic equation be x 4 +4a x 3 +6a x +4a 3 x+a 4 = 0. Now to remove the second term, we have to diminish the roots by h = a a 0 y = x h = x+ a a 0. or x = y a a 0. Substituting the value of x in the given equation, the transformed equation becomes a 4 0y 4 +6a 0y (a 0 a a )+4a 0 y(a 3a 0 a a +a 0a 3 )+(6a 0 a a 3a 4 4a 0a a 3 +a 3 0a 4 ) = 0 a 4 0y 4 +6a 0 y (a 0 a a )+4a 0 y(a 3a 0 a a +a 0a 3 )+[a 0(3a 4a a 3 +a 0 a 4 ) 3(a 0 a a ) ] = 0 Putting y = z a 0,H = a 0 a a,g = a 3 3a 0 a a +a 0a and 3a 4a a 3 +a 0 a 4, we get the transformed equation as z 4 +6Hz +4Gz +(a 0I 3H ) = 0, (0.) where a 0 x+a = z. Let the quadratic factor of (0.) be z +pz +q and z pz +q, so that (z +pz +q)(z pz +q ) = 0 (0.) is the same as (0.). Comparing the coefficients of the left hand side of (0.) and (0.), we get q +q p = 6H or q +q = 6H +p (0.3) p(q q) = 4G or q q = 4G p (0.4) Also, qq = a 0I 3H or 4qq = 4a 0I H. Since, we have 4qq = (q +q) (q q) (0.5) Putting p = t, we get 4a 0I H = p 4 +p H +36H 6G p or p 6 +p 4 H +(48H 4a 0I)p 6G = 0 t 3 +Ht +4(H a 0I)t 6G = 0 (0.6)

Equation (0.6) is the called Descartes Resolvent. Now, we can find t from (0.6) and so also p. From (0.3) and (0.4), we can find q and q. Thus, the two quadratic equations z + pz + q = 0 and z pz +q = 0 can be solved. Let z,z,z 3,z 4 be the roots of these equations, then corresponding to these we can find the four values of x from the relation a 0 x+a = z. Example. Solve Sol. Let the quadratic factors be x 4 6x +8x 3 = 0. (0.7) (x +px+q)(x px+q ) = 0 (0.8) = x 4 px 3 +q x +p x +pq x+qx pqx+qq = 0 = x 4 +(p+q p )x +p(q q)x+qq = 0 (0.9) Comparing the coefficients of the left hand side of (0.7) and (0.9), we get q +q p = 6 or q +q = p 6 (0.0) p(q q) = 8 or q q = 8 p (0.) Also, qq = 3 Since, we have (q +q) (q q) = 4qq ( ) 8 (p 6) = p = p 6 +36p p 4 64+p = 0 = p 6 p 4 ++48p 64 = 0 = (p ) 3 (p ) +48(p ) 64 = 0 Put p = t, we get t 3 t +48t 64 = 0 (0.) test the cubic for numbers which are perfect squares i,e. t =,4,9,6,5, t = 4, satisfies equation (0.). So that p = t = 4 = p = 4 = p =. Also, from (0.), we have q q = 8 q q = 4 (0.3)

3 From (0.0), we get q +q = 4 6 = q +q = (0.4) Adding (0.3) and (0.4), we get q = = q = Put q = in (0.4), we have q = = q = 3. Thus substitute p,q,q in (0.8), we get (x +x 3)(x x+) = 0 = (x +x 3) = 0 or (x x+) = 0 = (x+3)(x ) = 0 or (x )(x ) = 0 = x =, 3 or x =, Hence the four roots are x =,,, 3. Example. Solve x 4 4x 3 +6x +x 0 = 0. Sol. we first remove the second term, so we diminish roots of the given equation by h, where the transformed equation is 4 Sum of roots h = Degree of equation = = 4 4 6 0 3 3 4 3 3 4 6 5 0 0 Let the quadratic factors of the above equation be y 4 +5y 6 = 0 (0.5) (y +py +q)(y py +q ) = 0 (0.6)

4 Comparing the coefficients of the left hand side of (0.5) and (0.6), we get q +q p = 0 or q +q = p (0.7) p(q q) = 5 or q q = 5 p (0.8) Also, qq = 6 Since, we have (q +q) (q q) = 4qq ( ) 5 (p ) = 4 p = p 6 +4p 5 = 0 = (p ) 3 +4(p ) 5 = 0 Put p = t, we get t 3 +4t 5 = 0 (0.9) test the cubic for numbers which are perfect squares i,e. t =,4,9,6,5, t =, satisfies equation (0.9). So that p = t = = p = = p =. Also, from (0.7), we have q +q = (0.0) Adding (0.0) and (0.), we get Put q = 3 in (0.), we have Thus substitute p,q,q in (0.6), we get Which gives y =,, ±i. Since x = y +, gives x =,,, 3±i q q = 5 q q = 5 (0.) q = 3 q = (y +y )(y y +3) = 0 are the required roots.

Lecture 5 SOLUTION OF CUBIC EQUATIONS BY USING CARDEN S METHOD Let the cubic equation be a 0 x 3 +3a x +3a x+a 3 = 0 (0.) We first remove the second term of equation (0.), so setting h = a. a 0 Put y = x h, gives y = x+ a, or x = y a. a 0 a 0 Substituting this value of x in (0.), we get ( a 0 y a ) 3 ( +3a y a ) ( +3a y a ) +a 3 = 0 a 0 a 0 a 0 Putting y = z a 0 in this equation, we get or a 3 0y 3 +3a 0 y(a 0 a a )+(a 3 3a 0 a a +a 0a 3 ) = 0 z 3 +3z(a 0 a a )+(a 3 3a 0 a a +a 0a 3 ) = 0 or z 3 +3Hz +G = 0 (0.) where H = a 0 a a and G = a 3 3a 0 a a +a 0a 3. Now to solve (0.), we put Cubing it, we get z = u+v. z 3 = (u+v) 3 or z 3 3uvz (u 3 +v 3 ) = 0. (0.3) On equating coefficients of z and constant terms of (0.) and (0.3), we have and u 3 +v 3 = G u 3 v 3 = H 3. Now a quadratic equation in t whose roots are u 3 and v 3 is uv = H t (u 3 v 3 )t+u 3 v 3 = 0 or t Gt+ H 3 = 0. This gives t = G± G +4H 3, which means that u 3 = G+ G +4H 3 and v 3 = G G +4H 3. [ G+ ] [ G u = +4H 3 3 G+ ] [ G +4H 3 3 G+ ] G +4H 3 3, ω, ω [ G ] [ G and v = +4H 3 3 G ] [ G +4H 3 3 G ] G +4H 3 3, ω, ω, where ω,ω the complex roots of unity.

[ G+ ] [ G Putting +4H 3 3 G ] G +4H 3 3 = α and = β, we obtain three values of z as α+β,ωα+ω β,ω α+ωβ and then the values of x can be obtained from z = a 0 x+a or x = z a a 0. Example. Solve the cubic equation x 3 6x 9 = 0, by Carden s method. Sol. The given cubic equation is x 3 6x 9 = 0. (0.4) Here nd term is missing. Put x = u+v = x 3 = (u+v) 3 = u 3 +v 3 +3uv(u+v) Equating coefficients of (0.4) and (0.5), we get u 3 +v 3 = 9 and 3uv = 6 = uv = = u 3 v 3 = 8. Now take u 3 and v 3 as roots of quadratic in t, we have = x 3 3uvx (u 3 +v 3 ) = 0 (0.5) t (u 3 +v 3 )t+u 3 v 3 = 0 = t 9t+8 = 0 = (t )(t 8) = 0 = t =,8 u 3 = and v 3 = 8 = u = and v = is the one root. Now using synthetic division we have x +3x+3 = 0 = x = 3± 9 = x = 3± 3 = x = 3±i 3 Hence the roots are x = 3, 3±i 3 x = + = 3, 3 0 6 9 3 9 9 3 3 0 Example. Solve the cubic equation x 3 +3x +x 6 = 0. Sol. Here second term is present, so we first remove this second term and hence diminish it by h where 3 Sum of roots h = Degree of equation = 3 =

3 3 6 0 0 6 9 0 the transformed equation is Put y = u+v y 3 +9y 6 = 0 (0.6) = y 3 3uvy (u 3 +v 3 ) = 0 (0.7) Equating coefficients of (0.6) and (0.7), we get u 3 + v 3 = 6 and 3uv = 9 = uv = 3 = u 3 v 3 = 7. Now take u 3 and v 3 as roots of quadratic in t, we have t (u 3 +v 3 )t+u 3 v 3 = 0 = t 6t 7 = 0 = (t+)(t 7) = 0 = t =,7 u 3 = and v 3 = 7 = u = and v = 3 is the one root. Now using synthetic division we have y +y +3 = 0 = y = ± 4 5 = y = ± 48 = y = ±i 3 Since y = x h = x = y +h = y x = ( ), ±i 3 = x =, ±i 3. Hence the roots are x =, ±i 3. y = +3 =, 0 9 6 4 6 3 0 Nature Of The Roots Of The Cubic Equation z 3 +3Hz +G = 0

4 We know that the roots of the above cubic equation are given by z = x + y, where u 3 and v 3 are the roots of the quadratic equation t +Gt+H = 0. The discriminant of this quadratic equation is G +4H 3. Therefore, we have the following three cases. (i). G +4H 3 > 0. In this case both u 3 and v 3 are real, so u and v are also real. Therefore, the three values of z are u+v,uω +vω,uω +vω. The first root is real and the other two roots are complex. Hence in this case one root is real and other two are complex. (ii). G +4H 3 = 0. In this case both u 3 and v 3 are real and equal. Hence, the three roots of the cubic equation are u+v = u+u = u uω +vω = u(ω +ω ) = u uω +vω = u(ω +ω) = u. Thus in this case, all the roots are real and rational but two of them are equal. (iii). G +4H 3 < 0. In this case u 3 and v 3 are conjugate quantities and as such u and v are also conjugate complex quantities of the a±ib, where a and b are real. Hence, the roots of the cubic equation are and u+v = (a+ib)+(a ib) = a uω +vω = (a+ib)ω +(a ib)ω = a(ω +ω )+ib(ω ω ) = a b 3 using +ω +ω = 0 uω +vω = (a+ib)ω +(a ib)ω = a(ω +ω)+ib(ω ω) = a+b 3 using +ω +ω = 0. Thus, in this case all the three roots are real and distinct but unequal. Example 3. Discuss nature of roots of the equation x 3 +5x 4 = 0. Sol. Comparing it with z 3 +3Hz +G = 0, we get H = 5, G = 4 G +4H 3 = ( 4) +4(5) > 0. By case (i), the given equation has one real and two complex roots. Example 4. Discuss nature of roots of the equation x 3 +3x 4 = 0. Sol. Comparing it with z 3 +3Hz +G = 0, we get H =, G = 4 G +4H 3 = ( 4) +4() > 0. Again by case (i), the given equation has one real and two complex roots.

Unit-I: Differential Equations By Aijaz A. Najar LECTURE- Differential Equations: An equation involving derivatives of one or more dependent variables with respect to one or more independent variables is called a differential equation. Example.: The equations ( ) d y dy dx +xy =0 dx () d 4 x x dt 4 +5d +3x =sint dt () v t + v s v x + v y + v z =v (3) =0 (4) are all examples of differential equations. From the brief list of differential equations in Example., it is clear that the various variables and derivatives involved in a differential equation can occur in a variety of ways. Clearly some kind of classification must be made.

To begin with, we classify differential equations according to whether there is one or more than one independent variable involved. Definition : A differential equation involving ordinary derivatives of one or more dependent variables with respect to a single independent variable is called an ordinary differential equation. In Example., Equations () and () are ordinary differential equations. In Equation () the variable x is the single independent variable, and is a dependent variable. In Equation () the independent variable is t, whereas x is dependent. Definition : A differential equation involving partial derivatives of one or more dependent variables with respect to more than one independent variable is called a partial differential equation. For Example, Equations(3) and(4) in the Example. are partial differential equations. In Equation (3) the variables s and t are independent variables and v is a dependent variable. In Equation (4) there are three independent variables: x, y, and z in this equation v is dependent. We further classify differential equations, both ordinary and partial, according to the order of the highest derivative appearing in the equation. For this purpose we give the following definition. Definition 3: The order of the highest ordered derivative involved in a differential equation is called the order of the differential equation. The ordinary differential equation () is of the second order, since the highest derivative involved is a second derivative. Equation () is an ordinary differential equation of the fourth order. The partial differential equations (3) and (4) are of the first and second orders respectively. Definition 4: A linear ordinary differential equation of order n in the dependent variable and the independent variable x is an equation that is in, or can be expressed in, the form a 0 (x) dn y dx n +a (x) dn y dx n +a (x) dn y dx n +...+a n (x) dy dx +a n(x)y = b(x) where a 0 is not identically zero. From the above definition, it is clear (5)

(a) that the dependent variable and its various derivatives occur to the first degree only (b) that no products of and/or any of its derivatives are present and (c) that no transcendental functions of and/or its derivatives occur. Example.: The following ordinary differential equations are both linear. In each case y is the dependent variable. Observe that and its various derivatives occur to the first degree only and that no products of and/or any of its derivatives are present. ( ) d y dy dx +5 +6y =0 dx (6) d 4 y y dx 4 +xd3 dx +x3dy 3 dx =xex (7) Definition 5: A non linear ordinary differential equation is an ordinary differential equation that is not linear. Example.3: The following ordinary differential equations are all nonlinear: ( ) d y dy dx +5 +6y =0 (8) dx ( ) d y dy dx +5 +6y =0 (9) dx ( ) d y dy dx +5y +6y =0 (0) dx. Origin and Application of Differential Equations: Having classified differential equations in various ways, let us now consider briefly where, and how, such equations actually originate. In this way we shall obtain some indication of the great variety of subjects to which the theory and methods of differential equations may be applied. Differential equations occur in connection with numerous problems that are encountered in the various branches of science and engineering. We indicate a few such problems in the following list, which could easily be extended to fill many pages.. The problem of determining the motion of a projectile, rocket, satellite, or planet. 3

. The problem of determining the charge or current in an electric circuit. 3. The problem of the conduction of heat in a rod or in a slab. 4. The problem of determining the vibrations of a wire or a membrane. 5. The study of the rate of decomposition of a radioactive substance or the rate of growth of a population. 6. The study of the reactions of chemicals. 7. The problem of the determination of curves that have certain geometrical properties. The mathematical formulation of such problems give rise to differential equations. But just how does this occur? In the situations under consideration in each of the above problems the objects involved obey certain scientific laws. These laws involve various rates of change of one or more quantities with respect to other quantities. Let us re- call that such rates of change are expressed mathematically by derivatives. In the mathematical formulation of each of the above situations, the various rates of change are thus expressed by various derivatives and the scientific laws themselves become mathematical equations involving derivatives, that is, differential equations. LECTURE- Exact Differential Equations and integrating factors: The first-order differential equations to be studied in this chapter may be expressed in either the derivative form or the differential form dy dx = f(x,y) () M(x,y)dx+N(x,y)dy = 0 () An equation in one of these forms may readily be written in the other form. Definition 6: Let F be a function of two real variables such that F has 4

continuous first partial derivatives in a domain D. The total differential df of the function F is defined by the formula df(x,y) = F(x,y) x Definition 7: The expression dx+ F(x,y) dy forall (x,y)ǫd (3) y M(x,y)dx+N(x,y)dy (4) is called an exact differential in a domain D if there exists a function F of two real variables such that this expression equals the total differential df(x, y) for all (x,)ǫd. That is, expression (4) is an exact differential in D if there exists a function F such that F(x,y) x dx = M(x,y) and F(x,y) dy = N(x,y) (5) y If M(x, y)dx+n(x, y)dy is an exact differential, then the differential equation M(x,y)dx+N(x,y)dy = 0 is called an exact differential equation. Rule for solving Mdx+Ndx = 0 when it is exact:. First integrate w.r.t. y the terms in Mdx but treating y as a constant.. Then integrate w.r.t. y only those terms of Ndy which do not contain x. 3. Equate to some constant the sum of the results of () and (). That is, Mdx+ Ndy = c 4. If N has no term which is free from x, then Mdx = c, y constant. Example.: Solve the differential equation Solution: Here (x 3 +3xy )dx+(3x y +y 3 )dy = 0 M y = 6xy = N x Therefore, the equation is exact. Therefore, the solution is (x 3 +3xy )dx+ y 3 dy =c y constant x 4 +y 4 +6x y =k 5

LECTURE-3 Theorem (Necessary and Sufficient Condition for Exactness): Consider the differential equation M(x,y)dx+N(x,y)dy = 0 (6) where M and N have continuous first partial derivatives at all points (x,y) in a rectangular domain D. If the differential equation (6) is exact in D, then M(x,y) x = N(x,y) y for all (x,y) ǫ D (7) Conversely, if M(x,y) x = N(x,y) y for all (x,y) ǫ D then the differential equation (6) is exact in D. Proof: Necessary Part: If the differential equation (6) is exact in D, thenmdx + Ndy is an exact differential in D. By definition of an exact differential, there exists a function F such that F(x,y) x for all (x,y)ǫd. Then dx = M(x,y) and F(x,y) dy = N(x,y) y F(x,y) y x = M(x,y) y and F(x,y) x y = N(x,y) x Now, using the continuity of the partial derivatives, we have F(x,y) y x = F(x,y) y x and therefore M(x,y) y = N(x,y) x 6

Sufficient Part: we start with the hypothesis that F(x,y) y x = F(x,y) y x for all (x,y) ǫ D, and show that dx + Ndy = 0 is exact in D. This means that we must prove that there exists a function F such that F(x,y) x = M(x,y) and (8) F(x,y) y = N(x,y) (9) for all (x,y) ǫ D. We can certainly find some F(x,y) satisfying either (8) or (9), but what about both? Let us assume that F satisfies (8) and proceed. Then F(x,y) = M(x,y) x+φ(y) (0) where M(x,y) x indicates a partial integration with respect to x, holding y constant and φ is an arbitrary function of y only. This φ(y) is needed in (0) so that F(x,y) given by (0) will represent all solutions of (8). It corresponds to a constant of integration in the one-variable case. Differentiating (0) partially with respect to y, we obtain F(x,y) y = y M(x,y) x+ dφ(y) dy Now if (9) is to be satisfied, we must have N(x,y) = M(x,y) x+ dφ(y) y dy () and hence dφ(y) dy = N(x,y) y M(x, y) x () 7

Since φ is a function of y only, the derivative dφ must also be independent of dy x. That is, in order for () to hold, N(x,y) M(x, y) x (3) y must be independent of x. We show [ N(x,y) x y ] M(x,y) x = 0 we have x [ N(x,y) y ] M(x,y) x = N(x,y) x x y M(x,y) x and hence x [ N(x,y) y ] M(x,y) x = N(x,y) M(x,y) x y But by hypothesis, N(x,y) x = M(x,y) y Thus, [ N(x,y) x y ] M(x,y) x = 0 Hence, we can write φ(y) = [ N(x,y) y ] M(x,y) x y Substituting this into Equation (0), we have [ F(x,y) = M(x,y) x+ N(x,y) y ] M(x,y) x y (4) This F(x,y) thus satisfies both (8) and (9) for all (x,y)ǫd, and so Mdx+ Ndy = 0 is exact in D. 8

Example.: considertheequationy dx+xydy = 0,wehaveM(x,y) = y and N(x,y) = xy. Also N(x,y) x = y = M(x,y) y Thus the given differential equation is exact in any rectangular region D. Now, Consider ydx+xdy = 0, we have M(x,y) = y and N(x,y) = x. Also N(x,y) x = = M(x,y) y Therefore, this second differential equation is not exact in any rectangular region D. LECTURE-4 3 Integrating factor: IfthedifferentialequationM(x,y)dx+N(x,y)dy = 0isnotexactinadomain D but the differential equation µ(x,y)m(x,y)dx+µ(x,y)n(x,y)dy = 0 is exact, then µ(x, y) is called an integrating factor of the differential equation M(x,y)dx+N(x,y)dy = 0. In Example., the second differential equation was not exact, but by multiplying it with y becomes an exact equation in any rectangular domain. Thus y is the integrating factor of the differential equation ydx+xdy = 0 3. Some rules for finding integrating factors of the equation Mdx+Ndy = 0 to make it exact Rule-: If Mdx+Ndy 0 and the equation is homogeneous, then Mx+Ny is an integrating factor of Mdx+Ndy = 0 Rule-: If Mx + Ny 0 is not exact, but is of the form f (xy)ydx + f (xy)xdy = 0, then is an integrating factor of Mdx + Ndy = 0 Mx Ny 9

provided Mx Ny 0. Example 3.: (x y xy )dx+(3x y x 3 )dy = 0 The equation is not exact. but the given equation is homogeneous and by Rule-, is the integrating factor. Mx+Ny Now Therefore, the integrating factor is Mx = x 3 y x y,ny = x 3 y +3x y. I.F = Mx+Ny = x y Multiplying the given equation by the I.F., we have ( y ) ( 3 dx+ x y x ) dy = 0 y This equation is now exact because, Example 3.: The equation is of the form Now M y = y = N x y(xy +x y )dx+x(xy x y )dy = 0 yf (xy)ydx+xf (xy)xdy = 0 Mx = xy(xy +x y ),Ny = yx(xy x y ). Therefore, the integrating factor is I.F = Mx Ny = 3x 3 y 3 0

Multiplying the given equation by the I.F., we have This equation is now exact because, Rule-3: When M y N x y(xy +x y ) 3x 3 y 3 dx+ x(xy x y ) 3x 3 y 3 dy = 0 N M y = 3x y = N x is a function of x alone say f(x), then I.F = e f(x)dx Rule-4: When N x M y is a function of y alone say f(y), then I.F = e f(y)dy M Example 3.3: Consider the differential equation (x +y +x)dx+ydy. Here But M y = y, and N x = 0 M N y x N = y 0 y which may be regarded as a function of x. Therefore, I.F. = e dx = e x. Exercises for the reader:. (x 3 +xy 4 )dx+y 3 dy = 0. (x +y )dx xydy = 0 3. (xy 3 +y)dx+(xy +x+y 4 )dy = 0 = LECTURE- 5 4 Linear and Bernoulli Equation: Definition 8:A first-order ordinary differential equation is linear in the dependent variable y and the independent variable x if it is, or can be, written in the form dy +p(x)y = Q(x) (5) dx

For example, the equation dy dx +(+ x )y = x is a first-order linear differential equation. The integrating factor for the linear differential equation is given by dy +p(x)y = Q(x) dx µ(x,y) = e p(x)dx (6) Example 4.: Consider the differential equation dy dx + y x = x Then, the integrating factor is µ = e p(x)dx = e x dx = e logx = x 4. Bernoulli equation: Definition 9: An equation of the form is called a Bernoulli differential equation. dy dx +p(x)y = Q(x)yn (7) We observe that if n = 0 or, then the Bernoulli equation (7) is actually a linear equation and is therefore readily solvable as such. However, in the general case in which n = 0 or, this simple situation does not hold and we must proceed in a different manner. We now state and prove Theorem, which gives a method of solution in the general case. Theorem : Suppose n 0 or. Then the transformation v = y n reduces the Bernoulli equation dy +p(x)y = Q(x)yn dx to a linear equation in v. proof: We first multiply Equation (7) by y n, thereby expressing it in the equivalent form y ndy dx +p(x)y n = Q(x) (8)

If we let, v = y n, then and equation (8) reduces to Or dv dx = ( n)y ndy dx dv +p(x)v = Q(x) ndx dv +( n)p(x)v = ( n)q(x) dx Let, ( n)p(x) = p (x) and ( n)q(x) = Q (x). Then, we have dv dx +p (x)v = Q (x) which is linear in v. Example 4.: Consider the Bernoulli differential equation dy +y = xy3 dx Here n = 3, We first multiply the equation throughout by y 3, thereby expressing it in the equivalent form y 3dy dx +y = x Let v = y n = y 3 = y, then the preceding equation transforms into the linear equation Or dv dx +v = x dv v = x dx The integrating factor of this equation is e dx = e x. LECTURE-6 3

5 Homogeneous Linear equations with constant Co-efficients: In this section we consider the special case of the nth-order homogeneous linear differential equation in which all of the coefficients are real constants. That is, we shall be concerned with the equation d n y a 0 dx +a d n y n dx +a d n y n dx +...a dy n n dx +a ny = 0 (9) where a 0,a,a,...,a n are real constants. We shall show that the general solution of this equation can be found explicitly. In an attempt to find solutions of a differential equation we would naturally inquire whether or not any familiar type of function might possibly have the properties that would enable it to be a solution. The differential equation (9) requires a function having the property such that if it and its various derivatives are eachmultipliedbycertainconstants, thea i andtheresultingproducts,a i f n i, are then added, the result will equal zero for all values of x for which this result is defined. For this to be the case we need a function such that its derivatives are constant multiples of itself. Do we know of functions / having this property that d k dxk(f(x)) = cf(x) for all x. The answer is yes, the exponential function of the formy = e mx, where m is a constant is such that d k dx k(emx ) = m k e mx Thus we shall seek solutions of (9) of the form y = e mx, where the constant m will be chosen such that e mx does satisfy the equation. Assuming then 4

that y = e mx is a solution for certain m, we have: Substituting these in (9), we get dy dx =memx d y dx =m e mx... d n y dx n =mn e mx e mx( a 0 m n +a m n +...+a n m+a n = 0 ) Since e mx 0, therefore we have a 0 m n +a m n +...+a n m+a n = 0 (30) Three cases arise, according as the roots of (30) are real and distinct, real and repeated, or complex. Case-: Distinct real Roots Suppose the roots of (30) are the n distinct real numbers, m,m,m 3,...,m n. Then e m x,e m x,e m 3x,...,e mnx are n distinct solutions of (9). Note: Consider the nth-order homogeneous linear differential equation (9) with constant coefficients. If the auxiliary equation (30) has the n distinct real roots m,m,m 3,...,m n, then the general solution of (9) is y = c e m x +c e m x +...+c n e mnx Where, c,c,...,c n are arbitrary constants. Case-: Repeated real roots: Suppose the roots of (30) are the real numbers, m,m,m 3,...,m n, inwhichtworootsarerepeatedandtheremaining n roots are distinct. Then the linearly independent solution corresponding to the repeated root m is given by y = c e mx +xc e mx. Therefore the solution of (9) in this case is given by y = c e mx +xc e mx +c 3 e m 3x +...+c n e mnx 5

Case-3: Conjugate Complex Roots: Now suppose that the auxiliary equation has the complex numbera+ib as a non repeated root. Then, since the coefficients are real, the conjugate complex number a ib is also a non repeated root. The corresponding part of the general solution is k e (a+ib)x +k e (a ib)x where k and k are arbitrary constants. The solutions defined by k e (a+ib)x, k e (a ib)x are complex functions of the real variable x. It is desirable to replace these by two real linearly independent solutions. This can be accomplished by using Euler s formula, e iθ = cosθ+isinθ which holds for all θ. using this we have k e (a+ib)x +k e (a ib)x =e ax [k (cosbx+isinbx)+k (cosbx isinbx)] =e ax [(k +k )cosbx+(k ik )sinbx] =e ax [c cosbx+c sinbx] Thus the part of the general solution corresponding to the non repeated conjugate complex roots a±b is e ax [c cosbx+c sinbx] LECTURE-7 6 Symbolic Operators: The linear equation (9) can also be written as (a 0 D n +a D n +a D n +...a n D+a n )y =0 where D n = dn dx n (3) f(d)y =0 6

where f(d) = (a 0 D n +a D n +a D n +...a n D+a n ). The operator D n is called the symbolic operator. The auxiliary equation (30) is obtained from equation(3) by replacing D = m. The solution obtained for the auxiliary equation (30) through the different cases discussed above is called Complementary function (C.F). For the Non- Homogeneous equation, there is one more solution ( particular solution) called the Particular Integral (P.I) which contains no arbitrary constants. The complementary Function contains as many arbitrary constants as the order of the given differential equation. Example 6.: Solve the differential equation d y dx 7dy dx 44y = 0 Solution: Writing the given differential equation in symbolic form, we have (D 7D 44)y = 0 The auxiliary equation is m 7m 44 = 0 or (m )(m + 4) = 0 or m =,m = 4. Therefore, the general solution is y = c e x +c e 4x Example 6.: Find the general solution of d y dx +4y = 0 Solution: Writing the given differential equation in symbolic form, we have (D 7D 44)y = 0 The auxiliary equation is m +4 = 0 or m = ±i. Therefore, the general solution is y = (c cosx+c sinx) Now we discuss different cases for obtaining the Particular integral of the Non-Homogeneous differential equations. That is we consider the equation of the type f(d)y = X (3) Note : The operators D, D,... represent the differentiation while as the operators D, D,... represent the integration. 7

Case-: When X in equation (3 ) is of the form e ax The general solution in this case is sum of C.F and P.I. The Particular Integral (P.I) is obtained by P.I = X f(d) = eax f(a) provided f(a) 0 Example 6.3: Solve the equation ( D +5D+6 ) y = e x Solution: The auxiliary equation is m +5m+6 = 0 or m =, 3. Therefore, the C.F = c e x +c e 3x Now the P.I. is equal to (D+)(D+3)e x = (+)(+3)e x = 0 ex Hence, the general solution is c e x +c e 3x + 0 ex Case-: When X in equation (3 ) is of the form sinax or cosax, sinax or cosax then P.I= provided f( a ) 0 f( a ) Example 6.4: Solve the equation ( D +D+ ) y = sinx Solution: The auxiliary equation is m +m+ = 0 or m = ± 3. Therefore, the C.F. is Now the P.I. is equal to e x (c cos 3 3 +c sin ) sinx D +D+ = sinx 4+D+ = sinx D 3 = (D+3)sinx D 9 8

Thus, P.I. is Hence the general solution is D(sinx) + 3sinx 4 9 = cosx 3 e x (c cos 3 3 +c cosx sin ) 3 Case-3: When X in equation (3 ) is of the form x m, then P.I= xm f(d) Inthiscase, f(d)isevaluatedwith thehelpofbinomialtheorem aftertaking out the lowest degree term from f(d). Example 6.5: Solve the equation (D +5D+)y = 5+x Solution: The auxiliary equation is m +5m+ = 0 or m =,. The C.F is c e x +c e x The P.I is ( 5+x D +5D+ = + 5D ) +... (5+x) Or, P.I = (+ 5D ) +... (5+x) = (5+x 5 ). = x Hence the general solution is c e x +c e x +x Case-4: When X in equation (3 ) is of the form e ax V, where V is any function of x then P.I=e ax V f(d+a) Example 6.6: Solve the equation (D 4D+3)y = e x sin3x Solution: The auxiliary equation is m 4m+3 = 0 or m =,3. The C.F is c e x +c e 3x The P.I is (D )(D 3) ex sin3x =e x (D+ )(D+ 3) sin3x = (D ) sin3x = ex 9 sin3x 9

Thus P.I= e x sin3x 0 The general solution is c e x +c e 3x +e x sin3x 0 Case-5: When X in equation (3 ) is of the form e ax and f(a) = 0 In this case (D a) is a factor of f(d). Therefore, we can write f(d) = (D a)φ(d), where φ(a) 0. Therefore, the P.I is given by e ax P.I = (D a)φ(d) = ( φ(a) eax = x.eax φ(a) D+a a ) Example 6.7: Solve the equation (D D 6)y = e x Solution: The auxiliary equation is m m 6 = 0 or m = 3,. The C.F=c e 3x +c e x P.I= is equal to (D 3)(D+) e x = Example 6.8: Solve the differential equation ( 3)(D+) e x = 5 xe x (D 3 +D +D)y = e x +x +x Solution: The auxiliary equation is m 3 + m + m = 0. The roots are m =,,0. Therefore, the C.F=c +(c +xc 3 )e x Now, P.I. for e x is P.I. for (x +x) = e x D(D+) = 8 ex D(D+) (x +x) = D (D+) (x +x) = D ( D+3D...)(x +x) = D (x +x 4x +6) = x3 3 3 x +4x 0

Thus the general solution is c +(c +xc 3 )e x + x3 3 3 x +4x Example 6.9: Solve the differential equation (D +D )y = x+sinx Solution: Theauxiliaryequationism +m = 0. Therootsarem =,. Therefore, the C.F is equal to Now, P.I for x is P.I. for sinx is c e x +c e x ( x D +D = = sinx D +D = sinx +D = The general solution is D ) x (+ D ) +... x = 4 (x+) = D(sinx)+3sinx 9 = cosx+3sinx 0 c e x +c e x + cosx+3sinx 0 (D+3)sinx D 9 Case-6: When X in equation (3 ) is of the form sinax or cosax and f(d) = 0 when D = a In this case siax orcosax D +a

will not be evaluated by putting D = a. In such cases we shall calculate P.I for e iax = cosax+isinax. Therefore, real part of P.I for e iax = P.I. for cos ax and real part of P.I for e iax = P.I. for cos ax Now, e iax e iax D +a = (D+ia)(D ia) Thus Particular Integral of and of e iax e iax = (D+ia)(D ia) = (ia+ia)(d ia) ( ) = eiax ia D+ia ia. = eiax ia.. = x ia (eiax ) D = ix a cosax+ x a sinax cosax D +a = x a sinax sinax D +a = x a cosax Example 6.0: Solve the Differential equation (D +4)y = cosx+sinx Solution: If we replace D by, then D +4 vanishes. hence the P.I. is to be obtained by the above procedure discussed in Case-6.

e ix e ix D +4 = (D+i)(D i) e ix = (ia+ia)(d ia) ( ) = eix 4i D+i i. = eix 4i.. = x 4i (eix ) D = ix 4 cosx+ x 4 sinx Hence the P.I. of is cosx + sinx D +4 x 4 sinx x 4 cosx Example 6.: Solve the differential equation d y +4y = xsinx dx Solution: Here,the auxiliary equation is m +4 = 0 implies m = i,i. Therefore, the C.F. is c cosx+c sinx Now, P.I. for xsinx is xsinx D +4 = imaginary part of xe ix D +4 3

Hence the general solution is ( ) =Imaginary part of e ix x (D+i) +4 ( ) =Imaginary part of e ix x D +Di+3 ( e ix =Imaginary part of + Di ) 3 3... x ( e ix =Imaginary part of Di ) 3 3... x e ix =Imaginary part of ( 3 ) 3 i... x e ix =Imaginary part of (x 3 ) 3 i (cosx+isinx)(3x i) =Imaginary part of 9 = 3xsinx cosx 9 c cosx+c sinx+ 3xsinx cosx 9 Exercises for the reader: Solve the following differential equations:. (D 3 +4D +4D)y = 8e x +x. (D 3 +)y = 8+e x +5e x 3. (D D+5)y = e x sinx LECTURE-8 7 Equations reducible to homogeneous equations with constant coefficients The homogeneous linear equation of the form x ndn y dx n +a x n dn y dx n +a x n dn y dx n +...a n x dy dx +a ny = X 4

where a,a,...,a n are real constants and X is a function of x. By a little substitution, we show that the above linear equation can be transformed into linear equation with constant coefficients. Let x = e z so that z = logx and dz = dx x Now, dy dx =dy dz.dz dx = dy xdz d y =xdy. dz dy dz dx dz dx x Or, x d y y dx =d dz dy dz If we put d dz = D, then we have from above x dy dx =Dy x d y dx =D(D )y... x ndn y dx n =D(D )(D )...(D n+)y Example 7.: Solve the differential equation x d y dx 3xdy +4y = x dx Solution: Let x = e z so that z = logx and dz dx = x Then, [D(D ) 3D+4]y =e z [ D 4D+4 ] y = e z The auxiliary equation is (m ) = 0 or m =,. Therefore, C.F=(c +zc )e z 5

Now, Particular integral is P.I = ez (D ) =e z. (D+ ). =e z. D. = ez. z = z e z Hence the general solution is y = (c + zc )e z + z e z. Now substituting back z = logx and e x = x, we have y = (c +c logx)x +x (logx) Example 7.: Solve the differential equation x 3d3 y dx 3 +xd y dx xdy dx +y = x Solution: Let x = e z so that z = logx and dz dx = x and let D = d dz Then, [D(D )(D )+D(D ) D+]y = e z Or, [ D 3 D D+ ] y = e z The auxiliary equation is (m ) (m+) = 0 or m =,,. Therefore, C.F=(c +zc )e z +c 3 e z Now, Particular integral is P.I = e z (D ) (D+) = e z. ( ) (D+) = Hence the general solution is y = (c +zc )e z +c 3 e z + ze z 4 Thus, y = (c +c l ogx)x+ c 3 x + 4x (logx) e z ze z. = 4(D +) 4 Exercises for the reader: Solve the following differential equations. x 3d3 y x d y +x dy y = dx 3 dx dx x +3x. x d y +x dy 4y = dx x dx 3. x 3d3 y +3x d y +x dy +y = xlogx dx 3 dx dx 4. x d y 3x dy +5y = dx dx x sin(logx) 6

Differential equations, Integration of irrational functions & Reduction formulae Lecture Notes Mathematics S.P. College. Differential equations Bernoulli s equation Any differential equation of the form where P and Q are either constants or functions of x alone. dy dx +Py = Qyn (.) To solve such differential equation, we divide both sides of (.) by y n, we get Put y n = z then For (.), we get y ndy dx +Py n = Q. (.) ( n)y ndy dx = dz dx y ndy dx = n where P = P( n) and Q = ( z)q, is a linear differential equation. dz dx. dz ndx +Pz = Q dz +P( n)z = ( z)q dx dz dx +P z = Q, (.3) Therefore, I.F = e P dx, multiplying both sides of (.3) by e P dx, we get Integrate both sides, we get gives the solution of given differential equation. d ( e ) P dx z = Q e P dx. dx e P dx z = Q e P dx dx+c -

- Lecture Notes Example Solve the differential equation Solution. Divide both sides of (.4) by y, we get Put y = t, then we have dy dx + y x = y. (.4) y dy dx + x y =. From above equation, we get ( )y dy dx = dt dx y dy dx = dt dx. It is a linear differential. dt dx + x t = dt dx t =. (.5) x Therefore I.F = e x dx = e logx = e log x = /x. Multiplying both sides of (.5) by /x, we get Integrating both sides, we get ( ) d dx x t = x. t x = logx+c or t = xlogx+cx y = xlogx+cx y = x( logx+c) Exact differential equation The equation Mdx+Ndy = 0, where M & N are functions of x and y, is said to be an exact differential equation if Mdx+Ndy = 0 is the exact differential function of x and y i.e., Mdx+Ndy = du, where u is a function of x & y. Art: The necessary and sufficient condition for the equation Mdx+Ndy = 0 to be exact is M y = N x.

S.P. College -3 Proof. If Mdx+Ndy = 0 is exact, therefore Mdx+Ndy = du where u = f(x,y). Mdx+Ndy = u u dx+ x y dy M = u x, (i) N = u y (ii). Differential both sides of (i) partially with respect to y and (ii) partially with respect to x, we get M y = u y x, N x = u x y. But Conversely, suppose Let s = Mdx therefore M = s/ x and u y x = u x y. M y = N x. M y = N x. M y = s y x = s x y N x = x ( s y ). Integrating both sides, we get N = s +F(y), where F(y) is a function of y alone. y Now, Mdx+Ndy = s ( ) s x dx+ y +F(y) dy ( ) s s = dx+ x y dy +F(y)dy ( ) = dx+d F(y)dy = d ( s+ ) F(y)dy = dϕ, where ϕ = s+ F(y)dy. Mdx+Ndy is exact. To obtain its solution, we proceed as follows: Since Mdx+Ndy = 0 dϕ = 0 ϕ = constant. This implies Mdx+ F(y)dy = constant

-4 Lecture Notes Example Solve the differential equation ( +e x/y) dx+e x/y ( x/y)dy = 0. Solution. Comparing this differential equation with Mdx+Ndy = 0, we get M = +e x/y ; N = e x/y ( x/y). Therefore, and M ( y = xy ) ex/y N x = ex/y. ( x ( )+e x/y ) y y y = e x/y y x y ex/y e x/y y = x ex/y y. Therefore M/ y = N/ x, i.e., this differential equation is exact. Its solution is obtained as Mdx+ (terms in N independent of x)dy = k ( +e x/y) dx+ 0.dx = k x+ ex/y /y = k, or x+yex/y = k Linear Differential equations (LDE) with constant coefficients Operator: The part d/dx of the symbol dy/dx may be regarded as an operator, such that when it operates on y, the result is derivative of y with respect to x. Similarly d /dx,d 3 /dx 3,...,d n /dx n may be regarded as operator. In symbolic form, we have d dx = D; d dx = D ; d 3 dx 3 = d n D3 ;... ; dx n = Dn. LDE with constant coefficients A linear differential equation with constant coefficients is that in which the dependent variable and its differential coefficients occur only in the first degree and are not multiples together and the coefficients are all constants. Therefore, the equation of the form d n y a 0 dx n +a d n y dx n +...+a ny = X, (.6)

S.P. College -5 where a 0,a,...,a n are all constants and X is a function of x, is called LDE with constant coefficients. The above equation can also be written as or a 0 D n y +a D n y +...+a n y = X, ( a0 D n +a D n ) +...+a n y = X. where D = d dx Here a 0 D n + a D n +... + a n is a function of operator D. Thus a 0 D n + a D n +... + a n = f(d) is regarded as a single operator operating on y. NOTE If y = y,y = y,...,y = y n are n linearly independent solutions of ( a0 D n +a D n +...+a n ) y = 0 (.7) then, y = c y +c y +...+c n y n is the general solution or complete solution of (.7) where c,c,...,c n are arbitrary constants. Since given differential equation is has y = c y +c y +...+c n y n as its solutions. This implies a 0 D n y +a D n y +...+a n y = 0 (.8) a 0 D n y +a D n y +...+a n y = 0 a 0 D n y +a D n y +...+a n y = 0. (.9) a 0 D n y n +a D n y n +...+a n y n = 0 Substitute y = c y +c y +...+c n y n in right hand side of (.8), we have Thus equation (.8) is satisfied by a 0 D n (c y +c y +...+c n y n ) +a D n (c y +c y +...+c n y n ) +...+a n (c y +c y +...+c n y n ) =c ( a0 D n y +a D n y +...+a n y ) +c ( a0 D n y +a D n y +...+a n y ) +...+ ( a 0 D n y n +a D n y n +...+a n y n ) =c (0)+c (0)+...+c n (0) = 0 (by (.9)). y = c y +c y +...+c n y n. Hence it is the complete solution of given differential equation. NOTE : Auxiliary equation Consider the differential equation ( a0 D n +a D n +...+a n ) y = 0. Let y = e mx be its solution, then Dy = De mx = me mx

-6 Lecture Notes Substituting these values in (.8), we get D y = D e mx = m e mx D 3 y = D 3 e mx = m 3 e mx. D n y = D n e mx = m n e mx a 0 m n e mx +a m n e mx +...+a n e mx = 0 ( a0 m n +a m n +...+a n ) e mx = 0. Since e mx 0, m,x. Therefore a 0 m n +a m n +...+a n = 0. (.0) Hence e mx is a solution of (.8) if m satisfies (.0). Therefore equation (.0) is called the Auxiliary equation for the differential equation (.8). NOTE 3: Case I If all the roots of an auxiliary equation are real and distinct i.e., if m = m,m,...,m n are roots of auxiliary equation and are real & distinct, then y = e mx ; y = e mx ;...,y = e mnx are n-independent solutions of (.8). Therefore the complete solution of (.8) is given by y = c e mx +c e mx +...+c n e mnx. Case II When two roots of an auxiliary equation are equal and all others different say m = m then the roots of auxiliary equation are m,m,...,m n. The complete solution in this case is given by y = c e mx +xc e mx +c 3 e m3x +...+c n e mnx or y = (c +c x)e mx +c 3 e m3x +...+c n e mnx. Case III When two roots of an auxiliary equation are imaginary and conjugate to each other and rest are real and different, say m = α+iβ,m = α iβ then the complete solution is given by Thus y = c e (α+iβ)x +c e (α iβ)x +c 3 e m3x +...+c n e mnx = c e αx e iβx +c e αx e iβx +c 3 e m3x +...+c n e mnx = e αx( c e iβx +c e iβx) +c 3 e m3x +...+c n e mnx = e αx [c (cosβx+isinβx)+c (cosβx isinβx)] +c 3 e m3x +...+c n e mnx = e α [(c +c )cosβx+i(c c )sinβx] +c 3 e m3x +...+c n e mnx. y = e α [Acosβx+Bsinβx]+c 3 e m3x +...+c n e mnx, where A = (c +c ) & B = i(c c ), is the complete solution. NOTE 4: If in a linear differential equation a 0 D n y +a D n y +...+a n y = X

S.P. College -7 (i) X = 0 then the complete solution is given by y = complementary function=c.f (ii) X 0 then the complete solution is given by y = C.F+P.I where P.I= particular integral. Examples () Solve d y dx 3dy 4y = 0. dx Solution. The given equation in symbolic form is D y 3Dy 4y = 0 ( D 3D 4 ) y = 0 The auxiliary equation is m 3m 4 = 0 m 4m+m 4 = 0 m(m 4)+(m 4) = 0 (m 4)(m+) = 0. This gives m = 4 or m =, therefore roots of auxiliary equation are real and distinct. Hence complete solution is given differential equation is given by Complete solution = C.F i.e., () Solve d y dx dy +y = 0. dx y = c e 4x +c e x Solution. The equation in symbolic form can be written as D y Dy +y = 0 (D D +)y = 0. It s auxiliary equation is m m+ = 0 this gives m =,. Hence roots are identical or (repeated) then the complete solution is y = (c +xc )e x (3) Solve (D 3 D D )y = 0. Solution. Its auxiliary equation is m 3 m m = 0.

-8 Lecture Notes By inspection one can easily verify m = satisfies it. Therefore by synthetic division m = 0 This, auxiliary equation is equivalent to (m )(m +m+) = 0. This gives m =, ±i 3. Thus the complete solution is y = c e x +c e = c e x +e x ( ) +i 3 ( Acos x +c3 e( i 3 ) x 3 3 x+bsin x ) Exercise Solve ( D 4 +a 4) y = 0 As we mentioned earlier the differential equation of the form d n y a 0 dx n +a d n y dx n +...+a ny = X, where X 0 has complete solution of the form, complete solution= C.F+P.I. For particular integral of various differential equation we proceed as, D represents integral operator. Therefore D sinx = sinxdx = cosx; D emx = e mx dx = emx NOTE 5: Similarly De ax = ae ax m. D e ax = D(De ax ) = D(ae ax ) = ad(e ax ) = a e ax. D 3 e ax = a e ax ; D 4 e ax = a 4 e ax ;... ;D n e ax = a n e ax. In general if f(d) is a polynomial in D then f(d)e ax = f(a)e ax. This implies In case f(a) = 0 we proceed as follows: Case I f(d) eax = f(a) eax, provided f(a) 0.

S.P. College -9 operating both sides by (D a), we get D a eax = y say e ax = (D a)y ( ) d dx a y = e ax dy dx ay = eax (.) This is linear differential equation of the form dy/dx+py = Q. Here P = a and Q = e ax. Integrating factor= e Pdx = e adx = e ax. Multiplying both sides of (.) by e ax, we get ( ) dy e ax dx ay integrating both sides, we get = e ax e ax d ( ye ax ) = dx ye ax = x y = xe ax i.e., Case II D a eax = xe ax. (D a) eax = ( ) D a D a eax = D a (xeax ). operating both sides by (D a), we get Let D a (xeax ) = y (say) xe ax = (D a)y ( ) d dy a y = xe ax dy dx ay = xeax. (.) It is a linear differential equation and here integrating factor= e adx = e ax. Multiplying both sides of (.) by e ax, we get ( ) dy e ax dx ay = e ax (xe ax )

-0 Lecture Notes integrating both sides, we get i.e., Similarly, one can prove d ( ye ax ) = x dx ye ax = x y = x eax (D a) eax = x eax. (D a) 3eax = x3 3! eax (D a) 4eax = x4 4! eax. (D a) neax = xn n! eax. Rule II: f(d) eax where f(a) = 0. In such a case (D a) is a factor of f(d). The above expression can also be written as (D a)g(d) eax where g(d) is a polynomial in D of degree one less than that of f(d). Examples () () Rule III D ex = (D +)(D ) ex = D + ex = D ex = xex D 3 3D +3D ex = (D ) 3ex = x3 3! ex = x3 6 Dsinax = acosax

S.P. College - D sinax = D(Dsinax) = D(acosax) = a sinax D 3 sinax = D(D sinax) = D( a sinax) = a 3 cosax D 4 sinax = D(D 3 sinax) = D( a 3 cosax) = ( a ) sinax. In general, Therefore, D n sinax = ( a ) n sinax. f(d ) sinax = f( a ) sinax provided f( a ) 0. In case f( a ) = 0, we may proceed as in the following example. ( D +a sinax = e iax e iax ) (here f( a ) = 0) (D +ia)(d ia) i = [ ] i (D +ia)(d ia) eiax (D +ia)(d ia) e iax = [ ] i iad ia eiax ( ia)(d +ia) e iax = [ xe iax +xe iax] 4a = x ( e ax +e iax ) = x a a cosax. Similarly Rule IV D a cosax = x a sinax. ( ) (D a sinax = Im ) n D a eiax ( ) (D a cosax = Re ) n D a eiax Example Solve the differential equation Solution. The auxiliary equation is (D 3 +)y = 3+5e x. m 3 + = 0 (m+)(m +m+) = 0. This implies m = or m = ±i 3. Therefore complementary function is C.F = c e x +c e ( ) +i 3 x +c3 e( i 3 ) x

- Lecture Notes Next, the particular integral is given as Thus complete solution is Rule V Evaluate ( = c e x 3 +e x c e ii x +c e i 3 x) ( ) 3 3 = c e x +e x Acos x+bsin x. P.I = D 3 + (3+5ex ) = D 3 + (3)+5 D 3 + ex = 3 D 3 + e0.x +5 D 3 + ex ( ) = 3 0 3 + ()+5 e x = 3+ 5 ex. y = c e x +e x ( Acos ) 3 3 x+bsin x +3+ 5 ex. f(d) eax X where X is a function f x. Since if V is a function of x, therefore by successive differentiation, we have Next, Similarly, Now if f(d +a)v = X then Therefore D (e ax V) = D(D(e ax V)) D(e ax V) = e ax DV +ae ax V = e ax (D +a)v. = D(e ax (D +a)v)& = e ax D(D +a)v +ae ax (D +a)v = e ax (D +a)[d +a]v = e ax (D +a) V. D 3 (e ax V) = e ax (D +a) 3 V. f(d)e ax V = e ax f(d +a)v. f(d +a) X = V. [ ] f(d) e ax f(d +a) X = e ax X or e ax f(d +a) X = f(d) eax X or f(d) eax X = e ax f(d +a) X

S.P. College -3 Example () Solve the differential equation Solution. The equation can be written as The auxiliary equation is d y dx dy dx +4y = ex cosx. (D D +4)y = e x cosx. m m+4 = 0. This implies m = ±i 3, therefore complementary function is The particular integral is given by C.F = c e (+i 3)x +c e ( i 3)x = e x( Acos 3x+Bsin ) 3x. P.I = D D +4 ex cosx = e x (D +) (D +)+4 cosx = e x D +3 cosx = e x ( ) +3 cosx = ex cosx. Thus complete solution is y = e x( Acos 3x+Bsin ) 3x + ex cosx. () Solve (D 5D +6)y = xe 4x. Solution. The auxiliary equation of this differential equation is Thus complementary function is Next, particular integral is given by P.I = m 5m+6 = 0 (m )(m 3) = 0 m =, m = 3. C.F = c e x +c e 3x. D 5D +6 xe4x = e 4x (D +4) 5(D +4)+6) x = e 4x D +3D + x

-4 Lecture Notes = e 4x ( + 3D + )x D ) = (+ e4x 3D D + x ) = ( e4x 3D D... x = e4x ( x 3 ). Thus complete solution is y = c e x +c e 3x + e4x ( x 3 ) Show that Rule VI f(d) (xv) = x ( ) d f(d) V + V dd f(d) where V is a function of x. Solution. Let X be a function of x. Then by Leibnitz s theorem, we have By putting f(d)x = V, we have Since X is a function of x, so is V. From (.3) operating both sides by f(d), we get D n (xx) = (D n X)x+ ( ) n (D n X) = xd n X +nd n X ( ) d = xd n X + dd Dn X ( ) d f(d)(xx) = xf(d)x + dd f(d) X. (.3) X = f(d) V. ( f(d) x ) ( ) d f(d) V = xv + dd f(d) f(d) V, ( x ) f(d) V = f(d) (xv)+ f (D) (f(d)) V f(d) (xv) = x f(d) V f (D) (f(d)) V (xv) =x f(d) f(d) V + d dd ( f(d) ) V

S.P. College -5 Example Solve (D +4)y = xsinx. Solution. The auxiliary equation is Therefore, complementary function is Now, particular integral is given as Therefore, complete solution is m +4 = 0 m = ±i. C.F = c e ix +c e ix = Acosx+Bsinx. P.I = D +4 xsinx = x D +4 sinx D (D +4) sinx = x () +4 sinx D ( () +4) sinx = 3 xsinx 9 D(sinx) = 3 xsinx 9 cosx. y = Acosx+Bsinx+ 3 xsinx cosx 9. Homogeneous linear differential equations An equation of the form a 0 x ndn y dx n +a x n dn y dx n +...+a n x dy dx +a ny = X, where a 0,a,...,a n are all constants and X is a function of x, is called Homogeneous linear differential equation of nth order. Solution of Homogeneous linear differential equation Let x = e z then z = logx, therefore Now, or x dy dx = dy dz dy and dx =dy dz dz dx = dy xdz xdy = D y where D = d dx & D = d dz. dz dx = x d y dx = d ( ) dy dx xdz

-6 Lecture Notes Similarly, one can prove that = d y dz xdz dx dy x dz = d y x dz dy x dz or x d y dx = d y dz dy dz x D y = D (D )y. x 3 D 3 y = D (D )(D )y. ( dz dx = ) x Therefore by this suitable substitution the above homogeneous linear differential equation gets reduced to linear differential equation with constant coefficients Example Solve the following differential equation x d y dx y = x + x. Solution. In symbolic form the differential equation can be written as Put x = e z then This implies x D y y = x + x. xdy = D y & x D y = D (D )y. D (D )y y = e z + e z (D D )y = e z +e z. This is linear differential equation with constant coefficients. Hence its auxiliary equation is m m = 0 m =,. Therefore, complementary function is C.F = c e z +c e z = c (e z ) +c (e z ) = c (x) +c x. Now particular integral is ( P.I = e z e z) D D

S.P. College -7 = (D )(D +) ez + (D )(D +) e z = 3D ez + 3D + e z (z(e z ) z(e z ) ) = 3 = 3 ( x ) logx. x Thus complete solution is y = c (x) +c x + 3 ( x ) logx x.3 Equation solvable for p Suppose the equation of nth degree in p is f(x,y,p) = 0. As it is solvable for p. We can put it into the form [p F (x,y)][p F (x,y)]...[p F n (x,y)] = 0. Then, we solve each factors individually to get the solution of given differential equation Example Solve the differential equation p p = 6. Solution. We have i.e., Integrating both sides of these two equations, we get p +p 6 = 0 (p )(p+3) = 0 p = or p = 3 dy dx = or dy dx = 3. y = x+c and y = 3x+c. These two equations together gives the complete solution of given differential equation Equation solvable for y If the equation f(x,y,p) = 0 is solvable for y, we can express y explicitly in terms of x and p. Then, an equation solvable for y can be expressed as Differentiating both sides of this equation with respect to x, we get dy dx = d ( dx (ϕ(x,p)) = F x,p, dp ) dx y = ϕ(x,p). (.4)

-8 Lecture Notes i.e., p = F ( x,p, dp ) dx which is a differential equation involving two variables x and p. Let its solution be g(x,p,c) = 0. (.5) Therefore on eliminating p between (.4) and (.5), we get the required solution of (.4). Similar is the case for the differential equations solvable for x..4 Clairaut s equation A differential equation of the form y = px+f(p) (.6) is known as Clairaut s equation. To solve this differential equation we differentiate both sides of it with respect to x and we get Integrating, we get Eliminating p between (.6) and (.0), we get that is the solution of given differential equation dy dx = p+xdp dx +f (p) dp dx p = p+(x+f (p)) dp dx (.7) (.8) dp = 0. (.9) dx p = c, where c is a constant. (.0) y = cx+f(c).5 Legendre Polynomials Legendre Equation The differential equation of the form ( x )y xy +n(n+)y = 0 (.) is called Legendre s equation, where n is a positive integer. We solve this differential equation by the method of series solution (also called Frobenius method). In this method we find a series representation for the solution instead of the solution itself. You first saw something like this when you looked at Taylor series in your Calculus class. The basic idea to finding a series solution to a differential equation is to assume that we can write the solution as a power series in the form, y = a m x k m. (.) m=0