Answer to Question #302571 in Differential Equations for haru

Question

Use variation of parameter methods to find the particular solution of xy−(x+1)y+y = x2, given that y1(x) = ex and y2(x) = x + 1 form a fundamental set of solutions for the corresponding homogeneous differential equation.

Answer 1

$xy''-(x+1)y'+y=x^2$

$y=c_1e^x+c_2(x+1)$

$y'=c_1e^x+c_1'e^x+c_2+c_2'(x+1)$

Let $c_1'e^x+c_2'(x+1)=0.$ Then

$y'=c_1e^x+c_2$

$y''=c_1e^x+c_1'e^x+c_2'$

Substitute

$c_1xe^x+c_1'xe^x+c_2'x-(x+1)(c_1e^x+c_2)$

$+c_1e^x+c_2(x+1)=x^2$

$c_1'xe^x+c_2'x=x^2$

We have

$c_1'e^x+c_2'(x+1)=0$

$c_1'xe^x+c_2'x=x^2$

$c_1'e^x=-c_2'(x+1)$

$-c_2'x(x+1)+c_2'x=x^2$

$c_2'=-1$

$c_1'=e^{-x}(x+1)$

Then we can take

$c_1=−(x+1)e^{-x}-e^{-x}=−(x+2)e^{-x}$

$c_2=-x$

The particular solution is

$y_p=-x-2-x^2-x$

$y_p=-x^2-2x-2$