What is the solution for y^{''}-5y^'+6y=t^2 ?

The solution for y^{''}-5y^'+6y=t^2 is y=c_{1}e^{3t}+c_{2}e^{2t}+(t^2)/6+(5t)/(18)+19/108

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y^{''}-5y^'+6y=t^2

y^{^{''}} - 5 y^{^{'}} + 6 y = t^{2}

y = c_{1} e^{3 t} + c_{2} e^{2 t} + \frac{t ^{2}}{6} + \frac{5 t}{18} + \frac{19}{108}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 6 y = t^{2}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{2 t} + \frac{t ^{2}}{6} + \frac{5 t}{18} + \frac{19}{108}

y = c_{1} e^{3 t} + c_{2} e^{2 t} + \frac{t ^{2}}{6} + \frac{5 t}{18} + \frac{19}{108}

2 x y d x - (1 + y) d y = 0

\int \frac{3 x}{5 x ^{2} - 1} d x

\int 5 x d x

\frac{d}{d t} (6 e^{t})

cos (x) cos (y) d x + sin (x) sin (y) d y = 0

What is the solution for y^{''}-5y^'+6y=t^2 ?
The solution for y^{''}-5y^'+6y=t^2 is y=c_{1}e^{3t}+c_{2}e^{2t}+(t^2)/6+(5t)/(18)+19/108