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Popular >

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

Solution

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

Solution

y = - 2 e^{- t} + 2

Solution steps

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

Solve separable ODE:

y = - 2 e^{- t} + 2

y = - 2 e^{- t} + 2

Popular Examples

slope of (-5.3)(7.9)slope

(- 5.3) (7.9)

integral from 0 to 1 of 5xe^{-x^2}

\int_{0}^{1} 5 x e^{- x^{2}} d x

x^2y^{''}=2xy^'+(y^')^2

x^{2} y^{^{''}} = 2 x y^{^{'}} + (y^{^{'}})^{2}

tangent of f(x)=x^2+8,(1,9)tangent of

f (x) = x^{2} + 8, (1, 9)

derivative of (4x^3/(x^4+27))

\frac{d}{d x} (\frac{4 x ^{3}}{x ^{4} + 27})

Frequently Asked Questions (FAQ)

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