Second order ode with non constant coeffcients

[ice109]

Homework Statement

[tex] y''(x)-k y^2 y'(x)=0 [/tex]

The Attempt at a Solution

mathematica gives me this:

[tex]\left\{\left\{y(x)\to\text{InverseFunction}\left[-\frac{2\sqrt{3}\tan^{-1}\left(\frac{1}{\sqrt{3}}-\frac{2\sqrt[3]{k}\text{$\#$1}}{3^{5/6}\sqrt[3]{c_1}}\right)-2\log\left(3^{2/3}\sqrt[3]{k}\text{$\#$1}+3\sqrt[3]{c_1}\right)+\log\left(\sqrt[3]{3}k^{2/3}\text{$\#$1}^2-3^{2/3}\sqrt[3]{k}\sqrt[3]{c_1}\text{$\#$1}+3c_1^{2/3}\right)}{23^{2/3}\sqrt[3]{k}c_1^{2/3}}\&\right]\left[x+c_2\right]\right\}\right\}[/tex]

yes something so ugly and long that i broke the latex parser on the website. so how do i solve this? i have no idea where to begin because none of the methods in my ODE book talk about function with coeffIcients like that [itex]x^2[/itex]

 

[mighty2000]

Hello,

It seems like you are using Mathematica to solve the ODE. While Mathematica can provide a solution for certain types of ODEs, it may not always be the most intuitive or helpful solution. In order to better understand the solution, it may be helpful to break it down and try to solve it by hand.

First, let's rewrite the ODE in a more standard form:

y''(x) = k y^2 y'(x)

Now, let's try to solve this using the method of separation of variables. We can rewrite the equation as:

y''(x) = k y^2 (dy/dx)

We can then separate the variables and integrate both sides:

∫ y''(x) dx = ∫ k y^2 dy

Integrating the left side gives us:

y'(x) + C1 = ∫ k y^2 dy

We can then solve for y'(x):

y'(x) = ∫ k y^2 dy - C1

Now, we can solve for y(x) by integrating both sides again:

y(x) = ∫ ∫ k y^2 dy - C1 dx + C2

This is a general solution to the ODE. However, it is not in a very useful form as it involves double integration. In order to simplify it, we can make use of the initial conditions given in the problem. By plugging in the initial conditions, we can solve for the constants C1 and C2 and obtain a specific solution.

I hope this helps. Good luck with your problem!