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tracedinair
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Homework Statement
Find the general solution of the first-order differential equation,
y' + 3y = 2xe^(-3x)
Homework Equations
y' + P(x)y = Q(x)
Integrating factor = e^(∫P(x) dx)
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.