What is the solution for (d^2y)/(dt^2)-5(dy)/(dt)+6=0 ?

The solution for (d^2y)/(dt^2)-5(dy)/(dt)+6=0 is y=c_{1}e^{5t}+c_{2}+(6t)/5

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(d^2y)/(dt^2)-5(dy)/(dt)+6=0

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 6 = 0

y = c_{1} e^{5 t} + c_{2} + \frac{6 t}{5}

Solution steps

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 6 = 0

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} + \frac{6 t}{5}

y = c_{1} e^{5 t} + c_{2} + \frac{6 t}{5}

- 5 y^{^{''}} - 2 y^{^{'}} + 24 = 0

y^{^{''}} - 3 y^{^{'}} = cos (5 x), y (0) = 0, y^{^{'}} (0) = 0

\frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{3} e^{3 t}

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

y^{^{''}} + 2 y^{^{'}} + 392 y = cos (2 t)

What is the solution for (d^2y)/(dt^2)-5(dy)/(dt)+6=0 ?
The solution for (d^2y)/(dt^2)-5(dy)/(dt)+6=0 is y=c_{1}e^{5t}+c_{2}+(6t)/5