What is the solution for y^{''}-3y^'+7y=t^2+cos(t) ?

The solution for y^{''}-3y^'+7y=t^2+cos(t) is

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y^{''}-3y^'+7y=t^2+cos(t)

y^{^{''}} - 3 y^{^{'}} + 7 y = t^{2} + cos (t)

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2})) + \frac{t ^{2}}{7} + \frac{6 t}{49} + \frac{4}{343} - \frac{1}{15} sin (t) + \frac{2}{15} cos (t)

Solution steps

y^{''} (t) - 3 y^{'} (t) + 7 y = t^{2} + cos (t)

Solve linear ODE:

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2})) + \frac{t ^{2}}{7} + \frac{6 t}{49} + \frac{4}{343} - \frac{1}{15} sin (t) + \frac{2}{15} cos (t)

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2})) + \frac{t ^{2}}{7} + \frac{6 t}{49} + \frac{4}{343} - \frac{1}{15} sin (t) + \frac{2}{15} cos (t)

y^{^{'''}} - y^{^{''}} - 9 y^{^{'}} + 9 y = 2 - e^{x} + e^{3 x}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = 4 cos (x)

(d^{3} + d) y = 6 x (2 x + 1)

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

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{2 x} sin (3 x) - sin (x)

What is the solution for y^{''}-3y^'+7y=t^2+cos(t) ?
The solution for y^{''}-3y^'+7y=t^2+cos(t) is