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Popular >

y^'+y=2e^{-t}

Solution

y^{^{'}} + y = 2 e^{- t}

Solution

y = \frac{2 t}{e ^{t}} + \frac{c _{1}}{e ^{t}}

Solution steps

y^{'} (t) + y = 2 e^{- t}

Solve linear ODE:

y = \frac{2 t}{e ^{t}} + \frac{c _{1}}{e ^{t}}

y = \frac{2 t}{e ^{t}} + \frac{c _{1}}{e ^{t}}

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\int_{0}^{\frac{π}{2}} sin^{3} (t) d t

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\frac{d}{d x} (\frac{3 x ^{2} - 2}{2 x + 3})

Frequently Asked Questions (FAQ)

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