What is the solution for y^'-ky=a+be^{-kt},y(0)=0 ?

The solution for y^'-ky=a+be^{-kt},y(0)=0 is y=-a/k-b/(2ke^{kt)}+(a/k+b/(2k))e^{kt}

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y^'-ky=a+be^{-kt},y(0)=0

y^{^{'}} - k y = a + b e^{- k t}, y (0) = 0

y = - \frac{a}{k} - \frac{b}{2 k e ^{k t}} + (\frac{a}{k} + \frac{b}{2 k}) e^{k t}

Solution steps

y^{'} (t) - k y (t) = a + b e^{- k t}

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y = - \frac{a}{k} - \frac{b}{2 k e ^{k t}} + (\frac{a}{k} + \frac{b}{2 k}) e^{k t}

y = - \frac{a}{k} - \frac{b}{2 k e ^{k t}} + (\frac{a}{k} + \frac{b}{2 k}) e^{k t}

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

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What is the solution for y^'-ky=a+be^{-kt},y(0)=0 ?
The solution for y^'-ky=a+be^{-kt},y(0)=0 is y=-a/k-b/(2ke^{kt)}+(a/k+b/(2k))e^{kt}