What is the solution for y^{''}-7y^'=x^2 ?

The solution for y^{''}-7y^'=x^2 is y=c_{1}e^{7x}+c_{2}-(x^3)/(21)-(x^2)/(49)-(2x)/(343)

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y^{''}-7y^'=x^2

y^{^{''}} - 7 y^{^{'}} = x^{2}

y = c_{1} e^{7 x} + c_{2} - \frac{x ^{3}}{21} - \frac{x ^{2}}{49} - \frac{2 x}{343}

Solution steps

y^{''} (x) - 7 y^{'} (x) = x^{2}

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} - \frac{x ^{3}}{21} - \frac{x ^{2}}{49} - \frac{2 x}{343}

y = c_{1} e^{7 x} + c_{2} - \frac{x ^{3}}{21} - \frac{x ^{2}}{49} - \frac{2 x}{343}

x^{^{''}} + 3 x^{^{'}} + 2 x = 3 cos (2 t)

2 x^{^{''}} + 5 x^{^{'}} + 4 x = 10 r (t)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} - x^{3} + 16

x^{^{''}} + 6 x^{^{'}} - 4 x = 8 e^{t}

x^{^{''}} + 4 x^{^{'}} + 43 x = 9 cos (6 t)

What is the solution for y^{''}-7y^'=x^2 ?
The solution for y^{''}-7y^'=x^2 is y=c_{1}e^{7x}+c_{2}-(x^3)/(21)-(x^2)/(49)-(2x)/(343)