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

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

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3y^{''}-2y^'=t

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

y = c_{1} e^{\frac{2 t}{3}} + c_{2} - \frac{t ^{2}}{4} - \frac{3 t}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{2 t}{3}} + c_{2} - \frac{t ^{2}}{4} - \frac{3 t}{4}

y = c_{1} e^{\frac{2 t}{3}} + c_{2} - \frac{t ^{2}}{4} - \frac{3 t}{4}

(D^{4} - 2 D^{2} + 1) Y = cos (x)

D (D^{2} + 1) y = sin (x)

y^{^{'''}} - 2 y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{x} (1 - 6 x)

y^{^{''}} - y^{^{'}} - 2 y = 3 e^{2 x} - 4 e^{3 x}

x^{^{''}} + 3 x^{^{'}} + 33 x = 11 cos (5 t)

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