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Popular >

y^'=y-t,y(0)=0.5

Solution

y^{^{'}} = y - t, y (0) = 0.5

Solution

y = t + 1 - 0.5 e^{t}

Solution steps

y^{'} (t) = y - t

Solve linear ODE:

y = t + 1 - 0.5 e^{t}

y = t + 1 - 0.5 e^{t}

Popular Examples

integral of (4/(x^3))

\int (\frac{4}{x ^{3}}) d x

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\frac{d}{d x} (\frac{x ^{2} + 2}{x - 1})

derivative of f(x)=(x-2)^3+10 derivative of

f (x) = (x - 2)^{3} + 10

integral of (3x^2+2)sqrt(2x^3+4x+1)

\int (3 x^{2} + 2) 2 x^{3} + 4 x + 1 d x

derivative of f(x)=ln(xe^{-3x})derivative of

f (x) = ln (x e^{- 3 x})

Frequently Asked Questions (FAQ)

What is the solution for y^'=y-t,y(0)=0.5 ?
The solution for y^'=y-t,y(0)=0.5 is y=t+1-0.5e^t