some higher order differential equations

roldy

1. The problem statement, all variables and given/known data

a) s^2*t''+st*t'=s
b) y(dx/dy)^2=x^2+1
c) 17y''''-t^6*y"-4.2y^5=3cost(t)

2. Relevant equations



3. The attempt at a solution
For part a I thought about doing a reduction of order but I can't because I have the variable s present. Not sure what my other options are.

Part b I solved for dx/dy but now I'm stuck. Can't think clearly

Part c I'm not even remotely sure on how to do this. I was thinking integrating factor but I don't know.

HallsofIvy

You can solve (2) algebraically for dx/dy:
[tex]\frac{dx}{dy}= \pm \frac{\sqrt{x^2+ 1}}{\sqrt{y}}[/tex]
and have two separable differential equations.

As for the other two, they are badly non-linear and I have no clue!

Rainbow Child

If the 1st one is

[tex]s^2\,t''+s\,t\,t'=s\Rightarrow t\,x''+x\,x'=1[/tex]

then as HallsofIvy said it is non-linear. (I changed the notation a bit).
The general strategy for these ODE's are to search for Lie point symmetries. Unfortunately this one does not have any
A second approach is to search for a first integral of the equation, which is based on a search and a try method.

For this ODE if you try a first integral of the type

[tex]I=f(x,t)\,x'+g(x,t)=C[/tex]

you can find that

[tex]I=t\,x'+\frac{1}{2}x^2-x-t=C[/tex]

which is a Riccati type 1st order ODE. With the standard transformation you can convert it in to a 2nd order homogeneous linear ODE which can be solved the help of Bessel functions.

coomast

By carefully following Rainbow Child's instructions I think I was able to solve the first equation. The first part was to find the first integral, which was indeed as stated:

[tex]f(x,t)\cdot x' +g(x,t)=C[/tex]

The functions f and g can be found by taking the derivative with respect to t giving:

[tex]fx''+f'x'+g'=0[/tex]

Setting this equal to the original equation gives then:

[tex]f(x,t)=t[/tex]

and thus f'=0, and also

[tex]g(x,t)=\frac{x^2}{2}-x-t[/tex]

We have now:

[tex]t\cdot x'+ \frac{x^2}{2}-x-t=C[/tex]

or:

[tex]\frac{dx}{dt}-\frac{1}{t}\cdot x + \frac{1}{2t}x^2=\frac{C}{t}+1[/tex]

The Riccati equation:

[tex]\frac{dx}{dt}+Q(t)x+R(t)x^2=P(t)[/tex]

can be transformed by using:

[tex]x=\frac{u'}{Ru}[/tex]

Doing this transforms the equation into:

[tex]t^2\frac{d^2u}{dt^2}-\frac{1}{2}(t+C)u=0[/tex]

Which is a special case of the following extended Bessel equation:

[tex]t^2\frac{d^2u}{dt^2}+(2k+1)t\frac{du}{dt}- \left(\alpha^2t^{2r}+\beta^2 \right)u=0[/tex]

Which has the solution:

[tex]u(t)=t^{-k}\left[AI_{\frac{\gamma}{r}}\left(\frac{\alpha}{r}t^r\right) + BK_{\frac{\gamma}{r}}\left(\frac{\alpha}{r}t^r\right) \right][/tex]

with:

[tex]\gamma=\sqrt{k^2-\beta^2}[/tex]

The functions are the modified Bessel functions of the first and second kind of order [tex]\frac{\gamma}{r}[/tex].

In our case we have:

[tex]k=-\frac{1}{2} \qquad r=\frac{1}{2} \qquad \beta^2=\frac{C}{2} \qquad \alpha^2=\frac{1}{2}[/tex]

and:

[tex]\frac{\gamma}{r}=\sqrt{1-2C}=n \qquad \frac{\alpha}{r}=\sqrt{2}[/tex]

The solution is thus:

[tex]u(t)=\sqrt{t} \left[ A I_{n}\left(\sqrt{2t}\right) + B K_{n}\left(\sqrt{2t}\right) \right][/tex]

After transforming back to x(t) we get:

[tex]x(t) = 1+\sqrt{2t} \left[ \frac{\displaystyle \frac{A}{2} \left[I_{n-1}\left(\sqrt{2t}\right) + I_{n+1}\left(\sqrt{2t}\right) \right]- \frac{B}{2} \left[K_{n-1}\left(\sqrt{2t}\right) +K_{n+1}\left(\sqrt{2t}\right) \right]}{\displaystyle A I_{n}\left(\sqrt{2t}\right) + B K_{n}\left(\sqrt{2t}\right)} \right][/tex]

Rewriting a little bit gives now:

[tex]x(t) = 1+\sqrt{\frac{t}{2}} \left[ \frac{\displaystyle \left[I_{n-1}\left(\sqrt{2t}\right) + I_{n+1}\left(\sqrt{2t}\right) \right]- D \left[K_{n-1}\left(\sqrt{2t}\right) +K_{n+1}\left(\sqrt{2t}\right) \right]}{\displaystyle I_{n}\left(\sqrt{2t}\right) + D K_{n}\left(\sqrt{2t}\right)} \right][/tex]

This solution has indeed two parameters C (in n) and D. To see if it is the correct one, it must be validated using the original equation. I haven't done this yet, the algebra is somewhat involved.

Non-linear differential equations are indeed not easy. Thanks to Rainbow Child for the instructions, it was a pleasure doing this calculation.

@ roldy: Is this the result you were looking for? Where exactly do these equations come from?