0
$\begingroup$

I'm not looking for an answer to the whole question, but I'm just stuck on the beginning of the problem, finding the complementary solution. I'm almost certain that it wouldn't be $y_c = C_1\cos^3(x) + C_2\sin^3(x)$. Any ideas?

$\endgroup$
  • $\begingroup$ You are really asking about the complementary solution? You want help in solving the homogeneous equation $y''+4y=0$? Is that your question? $\endgroup$ – bof Nov 1 '14 at 21:25
  • $\begingroup$ Ah, thanks Amzoti. I also made a stupid technical error in denoting the auxiliary equation as: m^2 + 4m = 0. Regardless, I wasn't sure if the power of sin(x) would change the complimentary solution to C1cos^3(2x) + C2sin^3(x) $\endgroup$ – SE Student Nov 1 '14 at 21:39
1
$\begingroup$

Try it !

$$(\cos^3x)''=(-3\sin x\cos^2x)'=-3\cos^3x+6\sin^2x\cos x=6\cos x-9\cos^3x.$$ $$(\sin^3x)''=(3\cos x\sin^2x)'=-3\sin^3x+6\cos^2x\sin x=6\sin x-9\sin^3x.$$

So a term like $\sin^3x$ contributes to the solution. To get rid of the $\sin x$, a complementary $\sin x$ term will do.

$$y=C_1\sin^3x+C_2\sin x,$$ $$y''+4y=C_1(6\sin x-9\sin^3x+4\sin^3x)+C_2(-\sin x+4\sin x).$$ Solve $$6C_1+3C_2=0,\\-5C_1=1.$$

$\endgroup$
0
$\begingroup$

Note that:

$\sin 3x= \sin x \cos 2x + \cos x \sin 2x= \sin x (1-2\sin^2 x)+2\sin x (1-\sin^2 x)=3 \sin x - 4\sin^3 x$

Which is by way of saying $$\sin^3 x=\frac 34 \sin x -\frac 14 \sin 3x$$

You may be more familiar with the test functions for the form on the right hand side.

$\endgroup$
  • $\begingroup$ You are answering a different question. The question asks for help in finding the complementary solution. $\endgroup$ – bof Nov 1 '14 at 21:31
  • $\begingroup$ @bof I am responding to what OP put in the question, and assuming that a particular solution for the right-hand side was what was wanted (given the test form exhibited). I may be wrong, but I assumed this might be the help really wanted. $\endgroup$ – Mark Bennet Nov 1 '14 at 21:41
  • $\begingroup$ @Amzoti: and I was thinking I might be wrong! There is a little bit of confusion in the question, and it would likely have been better to wait, $\endgroup$ – Mark Bennet Nov 1 '14 at 21:53
  • $\begingroup$ @MarkBennet: This does confirm my solution to another problem that arose throughout the question. Though my derivation for that identity was different. Thank you. Not sure if this is the place to ask, but you made the assertion that $sin(3x) = 3sin(x) - 4sin^3(x)$, then with that logic, you can make the assertion that $\sin^3(x) = 3/4sin(x) - 1/4sin(3x)$ I'm just curious of that logic. $\endgroup$ – SE Student Nov 1 '14 at 22:26

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.