Question

Can anyone help me with solving this differential equaton:
y^4-y^3*y''=1
where y=y(x)

Answer 1

Tricky ... in that it needs some sort of trick of substitution, which I haven't seen yet.

In the meantime, here's the exact solution:
http://www.wolframalpha.com/input/?i=y%5E4-y%5E3*y%27%27%3D1

Answer 2

I observe that if we write \( u = (2y)^2 \), then from the Wolfram solution, we have
\[ u = (c_1 f - 1/f)^2 - 4f^2 \]
where \( f(x) = e^{x + c_2} \) and hence satisfies the equation \( f'/f = 1 \).

In my approach, the next step is to try and reverse engineer the method to solve the original equation. If I get any further with this later today I'll let you know.

Answer 3

Nothing? Because this to me seems like a real tough stuff.
I was trying to approach to it by trying to regroup the equation:
we can see, that (y^4)'=4*y^3*y' , although it has y^3, it has first derivative y', but we need y'', so this approach got me nowhere.

Answer 4

And basically saying \[y^3(y-y'')=1\] gives us nothing.

Answer 5

Finally thought of a way to get some kind of solution:
\[y^4-y^3*y''=1\]\[y''=(y^4-1)/y^3\] multiply both sides with 2y'
\[2y'*y''=2y'(y-1/y^3)\]\[2y'*y''=(y'^2)'\]\[(y'^2)'=2y'(y-1/y^3)\]
Integrate both sides:
\[(y')^2=2 \int\limits y'*(y-1/y^3)dy\]\[(y')^2=y^2+1/y^2+C_1\]
Now I guess we take square root of both sides and integrate one time.
Of course solving the equation like this, we can get solution in form of x=x(y), but for now I cannot think of anything else.

Answer 6

You've got it.

\[ y^2 (y')^2 = 1 + y^4 + C1.y^2 \]
and thus looking at the positive root,
\[ \frac{y}{\sqrt{1+C_1.y^2 + y^4}} dy = dx \]
Now substitute u = y^2, du = 2y dy
\[ \frac{du}{2\sqrt{(u + C_1/2)^2 + (1-C_1^2/4)}} = dx \]
which you can now integrate
\[ \ln(\sqrt{ (u + C_1/2)^2 + (1-C_1^2/4) } + (u + C_1/2)) = x + C_2 \]
and this you can now unwind to solve for u, then y.

Answer 7

...I dropped the 1/2

Answer 8

Hmm.. why only positive root?

Answer 9

You need both obviously. I just wanted to show it's straightforward.