y'-2y=t^2e^2t

Question

y'-2y=t^2e^2t

primeralph · Answer

What methods do you know?

primeralph · Answer

To explain I need to know what you already know.
I will leave this question in 2 minutes if you don't reply.

dumbcow · Answer

multiply equation by integrating factor....e^-2t
$$e^{-2t}y' -2e^{-2t}y = t^{2}$$
rewrite right side as derivative of product
$$(e^{-2t} y)' = t^{2}$$
integrate
$$\int\limits (e^{-2t} y)' = \int\limits t^{2}$$
$$e^{-2t} y = \frac{1}{3} t^{3} +C$$
$$y = \frac{e^{2t}}{3}(t^{3} +C)$$

anonymous · Answer

thanks dumbcow, sorry for the delay primeralph

anonymous · Answer

@dumbcow how did you go from step 1 to 2 in your method where you rewrite it as a product?

dumbcow · Answer

by recognizing that it was in the form of: $$fg' + f'g$$ which is the product rule for derivatives...you can say $$(fg)'$$ in general if the diff equ is in the form $$y' + p(t)y = q(t)$$ then multiplying by $$\large e^{\int\limits p(t) dt}$$ will always allow you to write right side as derivative of a product more in depth analysis can be found here: http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

dumbcow · Answer

@FutureMathProfessor

anonymous · Answer

Thanks Dumbcow! I don't know why you picked that name because you're sure not dumb!