Solve the equation $y''+y'=\sec(x)$

Question

Solve the equation $$y''+y'=\sec(x).$$ By solving the associated homogenous equation, I get complementary solution is $y_c(x)=A+Be^{-x}$. Then by using the method of variation of parameter, I let $y_p(x)=u_1+u_2e^{-x}$ where $u_1,u_2$ satisfy $$u_1'+u_2' e^{-x}=0, \quad -u_2' e^{-x}=\sec(x).$$ Then I stuck at finding $u_2$ as $u_2^{\prime}=-e^x \sec(x)$ and I have no idea on how to integrate this. Can anyone guide me?

Answer 1

$y''+y'=sec(x)$

$(y'e^x)'=\dfrac{d\int sec(x)e^xdx}{dx}$

$y'=e^{-x}\int sec(x)e^xdx +Ce^{-x}$

$y=\int e^{-x}\int sec(x)e^xdx dx+ C_1e^{-x}+C_2$

I do not know of any elementary form for the integrals. Wolfram gives an answer