When it comes to differential equations, what would be the correct procedure going about this problem?

When it comes to differential equations, what would be the correct procedure going about this problem?

I don't need answers just techniques.

Normally, there would be a preferred way based on what you're modeling.

If it's just homework, either way would be acceptable.

Shouldn't both ways result in the same general solution? Why would a certain model prefer one over the other?

op here; I believe the general solution is this, can someone confirm?

answer should be y(x)=Ae^(x^3)+Be^(x^2)+Ce^x+D

You exponentiate after the fourth integration. You can't exponentiate both sides when both sides are integral equations. And in the example you posted, you'd end up with the fourth integral of 1/y on one side and a degree 4 polynomial of x on the other, where all coefficients but the leading coefficient are constants of integration

how would I get rid of the ln y - y d^2? Something tells me the further I integrate the left hand side the further away I get from 1/y...

does y''''(x)=y(x) ?

I believe that it should, but when doing the integration, it seems...off

FEM

you dont seem to understand how to integrate properly. You can't just decide where the differential goes. If you write it with parenthesis, it will help you see.

[eqn]\iiiint \frac{1}{y}\; \mathrm{d}^4y=\iiint \ln y\;\mathrm{d}^3y=\iint (y\ln y-y)\;\mathrm{d}^2y=\int \frac{1}{4}y^2(2\ln y-3)\;\mathrm{d}y= \frac{1}{36}y^3 (6\ln y-11)[/eqn]
[eqn]=\iiiint \;\mathrm{d}^4x=\iiint x+A\;\mathrm{d}^3x=\iint \frac{x^2}{2} +Ax+B\;\mathrm{d}^2x=\int \frac{x^3}{6} + \frac{Ax^2}{2} +Bx+C\;\mathrm{d}x =\frac{x^4}{24}+\frac{Ax^3}{6}+\frac{Bx^2}{2} +Cx+D[/eqn]

jacob barnett is that you?

Anyways, I havent studied differential equations so I dont know if this method applies to equations above order 1

>what would be the correct procedure going about this problem?
Thinking for 2 seconds before starting to do all kinds of retarded calculations like .
The solution is clearly sines, cosines and exponents. It's not like you're going to algebraicly solve something like [math]y^3 \mathrm{ln}(y)=[/math] polynomial anyway.

yeah, now that I think about it, you'd solve this one as you would any other linear ODE (you typically see in your first ODE class up to order 2).
The roots to the auxiliary (?) equation are 1,-1,i,-i, so the solution would be evidently of the form [eqn]y(x)=Ae^x+Be^{-x}+C\cos x+D\sin x[/eqn]

I feel kind of stupid now

Fuck no. Separation of variable is pleb ass bullshit. You're secretly doing the inverse chain rule but they teach it as though you separate the dy and dx so retards can understand it. The dy and dx are not separate. dy^3 is not actual exponentiation. Fucking kys

hey, we're just having a friendly discourse here, no need for that moody tone.

anyway op, I'm pretty sure I can bet my money on 's answer

Separation of variables is not the correct method here. You just put in [math]e^{\gamma x}[/math] or whatever and see what you get. The general solution then is [math]c_1 e^{-x} + c_2 e^x + c_3 e^{-ix} + c_4 e^{ix}[/math]. It's very straight-forward.

how do you know you cannot separate the variables? is it because it's a higher order ode?

Separation of variables is a very specific "trick" to solve a small range of ODEs. Specifically this kind of ODE:
[eqn]\frac{\mathrm{d}y}{\mathrm{d}x} = f(x)g(y)[/eqn]

The solution is then given implicitely as

[eqn]\int \frac{\mathrm{d} y}{g(y)} = \int f(x) \mathrm{d} x[/eqn]

Don't mistake this for an example of some kind of algebra with differentials or a method any more general then the two equations above.

Are you ?

Because when you said to put e^gamma*x, I wasn't quite sure what you meant.

if you have a fourth order differential equation like this, you are supposed to convert it into a system of first order differential equations, and compute the change of basis matrix which converts your matrix into the (generalized) eigenvalue form. From there, it is easy to see that the solutions to the (generalized) eigenvalue form are of the form (x^k)*e^(lx). Applying the inverse change of basis, you get the solution to your system of first order differential equations, which corresponds to the solution of the fourth order equation.

When you evaluate a function like this, you'll get the same values either way, but often times, these describe physical systems. When looking at the system, having an exponential vs. Logs or hypertrig can be preferential for what measurements you're making

no, as long as you have sufficiently many independent solutions, you can take a linear combination of them and treat it like a general solution.

That only works for some linear n order diff. Eq.

Ops is clearly non linear

[math]\color{#b5bd68}{>\dfrac{\text{d}^4 y}{\text{d} x^4} = y \text{ is non-linear}}[/math]

>it's an user thinks derivatives are fractions episode

One word: CHARACTERISTICPOLYNOMIAL

...

>y^(4) - y = 0
>non linear
d^4/dx^4 of f(x) = f(x), d^4/dx^4 of g(x) = g(x)
d^4/dx^4 of a*f(x) + b*g(x) = d^4/dx^4 (a*f(x)) + d^4/dx^4 (b*g(x)) = a*(d^4/dx^4 of f(x)) + b*(d^4/dx^4 of g(x)) = a*f(x) + b*g(x), so d^4/dx^4 (a*f(x) + b*g(x)) = a*f(x) + b*g(x). Linear combinations of solutions to d^4/dx^4 of y = y are also solutions, so OP's differential equation is linear.