Choice of Particular Solution for an inhomogeneous ODE

Question

Some advice/guidance required...

Given the following problem:

$$ y'' - t^2 y = a\cos(tx)+b\cosh(t x), $$ where $$ y(0) = 1, y(1) = 2 $$

The method I am using is as follows:

1) Solve the related homogeneous equation (gives $ y_c = A\cosh(tx) + B\sinh(tx) $)

2) Use a particular solution to generate the general solution ($ y_g = y_c + y_p $). The particular solution I am using is $$ y_p = C\cos(tx) + D\cosh(tx) $$

The question I have is, am I using the correct $ y_p $ ?

I don't get the solution I require so I suspect I'm using an incorrect $ y_p $ but I'm not sure why?

Also, firstly, as $ \cosh(t x) $ is used within $ y_c $ do I need to 'adjust' it within $ y_p $ with another constant to allow the Math to work? This would give $$ y_p = C\cos(tx) + D w \cosh(tx) $$ Where I have added the constant $w$.

Or... should I be taking $ a\cos(tx) $ and $ b\cosh(t x) $ seperately and solving for each to give $ y_{p1} $ and $ y_{p2} $ and then adding to get $ y_g = y_c + y_{p1} + y_{p2} $ ?

Thank you in advance for any pointers.

Answer 1

Any correct particular solution is as good as any other. If I substitute yours into the equation $y_p''-t^2y_p=-Ct^2\cos(tx)+Dt^2\cosh(tx)=-Ct^2\cos(tx)-Dt^2\cosh(tx)=-2Ct^2\cos(tx)$ which doesn't solve the equation.

You can certainly find particular solutions for each term on the right. When I feed the first to Alpha I get $y_p=\frac a{2t^2}cos(tx)$ and the second gives $y_p=\frac {bx}{2t}\sinh(tx)$

Answer 2

Note that, since the Wronskian equals to $t$, then it is easier to use the variation of parameters method.