What is the solution for y^{''}-y^'=xe^x,y(0)=2,y^'(0)=1 ?

The solution for y^{''}-y^'=xe^x,y(0)=2,y^'(0)=1 is y=(e^xx^2)/2-e^xx+2e^x

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y^{''}-y^'=xe^x,y(0)=2,y^'(0)=1

y^{^{''}} - y^{^{'}} = x e^{x}, y (0) = 2, y^{^{'}} (0) = 1

y = \frac{e ^{x} x ^{2}}{2} - e^{x} x + 2 e^{x}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{x} x ^{2}}{2} - e^{x} x + 2 e^{x}

y = \frac{e ^{x} x ^{2}}{2} - e^{x} x + 2 e^{x}

y^{^{''}} + 4 y = e^{- 4 t}, y (0) = 2

y^{^{''}} + 25 y = 40 cos (5 t), y (0) = y

y^{^{''}} - 4 y^{^{'}} + 8 y = e^{2 x} + sin (2 x)

y^{^{''}} + 25 y = sec^{3} (5 t)

y^{^{''}} + 4 y^{^{'}} - 21 y = 5 e^{2 x} + 4

What is the solution for y^{''}-y^'=xe^x,y(0)=2,y^'(0)=1 ?
The solution for y^{''}-y^'=xe^x,y(0)=2,y^'(0)=1 is y=(e^xx^2)/2-e^xx+2e^x