1) The equations y1=x3 and y2=x5 form the basis of the second-order ODE x2y''-7xy'+15y=0 Solve the initial value problem for the ODE given y(1)=0.4 and y'(1)=1. 3 5 2 y ( x )=a x +b x y ( x )=3 a x +5 b x 4 y (1 ) =a+b=0.4 y ( 1 )=3 a+5 b=1 ( 5 a+5 b2 )( 3 a+5 b1 )=2 a1 2 a=1 a= 1 2 a+b=0.4 b=0.40.5=1 3 y=0.5 x 0.1 x 5 2) Find the general solution of the second-order ODE xy''+y'=x2 by reducing to first-order form. dxdy ' '' =x y + y ' dx d y ' = x 2 dx y'= x2 c + 3 x 2 x 3 +c ( ) x y= x3 y= +c ln x 9 3) Find the general solution for the homogeneous linear ODE y''+2.4y'+4y=0. Verify your answer by substitution and using MATLAB. 2 2 d +2.4 d + 4=0 a x + bx+ c=0 d= 2 b b 24 ac 2.4 2.4 4(1)(4 ) 2.4 10.24 2.4 3.2 i = = =1.2 1.6 i 2a 2 2 2(1) 1.6 x a sin 1.6 x+b cos y=eax ( a sin b +b cos b )=e1.2x 4) Solve the initial value problem for the ODE y''-9y=0 given the initial conditions y(0)=-2 and y'(0)=-12. d 29=0 d=3 eax y ( 0 )=c e 0+ c e0=2 y=c e3 x +c e3 x y' =3 c e3 x 3 c e3 x ' y ( 0 )=3 c3 c=12 cc=4 ( c +c2 ) + ( c c+ 4 )=1 ( cc + 4 ) ( c+ c2 )=3 y=c e3 x +c e3 x y=3 e3 x +(1) e3 x 5) Find the location of the maxima and minima of y=e-2t cos t. Round your answer to an accuracy of 2 decimal points. Hint: Recall from calculus that a maxima or minima of a function y occurs where the derivative of the function y'=0. In other words the slope of the function is 0. ' 2 t y =e 2 t sin t 2 e cos t=0 sin t+2 cos t=0 tan t +2=0 tan t=2 1 t=tan 2+ n=1.1+n 6) Find the general solution of a non-homogeneous ODE y''+9y=cos(x)+1/3 cos(3x). y=asin x +b cos x '' y =asin xb cos x y ' ' +9 y =asin xb cos x+ 9 ( a sin x+b cos x ) =cos x y=cos x 8 y=ax sin 3 x+ bx cos 3 x 3x 3x 9 x cos 3 x6 sin 3 x +6 cos + b 9 x sin y ' ' =a 3x 3x 9 x cos 3 x6 sin + 9 ax sin 3 x+ 9 bx cos 3 x=cos 3 x +6 cos +b 9 x sin y ' ' +9 y =a a= 1 x sin 3 x 18 y=bsin 3 x+ acos 3 x+ 1 1 xsin 3 x+ cos x 18 8 3x 3 7) Solve the initial value problem for the non-homogeneous ODE y''+10y'+25y=100 sin h(5x), given initial conditions y(0)=0 and y'(0)=0. 5x '' y=a e y =25 a e 5x y ' =25 a e 5 x +10 ( 5 a e5 x )+ 25 a e5 x =50 e 5 x y= 1 5 x e 2 y=a x 2 e5 x '' 5 x y =e 2 ( 25 a x 20 ax+2 a) y ' =e5 x ( 25 a x 220 ax+2 a )+ e5 x (50 a x 2+20 x ) +25 a x 2 e5 x =50 e5 x 2 5 x y=25 x e 5 x y=c e y= 5 x +cx e 1 5x 2 5 x + e 25 x e 2 1 5 x 1 e 5 x e5 x + e 5 x 25 x 2 e5 x 2 2 8) Find the steady state current in the RLC circuit given below where R=8, L=0.5H, C=0.1F and E=100sin(2t) V. =2 X L X C = 1 1 L= (2 0.5)=4 C 2 0.1 Z =r 2+( X L X C )2= 82 +4 2=8.94 =tan 1 X L X C 4 =tan1 =0.46 r 8 V 100 I= = =11.19 Z 8.94 I =11.19 sin 2 t cos 0.4611.19cos 2 t sin 0.46=10 sin 2t5 cos 2 t 9) Find the transient current (a general solution) in the RLC circuit given below given R=0.2, L=0.1H, C=2F, and E=754sin(0.5t) V. 2t a sin 2 t+b cos I =et I =a sin 0.5 t+b cos 0.5 t '' I =0.25 a sin 0.5 t0.25 b cos 0.5 t I ' ' +2 I ' + 5 I =3770 cos 0.5t (0.25 ab +5 a ) sin 0.5t + ( a0.25 b+5 b )=3770 sin 0.5 t ( 4.75 ab ) sin 0.5 t+ ( a+4.75 b ) cos 0.5t=3770 cos 0.5t 2t a sin 2 t+b cos t I =e 10) Find the general solution of the non-homogeneous linear ODE y''-2y'+y=ex sin (x) by variation of parameters. x y=e ( c+c) y=a e x sin x '' '' ' x y = y y + y=e sin x '' x x x x x y =2a e cos x2(a e cos x2 a e sin x )+ a e sin x=e sin x y=e x sin x x y=ay+ by=a e +bx e x e x (x e x + e x )x e 2 x =x e2 x + e2 x - x e2 x =e 2 x a '= x sin x e 2 x =x sin x e2x 2x sin x e b '= =sin x 2x e a=x cos x - sin x b=cos x y=x e x cos x - e x sin x - x e x cos x=e x sin x x x x y=c e +cx e e sin x