Answer to Question #239037 in Differential Equations for rish

Question

What is the value of y(π

2

)

y(π2) where y

′′

−2y

′

+y=xe

x

sinx

y″−2y′+y=xexsin⁡x ; y(0)=0

y(0)=0 and y

′

(0)=1

y′(0)=1 ?

Answer 1

$k^2-2k+1=0$

$k_{1,2}=1$

$Y=c_1e^x+c_2xe^x$

$y(x)=c_1(x)e^x+c_2(x)xe^x$

$y_1(x)\frac{dc_1(x)}{dx}+y_2(x)\frac{dc_2(x)}{dx}=0$

$\frac{dy_1(x)}{dx}\frac{dc_1(x)}{dx}+\frac{dy_2(x)}{dx}\frac{dc_2(x)}{dx}=xe^xsinx$

$e^x\frac{dc_1(x)}{dx}+xe^x\frac{dc_2(x)}{dx}=0$

$\frac{d}{dx}e^x\frac{dc_1(x)}{dx}+\frac{d}{dx}(xe^x)\frac{dc_2(x)}{dx}=xe^xsinx$

$\frac{d}{dx}e^x\frac{dc_1(x)}{dx}+\frac{d}{dx}(xe^x+e^x)\frac{dc_2(x)}{dx}=xe^xsinx$

$\frac{d}{dx}c_1(x)=-x^2sinx$

$\frac{d}{dx}c_2(x)=xsinx$

$c_1(x)=c_3+\int(-x^2sinx)dx$

$c_2(x)=c_4+\int xsinxdx$

$c_1(x)=c_3+x^2cosx-2xsinx-2cosx$

$c_2(x)=c_4-xcosx+sinx$

$y(x)=c_3e^x+c_4xe^x-xe^xsinx-2e^xcosx$

$y(0)=c_3-2=0\implies c_3=2$

$y'(x)=c_3e^x+c_4xe^x+c_4e^x-e^x(sinx+xcosx)-xe^xsinx-2e^xcosx+2e^xsinx$

$y'(0)=c_3+c_4-2=1\implies c_4=1$

$y(x)=2e^x+xe^x-xe^xsinx-2e^xcosx$

$y(\pi/2)=9.62$