Integrating Factor for Solving ODE with Linear Coefficients

[iRaid]

Problem:
xy'+2y=3x
Attempt:
Divide by x...
y'+\frac{2y}{x}=3
I think I find the integrating factor by doing:
e^{\int \frac{2}{x}dx}

Not sure if that's right but if it is then the solution to the integral is just 2x.

Any help is appreciated

 

[rock.freak667]

You will need to add in the constant of integration to the right side before you do any simplification.

 

[HallsofIvy]

Problem:
xy'+2y=3x
Attempt:
Divide by x...
y'+\frac{2y}{x}=3
I think I find the integrating factor by doing:
e^{\int \frac{2}{x}dx}

Not sure if that's right but if it is then the solution to the integral is just 2x.

Any help is appreciated

Yes, the integrating factor is e^{\int (2/x)dx} but then the integral is 2 ln(x) so the integrating factor is e^{2ln(x)}= e^{ln(x^2)}= x^2, not 2x.

Multiplying the equation by that gives x^2y'+ 2xy= (x^2y)'= 3x^2

 

[iRaid]

Yeah I figured it out ends up:
x^{2}y'+2xy=3x^{2}
Take integral/derivative of integrating factor:
\int \frac{d}{dx} x^{2}y=\int 3x^{2}dx
Simplifies to:
x^{2}y=x^{3}+C \implies y=\frac{x^{3}+C}{x^{2}}
(can be simplified a little more too)

Thanks for the help.