Question

Solve the differential equation by variation of parameters. y''+y=secx

Answer 1

Answer:y=

Acosx+Bsinx +cosx ln(cosx)+x sinx

Step-by-step explanation:

given  equation y''+y=secx

auxiliary  equation

$p^2+1=0\\p=\pm i$

so CF is y=Acosx+Bsinx

now

$y_1(x)=cosx \ \ \ \ y_2(x)=sinx\\{y_1}'(x)=-sinx \ \ \ \ {y_2}'(x)=cosx$

using wronskian formula

$W=\begin{vmatrix}cosx &sinx \\ -sinx & cosx\end{vmatrix}$

          =$cos^2x+sin^2x=1$

now f(x)=secx

$u=-\int \frac{f(x)y_2(x)}{W(x)}dx \ and \ v=\int \frac{f(x)y_1(x)}{W(x)}dx$

$u=-\int \frac{secx\times sinx}{1}dx \ and \ v=\int \frac{secx\times cosx}{1}dx$

$u=-\int tanx dx \ and \ v=\int {1}dx$

$u=ln(cosx) \ \ \ and \ \ v=x$

now particular integrals are

PI=cosx ln(cosx)+x sinx

total solution

y= C.F+P.I

y=Acosx+Bsinx +cosx ln(cosx)+x sinx