What is the solution for y^{''}-7y^'+12=0 ?

The solution for y^{''}-7y^'+12=0 is y=c_{1}e^{7t}+c_{2}+(12t)/7

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y^{''}-7y^'+12=0

y^{^{''}} - 7 y^{^{'}} + 12 = 0

y = c_{1} e^{7 t} + c_{2} + \frac{12 t}{7}

Solution steps

y^{''} (t) - 7 y^{'} (t) + 12 = 0

Solve linear ODE:

y = c_{1} e^{7 t} + c_{2} + \frac{12 t}{7}

y = c_{1} e^{7 t} + c_{2} + \frac{12 t}{7}

y^{^{''}} - 2 y^{^{'}} - 3 = 0

y^{^{''}} + 4 y^{^{'}} + 13 y = 5

y^{^{''}} + 4 y^{^{'}} + y = 2

y^{^{''}} + 5 y^{^{'}} + 6 y = 20 t e^{5 t}

y^{^{''}} - 6 y^{^{'}} + 8 y = - 3 e^{2 x} - 5 x e^{- 4 x} + 3 x

What is the solution for y^{''}-7y^'+12=0 ?
The solution for y^{''}-7y^'+12=0 is y=c_{1}e^{7t}+c_{2}+(12t)/7