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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?


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

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?

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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

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