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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Popular Examples

integral from 6 to 8 of (20)/((x-6)^3)

\int_{6}^{8} \frac{20}{( x - 6 ) ^{3}} d x

derivative of {f}(x(e^x))

\frac{d}{d x} (f (x) (e^{x}))

y^{^{'}} - 6 y^{^{'}} + 9 y = 0

sum from n=0 to infinity of 3(-1/3)^n

n = 0 \sum \infty 3 (- \frac{1}{3})^{n}

tangent of y=x^3-2x+1,(4,57)tangent of

y = x^{3} - 2 x + 1, (4, 57)

Frequently Asked Questions (FAQ)

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