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Popular Calculus >

y^'+y=te^t

Solution

y^{^{'}} + y = t e^{t}

Solution

y = \frac{e ^{t} ( 2 t - 1 )}{4} + \frac{c _{1}}{e ^{t}}

Solution steps

y^{'} (t) + y = t e^{t}

Solve linear ODE:

y = \frac{e ^{t} ( 2 t - 1 )}{4} + \frac{c _{1}}{e ^{t}}

y = \frac{e ^{t} ( 2 t - 1 )}{4} + \frac{c _{1}}{e ^{t}}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'+y=te^t ?
The solution for y^'+y=te^t is y=(e^t(2t-1))/4+(c_{1})/(e^t)