What is the solution for y^{''}-y^'=ex(10+x^2) ?

The solution for y^{''}-y^'=ex(10+x^2) is y=c_{1}e^x+c_{2}-(ex^4)/4-ex^3-8ex^2-16ex

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y^{''}-y^'=ex(10+x^2)

y^{^{''}} - y^{^{'}} = e x (10 + x^{2})

y = c_{1} e^{x} + c_{2} - \frac{e x ^{4}}{4} - e x^{3} - 8 e x^{2} - 16 e x

Solution steps

y^{''} (x) - y^{'} (x) = e x (10 + x^{2})

Solve linear ODE:

y = c_{1} e^{x} + c_{2} - \frac{e x ^{4}}{4} - e x^{3} - 8 e x^{2} - 16 e x

y = c_{1} e^{x} + c_{2} - \frac{e x ^{4}}{4} - e x^{3} - 8 e x^{2} - 16 e x

y^{^{''}} + 9 y = \frac{1}{9} sin (x)

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 7 + 7 sin (x)

y^{^{''}} + y^{^{'}} = 6 x + e^{x}

y^{^{''''}} - y = 13 t + cos (t)

y^{^{''}} + 4 y = 2 e^{3 x}

What is the solution for y^{''}-y^'=ex(10+x^2) ?
The solution for y^{''}-y^'=ex(10+x^2) is y=c_{1}e^x+c_{2}-(ex^4)/4-ex^3-8ex^2-16ex