1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $y_p=\frac {bx}{2t}\sinh(tx)$

$\endgroup$
1
  • $\begingroup$ Your a star - thankyou. $\endgroup$ – Mike Apr 3 '13 at 14:14
0
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Great - thankyou. $\endgroup$ – Mike Apr 3 '13 at 14:20
  • $\begingroup$ @MikeShaw: You are welcome. $\endgroup$ – Mhenni Benghorbal Apr 3 '13 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.