solve the initial value problem y y 2y x sinx y0 1920 y

solve the initial value problem y\'\' + y\' -2y = x + sinx , y(0) = 19/20, y\' (0) =0

Solution

First we solve the homogeneous ode

y\'\'+y\'-2y=0

Let, y=exp(kx),substituting gives

k^2+k-2=0

k^2+2k-k-2=0

k=-2,1

Hence solution to homogeneous ode is

y=A e^x+B e^(-2x)

For complementary solution let the guess be:

yp=Ax+B+C sin(x)+D cos(x)

Substituting gives

-C sin(x)-D cos(x)+A+Ccos(x)-D sin(x)-2Ax-2B-2C sin(x)-2D cos(x)=x +sin(x)

Comparing coefficients we get

-3C-D=1

C-3D=0,C=3D

Substituting gives

-9D-D=1

D=-1/10,C=-3/10

-2A=1,hence, A=-1/2

A=2B ,hence, B=-1/4

Hence, y=A e^x+B e^(-2x)-x/2-1/4-sin(x)/10-3cos(x)/10

y(0)=19/20=A+B-1/4-3/10=A+B-11/20

A+B=30/20=3/2

y\'(0)=0=A-2B-1/2-1/10

A-2B=3/5

Solving gives

A=19/15,B=1/6