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=(2-(1-2e))e^x+1-2e-(ex^3)/3-ex^2-2ex

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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 = (2 - (1 - 2 e)) e^{x} + 1 - 2 e - \frac{e x ^{3}}{3} - e x^{2} - 2 e x

Solution steps

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

Solve linear ODE:

y = (2 - (1 - 2 e)) e^{x} + 1 - 2 e - \frac{e x ^{3}}{3} - e x^{2} - 2 e x

y = (2 - (1 - 2 e)) e^{x} + 1 - 2 e - \frac{e x ^{3}}{3} - e x^{2} - 2 e x

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{2 x} + \frac{x}{2}

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 x

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 7 e^{- t} + 2

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=(2-(1-2e))e^x+1-2e-(ex^3)/3-ex^2-2ex