1. ## ODE help

Code:


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$(3x^2y-y^3)dx-(3xy^2-x^3)dy=0$

$
d^2y/dx^2 -4dy/dx +3y = e^3e^x
$

2. First one looks exact to me (divide out the $dx$). Then use the procedure involving potential functions for exact ODEs to find the solution. second is a linear non-homogeneous equation, so your general solution will be the general solution of the associated homogeneous equation plus any particular solution of the non-homogeneous equation. Since you have constant coefficients, you should use the method of undetermined coefficients.

3. ## trying out

I got $xy^3 - x^3y -2/x^2 =a$ where a is a constant
I use dz=partial z/partial x dx + partial z/ partial y dy method
dunno correct or not

by the way what is the definition of homogenous in layman terms thanks
As for the second Q i got y= C1e^3x +C2e^x

4. Originally Posted by tak
I got $xy^3 - x^3y -2/x^2 =a$ where a is a constant
I use dz=partial z/partial x dx + partial z/ partial y dy method
dunno correct or not

by the way what is the definition of homogenous in layman terms thanks
As for the second Q i got y= C1e^3x +C2e^x
The homogeneous solution is just what you get if the right side, typically written as $f(x)$ is zero. It is independent of the right side of the fuction, or the inhomogeneous part. In other words, you will get the same homogeneous solution, no matter what $f(x)$ is equal to.

Edit...It might help to know the standard form for a Second Order ODE is $Ay''+By'+Cy=f(x)$

Edit 2...This answer is correct so far, but you have only solved for the homogeneous solution. You need to set up, using the method of undetermined coefficients, a solution to the inhomogeneous part. It will be of the form: $Ate^x$ because you need a solution linearly independent of the solutions you have already gotten.

5. ## constant zero

I found my c3 to be zero; the general solution is just the complementry solution. correct?