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Popular >

y^'+y=0,y(0)=1

Solution

y^{^{'}} + y = 0, y (0) = 1

Solution

y = e^{- t}

Solution steps

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

Solve separable ODE:

y = e^{- t}

y = e^{- t}

Popular Examples

y d x + (1 - x) d y = 0

(e^x+x)ydy-e^ydx=0

(e^{x} + x) y d y - e^{y} d x = 0

cos(y)sin(t)(dy)/(dt)=sin(y)cos(t)

cos (y) sin (t) \frac{d y}{d t} = sin (y) cos (t)

(y^')/t =3y^2+2y+1

\frac{y ^{^{'}}}{t} = 3 y^{2} + 2 y + 1

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

Frequently Asked Questions (FAQ)

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