What is the solution for y^{''}-y^'=4xe^x,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-y^'=4xe^x,y(0)=0,y^'(0)=1 is y=e^x-5+2e^xx^2-4(e^xx-e^x)

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y^{''}-y^'=4xe^x,y(0)=0,y^'(0)=1

y^{^{''}} - y^{^{'}} = 4 x e^{x}, y (0) = 0, y^{^{'}} (0) = 1

y = e^{x} - 5 + 2 e^{x} x^{2} - 4 (e^{x} x - e^{x})

Solution steps

y^{''} (x) - y^{'} (x) = 4 x e^{x}

Solve linear ODE:

y = e^{x} - 5 + 2 e^{x} x^{2} - 4 (e^{x} x - e^{x})

y = e^{x} - 5 + 2 e^{x} x^{2} - 4 (e^{x} x - e^{x})

What is the solution for y^{''}-y^'=4xe^x,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-y^'=4xe^x,y(0)=0,y^'(0)=1 is y=e^x-5+2e^xx^2-4(e^xx-e^x)