General solution of first-order differential equation

[tracedinair]

1. The problem statement, all variables and given/known data

Find the general solution of the first-order differential equation,

y' + 3y = 2xe^(-3x)


2. Relevant equations

y' + P(x)y = Q(x)

Integrating factor = e^(∫P(x) dx)

3. The attempt at a solution

Since it's already in the form y' + P(x)y = Q(x),

the integrating factor is I(x) = e^(∫3 dx) = e^(3x)

Now multiplying both sides by the integrating factor,

e^(3x)*y' + 3ye^(3x) = 2xe^(-3x)e^(3x)

d/dx (e^(3x)y) = 2x

Finally, integrating both sides and solving for y,

e^(3x)y = x^2 + C

General solution: y = x^(2)e^(-3x) + Ce^(-3x)

Not to sure if this is right, this is the first time I've studied diff eqns..

Thanks for any help.

[rock.freak667]

Yes that should be correct.

[Unco]

General solution: y = x^(2)e^(-3x) + Ce^(-3x)

Not to sure if this is right, this is the first time I've studied diff eqns.. There is one sure-fire way to check that your expression for y is a solution to the DE: plug it in!