0
$\begingroup$

I'm trying to solve a below ODE problem with TI-nspire cas:

$$ y''-4y'+3y=cos^2x $$

If I solve the above problem by hand (or using the wolfram alpha), I get the

$$ y=c_1e^x+c_2e^{3x}+\frac16-\frac{1}{130}(cos 2x+8sin2x) $$ However, when I solve this with Ti-nspire cas, gives the following answer: $$ y=c_1e^x+c_2e^{3x}+\frac{31}{195}-\frac{8sin(x)cos(x)}{65}+\frac{sin(x)^2}{65} $$ I type in the calculator as follows: $$ desolve(y''-4y'+3y=(cos(x))^2,x,y) $$ Is there any problem with my input? Or is the calculator failing to solve this?

I look forward to some help. Thanks.

$\endgroup$
  • 1
    $\begingroup$ I think you meant to write $e^{x}$ and $e^{3x}$ instead of $x^x$ and $x^{3x}$. $\endgroup$ – Dave Jan 1 '17 at 5:32
  • $\begingroup$ And I think you typo'd one off the signs there (on $\cos 2 x$) $\endgroup$ – Batman Jan 1 '17 at 5:33
  • $\begingroup$ Oh yeah, Dave, you're completely right, I've edited the text. Thank you. $\endgroup$ – James Jan 1 '17 at 5:42
  • $\begingroup$ Thanks Batman, I think it 's all right now. $\endgroup$ – James Jan 1 '17 at 5:46
  • $\begingroup$ MathJax hint: if you put backslashes before functions they come out in the correct font and spacing, so \sin x gives $\sin x$ instead of sin x giving $sin x$ $\endgroup$ – Ross Millikan Jan 1 '17 at 6:37
2
$\begingroup$

Use the trigonometric identities $\cos(2x) = 1-2 \sin^2 x $ and $\sin(2x) = 2 \sin x \cos x$.

You can check your answer by plugging each solution into the original ODE and seeing if it is true.

$\endgroup$

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.