Differentiation exercise show differential equation

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1 Differentiation exercise show differential equation 1. If y x sin 2x, prove that x d2 y y x + 4xy 0 y x sin 2x sin 2x + 2x cos 2x 2 2cos 2x + (2 cos 2x 4x sin 2x) x d2 y y x + 4xy (2x cos 2x + 2x cos 2x 4x 2 sin 2x) (2sin 2x + 4x cos 2x) + 2x sin 2x x + 4x 2 sin 2x 0 2. Given that y e x e x, show that ( )2 y y e x e x ex + e x ( )2 y 2 4 (e x + e x ) 2 (e x e x ) 2 4 (e 2x e 2x ) (e 2x 2 + e 2x ) Given that v sin u, show that 4v 3 d2 v du 2 + v dv du d 2 v du cos u 2 sin u sin u( sin u) cos u cos u 2 sin u 1 2sin 2 u cos 2 u sin2 u+1 v4 +1 sin u 4 sin u sin u 4 sin u sin u 4v 3 4v 3 d2 v du 2 + v Given y e x cos x, show that y 0. 2 y e x cos x e x sin x e x cos x 2 (e x sin x e x cos x ) + (e x sin x + e x cos x ) 2e x sin x d2 y y (2e x sin x ) 2(e x sin x + e x cos x ) + 2e x cos x 0 1

2 Method 2 y e x cos x ln y x + ln(cos x) 1 y 1 sin x cos x y y tan x + 2y y(1 tan x) we get: + 2 (1 tan x) y 2 sec2 x ( y y tan x)(1 tan x) y (1 + tan 2 x) 2y d2 y y 0 Method 3 y e x cos x y e x cos x (1) ex + ye x sin x ( d2 y 2 ex + ex ) + ( ex + ye x ) cos x 2 ex + 2 ex + ye x ye x, by (1) d2 y y 0 5. Given that y, where k is a positive integer, show that d2 y 1+cos kx 2 k2 y 2. (1+cos kx)(k cos kx) ( k ) (1+cos kx) 2 k ky 1+cos kx d2 y 2 + kcos kx k d2 y k 2 k[ky]y k 2 y 2 k cos kx+k cos2 kx+k sin 2 kx (1+cos kx) 2 (1 cos kx)(1+cos kx) (1 cos kx) k (1+cos kx) k (1+cos kx) k cos kx+k k(1+cos kx) (1+cos kx) 2 1+cos kx sin 2 kx k [ ] 1+cos kx k 1+cos kx 2

3 Method 2 Given that y Note that : y, where k is a positive integer, show that d2 y 1+cos kx 2 k2 y 2. 1+cos kx 1 cos kx (k ) (1 cos kx)(k cos kx) sin 2 kx ky d2 y 2 + kcos kx k d2 y k(1 cos kx) 2 ksin2 kx+k cos 2 kx k cos kx sin 2 kx k k cos kx sin 2 kx kx k (1 cos ) ( ) k(y)(ky) k2 y 2 k 1 cos kx ky 6. Given y(2 x) 3, show that 3 d2 y 2 2y 0. (2 x) y 0 y 2 x Differentiate again, (2 x) y( 1) (2 x) y 2 x +y 2y 2 (2 x) 2 (2 x) 2 (2 x) 2 3 d2 y 2y 3 2 Method 2 2y 2y y (2 x) 2 2 x 6y (2 x) 2 2y2 2 x 6y 2y[y(2 x)] 6y 2y[3] 0 (2 x) 2 (2 x) 2 (2 x) 2 (2 x) 2 Differentiate again, (2 x) y 0 [(2 x) d2 y 2 ] 0 or (2 x) d2 y Multiply by y, y(2 x) d2 y d2 y 2y Given y (1 + 4x)e 2x, prove that y 0 2 2e 2x 8xe 2x 2 16xe 2x 12e 2x 3

4 y 2 (16xe 2x 12e 2x ) + 4(2e 2x 8xe 2x ) + 4(1 + 4x) Method 2 e 2x y (1 + 4x) 2x e + 2ye2x 4 (16xe 2x 12e 2x ) + (8e 2x 32xe 2x ) + 4e 2x + 16xe 2x 0 (e 2x d2 y + 2e2x 2 ) + (2e2x + 4e2x y) 0 e 2x ( d2 y y) y Let y cos x, show that 4y 3 d2 y 2 + y y cos x y 2 cos x 2y sin x (1) Differentiate again, 2y d2 y cos x y2 2y d2 y ( )2 y 2 Multiply by 2y 2, 4y 3 d2 y + (2y 2 )2 2y 4 By (1), 4y 3 d2 y 2 + ( sin x)2 2y 4, 4y 3 d2 y cos2 x 2y 4 4y 3 d2 y y4 2y 4 4y 3 d2 y 2 + y Given (1 + x 2 )y 2 1 x 2, show that ( )2 1 y4 1 x 4. (1 + x 2 )y 2 1 x 2 (1) y 2 1 x2 1+x 2, 1 y2 1 1 x2 1+x 2 2x2 1+x 2, 1 + y x2 1+x x 2 1 y 4 (1 y 2 )(1 + y 2 ) 4x2 (1+x 2 ) 2 Differentiate (1), 2xy 2 + (1 + x 2 ) [2y ] 2x 4

5 2x(1+y2 ), 2y(1+x 2 ) ( )2 4x2 (1+y 2 2 ) (1+x 2 ) 2 4y 2 (1 y 4 ) ( 2 1+x 2)2 4( 1 x2 1+x 2) 1 y 4 (1+x 2 )(1 x 2 ) 1 y4 1 x 4 Method 2 Let x tan θ 1 x2, cos θ 2 1+x 2, 1 (1 + dθ 2 x2 ) y 2 1 x2 1+x2 cos θ, y cos θ sin θ, dθ 2 cos θ ( dθ )2 sin2 θ 1 cos2 θ 4 cos θ ( )2 ( dθ )2 ( dθ )2 1 y 4 4( 1 x2 1+x 2) [ 1 2 (1+x2 )] 2 1 y4 4 cos θ 4( 1 x2 1+x 2) 1 y 4 (1+x 2 )(1 x 2 ) 1 y4 1 x Form a differential equation from y Ax 3 + B x 2 6x, x > 0. yx 2 Ax 5 + B 6x 3 Divide by x, x 2 + 2xy 5Ax4 18x 2 x + 2y 5Ax3 18x (1) Differentiate (1), (x d2 y + 2 x d2 y Ax2 18 ) + (2 ) 15Ax2 18 Multipy by x, 2 + 3x 15Ax3 18x 2 + 3x 3(5Ax3 18x) + 36x 3 (x + 2y) + 36x, by (1) 6xy 36x 2 x d2 y 2 6y Form a differential equation from y Ax 3 + B x 2 6, x > 0. yx 2 Ax 5 + B 6x 2 x 2 + 2xy 5Ax4 12x 5

6 Divide by x, x + 2y 5Ax3 12 (1) Differentiate (1), (x d2 y + 2 x d2 y Ax2 ) + (2 ) 15Ax2 Multipy by x, 2 + 3x 15Ax3 + 3x 2 3(5Ax3 12) (x + 2y) + 36, by (1) 2 6y 36 x 3 d2 y 2 6xy 36x 12. y sin 1 3x 1 9x 2, show that (1 9x 2 ) + 3y2 + 9xy 0 y sin 1 3x 1 9x 2 sin 1 3x + y 3 1 9x 2 ysin 1 3x + y x 2 1 9x 2 + 3y2 1 9x x 2 1 9x 2 9xy 1 9x 2 9xy 1 9x 2 0 (1 9x 2 ) + 3y2 + 9xy Given that y x n [A cos(ln x) + B sin(ln x)], where A and B are constants, show that 2 + (1 2n)x + (1 + n2 )y 0 y x n [A cos(ln x) + B sin(ln x)] (1) nxn 1 [A cos(ln x) + B sin(ln x)] + x n { A sin(ln x) + B cos(ln x)} x x Multiply by x, we have, x nxn [A cos(ln x) + B sin(ln x)] + x n { A sin(ln x) + B cos(ln x)} By (1), x ny + xn { A sin(ln x) + B cos(ln x)} (2) Differentiate again, 6

7 x d2 y + n + 2 nxn 1 { A sin(ln x) + B cos(ln x)} x n { A cos(ln x) + B sin(ln x)} x x Multipy by x, + x nx + 2 nxn { A sin(ln x) + B cos(ln x)} x n {A cos(ln x) + B sin(ln x)} + x nx + n [x ny] y, by (1) and (2) (1 2n)x + (1 + n2 )y Given that y sin 1 x, show that (1 x 2 ) d2 y 2 x 0 y sin 1 x sin y x cos y 1 (1) Differentiate (1), Multiply by cos y, cos y d2 y 2 (sin y ) 0 cos 2 y d2 y (sin y ) (cos y ) 0 2 (1 x 2 ) d2 y x (1) 0, by (1) 2 (1 x 2 ) d2 y 2 x Let y 5e ( 3 2)x + 3e ( 3+2)x, show that d2 y y 0. Let y u + v, where u 5e ( 3 2)x, v 3e ( 3+2)x. du 5( 3 2)e( 3 2)x, d 2 u 2 5( 3 2)2 e ( 3 2)x 5(7 4 3)e ( 3 2)x Hence, d 2 u + 4 du + u 5( )e( 3 2)x + 20( 3 2)e ( 3 2)x + 5e ( 3 2)x 5e ( 3 2)x [(7 4 3) + 4( 3 2) + 1] 0 dv 5( 3 + 2)e ( 3+2)x, d 2 v 2 5( 3 + 2)2 e ( 3+2)x 5( )e ( 3+2)x Hence, d 2 v + 4 dv + v 5( )e ( 3+2)x 20( 3 2)e ( 3+2)x + 5e ( 3+2)x 5e ( 3+2)x [( ) 4( 3 + 2) + 1] y d2 (u+v) + 4 d(u+v) + (u + v) ( d2 u + 4 du + u) + v (d2 + 4 dv + v) 0 2 7

8 Method 2 y 5e ( 3 2)x + 3e ( 3+2)x e 2x y 5e 3x + 3e 3x (1) 2x e + 2e2x y 5 3e 3x 3 3e 3x (e 2x d2 y + 2e2x 2 ) + (2e2x + 4e2x y) 15e 3x + 9 3e 3x e 2x d2 y 2 + 4e2x + 4e2x y 3e 2x y, by (1) e 2x d2 y 2 + 4e2x + e2x y 0 Method 3 Let α 3 2, β ( 3 + 2). α + β 4, α β ( 2) α, β are roots of u 2 + 4u (1) Hence { α2 + 4α β 2 + 4β (2) y 5e ( 3 2)x + 3e ( 3+2)x 5e αx + 3e βx 5αeαx + 3βe βx, 2 5α2 e αx + 3β 2 e βx y 2 5eαx (α 2 + 4α + 1) + 3e βx (β 2 + 4β + 1) 0, by (2). Comment (a) The general solution of the differential equation: d2 y y 0 is y c 1 e ( 3 2)x + c 2 e ( 3+2)x, c 1, c 2 are integrating constants. So y 5e ( 3 2)x + 3e ( 3+2)x is a special solution of the differential equation. (b) It is interesting that solving d2 y y 0 is just to find the roots of the auxiliary 2 equation u 2 + 4u + 1 0, that is α 3 2, β ( 3 + 2) and then form the general solution y c 1 e ( 3 2)x + c 2 e ( 3+2)x. Solving this kind of second order differential equation becomes solving a quadratic equation. The story is longer since quadratic equation may have equal roots or complex roots and is not discussed here. (c) Showing differential equation is just learning how to dive in the sea of calculus and solving differential equation is to get some pearls in the sea. 19/1/2018 Yue Kwok Choy 8