What is the solution for y^{''}-4y^'+3y=e^t+t^2 ?

The solution for y^{''}-4y^'+3y=e^t+t^2 is y=c_{1}e^{3t}+c_{2}e^t-(e^tt)/2+(t^2)/3+(8t)/9+26/27

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y^{''}-4y^'+3y=e^t+t^2

y^{^{''}} - 4 y^{^{'}} + 3 y = e^{t} + t^{2}

y = c_{1} e^{3 t} + c_{2} e^{t} - \frac{e ^{t} t}{2} + \frac{t ^{2}}{3} + \frac{8 t}{9} + \frac{26}{27}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 3 y = e^{t} + t^{2}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{t} - \frac{e ^{t} t}{2} + \frac{t ^{2}}{3} + \frac{8 t}{9} + \frac{26}{27}

y = c_{1} e^{3 t} + c_{2} e^{t} - \frac{e ^{t} t}{2} + \frac{t ^{2}}{3} + \frac{8 t}{9} + \frac{26}{27}

y^{^{''}} + y^{^{'}} - 2 y = 4, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} + y = x^{2} + 2

y^{^{''}} + y = sin (3 t)

y^{^{''}} + 2 y^{^{'}} + 2 y = 1 + x

y^{^{''}} - 8 y^{^{'}} + 16 y = t, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}-4y^'+3y=e^t+t^2 ?
The solution for y^{''}-4y^'+3y=e^t+t^2 is y=c_{1}e^{3t}+c_{2}e^t-(e^tt)/2+(t^2)/3+(8t)/9+26/27