# an ODE question

• Feb 14th 2010, 04:56 PM
renlok
an ODE question
ok i have a question im really stuck on

$x^3\frac{dy}{dx} + 3x^2y = \frac{1}{x+2}$
for this i used the substitution u = x^3y and end up with $(x+2)\frac{du}{dx} + 3u + \frac{6u}{x} = 1$ but no matter how i rearrange it i cant separate the variables
• Feb 14th 2010, 05:12 PM
Jhevon
Quote:

Originally Posted by renlok
ok i have a question im really stuck on

$x^3\frac{dy}{dx} + 3x^2y = \frac{1}{x+2}$
for this i used the substitution u = x^3y and end up with $(x+2)\frac{du}{dx} + 3u + \frac{6u}{x} = 1$ but no matter how i rearrange it i cant separate the variables

overkill. a substitution is not necessary. note that the left hand side of the original equation is $\left( x^3y \right)'$
• Feb 14th 2010, 05:18 PM
ANDS!
Oh dear. Well, at least you recognize this as a linear first order differential. It is the confusion on what your integrating factor is that is the problem. Remember, a differential is in linear form when it is written as:

$y'+P(x)+Q(x)=0$, with an integrating factor equal to $\mu = e^{\int P(x) dx}$.

Rewriting the equation so that it is in linear form we get:

$y'+\frac{3y}{x}=\frac{1}{x^{3}(x+2)}$

Thus, our integrating factor is $\mu = e^{\int \frac{3}(x) dx}$. From here you should see how things start to cancel out. If you need more help, post here.
• Feb 15th 2010, 08:58 AM
Jhevon
Quote:

Originally Posted by ANDS!
Oh dear. Well, at least you recognize this as a linear first order differential. It is the confusion on what your integrating factor is that is the problem. Remember, a differential is in linear form when it is written as:

$y'+P(x)+Q(x)=0$, with an integrating factor equal to $\mu = e^{\int P(x) dx}$.

Rewriting the equation so that it is in linear form we get:

$y'+\frac{3y}{x}=\frac{1}{x^{3}(x+2)}$

Thus, our integrating factor is $\mu = e^{\int \frac{3}(x) dx}$. From here you should see how things start to cancel out. If you need more help, post here.

the ODE is already in the required form. if we do what you did, we'd end up with $\mu = x^3$, and multiplying through by that brings us back to the original ODE! we don't need an integrating factor here, everything is already in a nice form. just continue as if you already found the integrating factor