Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have done very little on this problem. Sorry if I don't have even begun a solution. The reason is that I am unsure how to do it.

Here's a differential equation: $$ (y'')^3sin(y')+cos(xy')=tan((y')^4) $$

The question is : Which variable substitution would you do to bring this to a differential equation of order 1?

The answer is : p(x) = y'

Considering I'm not supposed to use trigonometric identities, I have difficulty seeing how I can achieve it (but it's ok if you use one... willing to take any solution)

I mean... it would give

$$ (p')^3sin(p) + cos(xp)= tan(p^4) $$

I still don't get it.

How to solve this?


share|cite|improve this question

1 Answer 1


$$(p')^3 \sin(p) + \cos(xp)= \tan(p^4)$$

This gives us:

$$p' = \left(\dfrac{\tan(p^4) - \cos(xp)}{\sin(p)}\right)^{1/3}$$

It is now a first-order, non-linear equation.

Can you see how to proceed now?

share|cite|improve this answer
Yes, you mean there's only p' and p. But still, I get you, but... from that point... how would you solve the equation? By using euler's equations maybe? Using tan(x) = sin(x)/cos(x)? – Yannick Dec 9 '13 at 18:18
I do not think they intended for you to solve it, if I read the question properly. Just to make a substitution to convert from second to first order. – Amzoti Dec 9 '13 at 18:47
I'm still curious. Would it even be possible to solve this? :) – Yannick Dec 10 '13 at 8:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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