What is the solution for y^{''}-7y^'=t^2 ?

The solution for y^{''}-7y^'=t^2 is y=c_{1}e^{7t}+c_{2}-(t^3)/(21)-(t^2)/(49)-(2t)/(343)

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

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

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

Solution steps

y^{''} (t) - 7 y^{'} (t) = t^{2}

Solve linear ODE:

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

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

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

y^{^{''}} - a^{2} y = sech (a x)

y^{^{''}} - 5 y^{^{'}} + 6 y = - 4 e^{3 x} + 10 e^{x} - 12

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

y^{^{''}} + y = cos (- t + \frac{π}{2})

What is the solution for y^{''}-7y^'=t^2 ?
The solution for y^{''}-7y^'=t^2 is y=c_{1}e^{7t}+c_{2}-(t^3)/(21)-(t^2)/(49)-(2t)/(343)