What is the solution for y^{''}-y^'+4=3x^2+4x ?

The solution for y^{''}-y^'+4=3x^2+4x is y=c_{1}e^x+c_{2}-x^3-5x^2-6x

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

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

y = c_{1} e^{x} + c_{2} - x^{3} - 5 x^{2} - 6 x

Solution steps

y^{''} (x) - y^{'} (x) + 4 = 3 x^{2} + 4 x

Solve linear ODE:

y = c_{1} e^{x} + c_{2} - x^{3} - 5 x^{2} - 6 x

y = c_{1} e^{x} + c_{2} - x^{3} - 5 x^{2} - 6 x

y^{^{''}} - y^{^{'}} + 4 y = 5 t^{2} + 9 t + 4

y^{^{'''}} + 3 y^{^{'}} + 2 = 0

y^{^{''}} + 4 y^{^{'}} + 12 y = 80 sin (2 x)

y^{^{''}} - y^{^{'}} + y = 3 sin (t) - 27 e^{2 t}

\frac{d ^{2} y}{d t ^{2}} + \frac{3}{100} \frac{d y}{d t} = 12

What is the solution for y^{''}-y^'+4=3x^2+4x ?
The solution for y^{''}-y^'+4=3x^2+4x is y=c_{1}e^x+c_{2}-x^3-5x^2-6x