First Order Differential equation

**p00ndawg** · February 11th 2009, 07:16 PM

$(9x^2 + y -1)dx - (4y -x)dy = 0, y(0) = 1$

solve the given initial value problem and determine at least approximately where the solution is valid.

Can someone help me with this? im not quite sure how to really start this problem, I got that after integrating

$3x^3 -x = 2y^2$

but im not sure where to go. The equations came out as exact after partial differentiating both with respect to each other x and y.

**Jhevon** · February 11th 2009, 07:19 PM

Originally Posted by p00ndawg

$(9x^2 + y -1)dx - (4y -x)dy = 0, y(0) = 1$

solve the given initial value problem and determine at least approximately where the solution is valid.

Can someone help me with this? im not quite sure how to really start this problem, I got that after integrating

$3x^3 -x = 2y^2$

but im not sure where to go. The equations came out as exact after partial differentiating both with respect to each other x and y.

you have an exact equation here. now do you know how to proceed?

see post #2 here

**p00ndawg** · February 11th 2009, 07:21 PM

Originally Posted by Jhevon

you have an exact equation here. now do you know how to proceed?

no, like what do i do with the initial value?

**p00ndawg** · February 11th 2009, 07:26 PM

Originally Posted by Jhevon

you have an exact equation here. now do you know how to proceed?

see post #2 here

oh so i just plug it into the general solution?

**Chris L T521** · February 11th 2009, 07:44 PM

Originally Posted by p00ndawg

oh so i just plug it into the general solution?

Yes. Since the general solution has the form $f\!\left(x,y\right)=C$ , plugging in the initial condition yields $f\!\left(0,1\right)=C$ . So you're particular solution would be $f\!\left(x,y\right)=f\!\left(0,1\right)$

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