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Popular >

y^'+2y=0,y(1)=1

Solution

y^{^{'}} + 2 y = 0, y (1) = 1

Solution

y = e^{- 2 t + 2}

Solution steps

y^{'} (t) + 2 y = 0

Solve separable ODE:

y = e^{- 2 t + 2}

y = e^{- 2 t + 2}

Popular Examples

yy^'= x/(x^2+1)

y y^{^{'}} = \frac{x}{x ^{2} + 1}

y^{^{'}} + y = e^{x}

laplacetransform e^tt laplace transform

e^{t} t

integral of (1-cos(x))/(csc^2(x))

\int \frac{1 - cos ( x )}{csc ^{2} ( x )} d x

derivative of (8x+3^2)

\frac{d}{d x} ((8 x + 3)^{2})

Frequently Asked Questions (FAQ)

What is the solution for y^'+2y=0,y(1)=1 ?
The solution for y^'+2y=0,y(1)=1 is y=e^{-2t+2}