What is the solution for y^{''}-2y^'=x+1 ?

The solution for y^{''}-2y^'=x+1 is y=c_{1}e^{2x}+c_{2}-(x^2)/4-(3x)/4

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y^{''}-2y^'=x+1

y^{^{''}} - 2 y^{^{'}} = x + 1

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{2}}{4} - \frac{3 x}{4}

Solution steps

y^{''} (x) - 2 y^{'} (x) = x + 1

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{2}}{4} - \frac{3 x}{4}

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{2}}{4} - \frac{3 x}{4}

y^{^{'''}} + y^{^{'}} = 3 + cos (x) - sin (x) + x

y^{^{''}} - y = e^{- 2 x} sin (e^{- x})

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 91 x e^{x}

x^{^{''}} + x = sin^{2} (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- x}

What is the solution for y^{''}-2y^'=x+1 ?
The solution for y^{''}-2y^'=x+1 is y=c_{1}e^{2x}+c_{2}-(x^2)/4-(3x)/4