Question

Still having trouble solving this differential equation:
y^4-y^3*y''=1
Anyone can help?
[ previous post here : http://openstudy.com/#/updates/4edc8f26e4b01b4ec492aced ]

Answer 1

\[y^4-y^3*y''=1\]

Answer 2

is that y^n (an exponent ?) or the nth differential of y?

Answer 3

or the second differential of y?

Answer 4

second differential of y

Answer 5

OK, I will try it

Answer 6

wait is it with respect to x or y?

Answer 7

Yes, y=y(x)
x - independent argument.

Answer 8

\[y^4 - y^3(y'') = 1\]
\[\therefore y'' = (y^4 - 1)/y^3\]
\[\therefore y' = \int\limits (y^4 - 1)/y^3 dx = \int\limits y -1/y^3 dx\]

Answer 9

You know how to take it from there?

Answer 10

Hmmm... \[\int\limits ydx\] and \[\int\limits -1/y^3 dx\]
is what then?
I don't know if it is legit to say that, for example, \[\int\limits ydx = yx+C_1\] because y is function of x?

Answer 11

yx differentiates to y + xdy/dx which is not what we want.
But using integration by parts:
\[\int\limits y dx = yx - \int\limits x dy\]
but that leaves us with the same problem.
We will probably need to find some substitution

Answer 12

Normally with a second order differential equation of the type y'' +5y' +6y. I would use the substitution \[y = e^{rx}\]
then \[y' = re^{rx}\]
and \[y'' = r^2 e^{rx} \]
and you substitute and solve for r.
This does not work in this case, because this is not a linear d.e.
I can't think of anything else. Will keep looking.

Answer 13

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.