What is the solution for y^{''}+4y^'-5y=10x^2+9e^x ?

The solution for y^{''}+4y^'-5y=10x^2+9e^x is y=c_{1}e^x+c_{2}e^{-5x}-2x^2-(16x)/5-84/25+(3e^xx)/2

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y^{''}+4y^'-5y=10x^2+9e^x

y^{^{''}} + 4 y^{^{'}} - 5 y = 10 x^{2} + 9 e^{x}

y = c_{1} e^{x} + c_{2} e^{- 5 x} - 2 x^{2} - \frac{16 x}{5} - \frac{84}{25} + \frac{3 e ^{x} x}{2}

Solution steps

y^{''} (x) + 4 y^{'} (x) - 5 y = 10 x^{2} + 9 e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 5 x} - 2 x^{2} - \frac{16 x}{5} - \frac{84}{25} + \frac{3 e ^{x} x}{2}

y = c_{1} e^{x} + c_{2} e^{- 5 x} - 2 x^{2} - \frac{16 x}{5} - \frac{84}{25} + \frac{3 e ^{x} x}{2}

What is the solution for y^{''}+4y^'-5y=10x^2+9e^x ?
The solution for y^{''}+4y^'-5y=10x^2+9e^x is y=c_{1}e^x+c_{2}e^{-5x}-2x^2-(16x)/5-84/25+(3e^xx)/2