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Popular >

y^'+y=0

Solution

y^{^{'}} + y = 0

Solution

y = e^{- t + c_{1}}

Solution steps

y^{'} (t) + y = 0

Solve separable ODE:

y = e^{- t + c_{1}}

y = e^{- t + c_{1}}

Popular Examples

(ln (y))^{^{'}}

(3x^2+y^2)/(y^2)dx-(2x^3+5y)/(y^3)dy=0

\frac{3 x ^{2} + y ^{2}}{y ^{2}} d x - \frac{2 x ^{3} + 5 y}{y ^{3}} d y = 0

derivative of 1/(sin(1/x))

\frac{d}{d x} (\frac{1}{sin ( \frac{1}{x} )})

derivative of (4x/(5-cot(x)))

\frac{d}{d x} (\frac{4 x}{5 - cot ( x )})

integral of (e^t+2e^{2t})/(e^{3t)}

\int \frac{e ^{t} + 2 e ^{2 t}}{e ^{3 t}} d t

Frequently Asked Questions (FAQ)

What is the solution for y^'+y=0 ?
The solution for y^'+y=0 is y=e^{-t+c_{1}}