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

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

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y^{''}-y^'=x[e^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^{^{''}} - 2 y^{^{'}} - 8 y = - 2 e^{2 x} - 5 x e^{- 4 x} - 2 x

D, D (t) = 2000 sin (\frac{π t}{90}) + 4000

y^{^{''}} - 16 y = 48 t - 24 e^{- 4 t}, y (0) = 0, y^{^{'}} (0) = 16

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 x + 4

y^{^{''}} + 6 y^{^{'}} - 7 y = 3 e^{2 x} - 2 e^{3 x}

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