Please help solve the aboveCode:$\displaystyle (3x^2y-y^3)dx-(3xy^2-x^3)dy=0 $

$\displaystyle

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

$

Please solve both

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- Apr 18th 2008, 11:27 AM #1

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- Apr 18th 2008, 11:56 AM #2

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First one looks exact to me (divide out the $\displaystyle dx$). 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.

What have you already tried?

- Apr 18th 2008, 01:32 PM #3

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## trying out

I got $\displaystyle xy^3 - x^3y -2/x^2 =a$ 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

- Apr 18th 2008, 02:09 PM #4

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The homogeneous solution is just what you get if the right side, typically written as $\displaystyle f(x)$ 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 $\displaystyle f(x)$ is equal to.

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

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: $\displaystyle Ate^x$ because you need a solution linearly independent of the solutions you have already gotten.

- Apr 19th 2008, 01:24 AM #5

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- Apr 19th 2008, 01:46 AM #6

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