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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve linear ODE:

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

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

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(dy)/(dx)=y^2e^{-x}

\frac{d y}{d x} = y^{2} e^{- x}

integral of x*x^{20}

\int x \cdot x^{20} d x

Frequently Asked Questions (FAQ)

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