What is the solution for y^{''}-4y^'+13=0 ?

The solution for y^{''}-4y^'+13=0 is y=c_{1}e^{4t}+c_{2}+(13t)/4

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y^{''}-4y^'+13=0

y^{^{''}} - 4 y^{^{'}} + 13 = 0

y = c_{1} e^{4 t} + c_{2} + \frac{13 t}{4}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 13 = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} + \frac{13 t}{4}

y = c_{1} e^{4 t} + c_{2} + \frac{13 t}{4}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x} + cos (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e x - 12

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 6 y = 18

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

y^{^{''}} + 49 y = sec (7 x)

What is the solution for y^{''}-4y^'+13=0 ?
The solution for y^{''}-4y^'+13=0 is y=c_{1}e^{4t}+c_{2}+(13t)/4