Question

Hello!

How do I solve the differential equation:

y''+y=1/(cos(x))

Thanks in advance!

Answer 1

In more detail: How can I solve the particular integral 1/cos x ?

Answer 2

\[\frac{1}{\cos(x)}=\sec{x}\]

\[\int\sec(x)dx=\int\sec(x)\frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}dx\]

\[\int\frac{\sec^2(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}dx\]

let \[u=\sec(x)+\tan(x)\]

answer should be easy to get from there

Answer 3

Thank you for the integral, however I do not need solely integrate 1/cos(x), I have to find a trial function y for which y''+y=1/cos (x)

The homogeneous solution is easy, given roots of the auxiliary equation i and -i, it is:

y=Acos(x)+Bsin(x)

The general solution is then y=Acos(x)+Bsin(x)+PI

and PI is then the function y, which satisfies y''+y=1/cos(x)

Answer 4

Use variation of parameters technique

Answer 5

Thanks, found a good tutorial for that technique and I think I might find the solution with that