Answer to Question #191326 in Differential Equations for Akshay Rajesh Yems

Question

y'' − y' + 1/4 y = 8 + e^x/2

Answer 1

y'' − y' + 1/4 y = 8 + e^x/2

$= y''(x)-y'(x)+\frac{1}{4}y(x)=8+e^{x\over2}$

the general solution will be sum of the complementary solution and particular solution.

we find the complementary solution by solving $\frac{d^2y(x)}{dx^2}-\frac{dy(x)}{dx}+\frac{y(x)}{4}=0$

comes out to be

$y(x)=y_1(x)+y_2(x)$

$y(x)=c_1e^{x\over 2}+c_2e^{x\over 2}$ .........(1)

now find particular solution

$\frac{d^2y(x)}{dx^2}-\frac{dy(x)}{dx}+\frac{y(x)}{4}=e^{x\over 2}$

comes out to be

$y_p(x)=\frac{1}{2}e^{x\over 2}x^2+32$ ............(2)

combining both (1) and (2) we get the answer

$\boxed{y(x)=c_2e^{x\over 2}x+c_1e^{x\over 2}+\frac{1}{2}e^{x\over 2}x^2+32}answer$