What is the solution for y^{''}-10y^'+26y=4 ?

The solution for y^{''}-10y^'+26y=4 is y=e^{5t}(c_{1}cos(t)+c_{2}sin(t))+2/13

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y^{''}-10y^'+26y=4

y^{^{''}} - 10 y^{^{'}} + 26 y = 4

y = e^{5 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2}{13}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 26 y = 4

Solve linear ODE:

y = e^{5 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2}{13}

y = e^{5 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2}{13}

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\frac{d y}{d x} (x^{2} + 1) + 7 x (y - 1) = 0, y (0) = 4

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What is the solution for y^{''}-10y^'+26y=4 ?
The solution for y^{''}-10y^'+26y=4 is y=e^{5t}(c_{1}cos(t)+c_{2}sin(t))+2/13