Answer to Question #317946 in Differential Equations for Sharlene

Question

Solve by method of variation of parameters.

d^y/dx^2 + dy/dx + 1 = e^x

Answer 1

First let’s find a fundamental set of solutions for

$y’’+y’=0$

$y_1=1$; $y_2=e^{-x}$

Then a general solution to the nonhomogeneous differential equation is

$y(x)=-y_1\int\frac{y_2(e^x-1)dx}{W(y_1,y_2)}+$$y_2\int\frac{y_1(e^x-1)dx}{W(y_1,y_2)}=$

$-\int\frac{(1-e^{-x})dx}{y_1y_2’-y_2y_1’}+$$e^{-x}\int\frac{(e^{x}-1)dx}{y_1y_2’-y_2y_1’}=$$\int\frac{(1-e^{-x})dx}{e^{-x}}-$$e^{-x}\int\frac{(e^{x}-1)dx}{e^{-x}}=$

$e^x-x+c_1-e^{-x}(\frac12e^{2x}-e^x-c_2)=$

$\frac12e^{x}-x+1+c_1+c_2e^{-x}$

Answer: $y(x)=c_1+c_2e^{-x}+\frac12e^{x}-x+1$.