What is the solution for y^{''}-14y^'+48y=0 ?

The solution for y^{''}-14y^'+48y=0 is y=c_{1}e^{8t}+c_{2}e^{6t}

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y^{''}-14y^'+48y=0

y^{^{''}} - 14 y^{^{'}} + 48 y = 0

y = c_{1} e^{8 t} + c_{2} e^{6 t}

Solution steps

y^{''} (t) - 14 y^{'} (t) + 48 y = 0

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} e^{6 t}

y = c_{1} e^{8 t} + c_{2} e^{6 t}

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\int_{0}^{\frac{π}{2}} cos (5 t) cos (10 t) d t

What is the solution for y^{''}-14y^'+48y=0 ?
The solution for y^{''}-14y^'+48y=0 is y=c_{1}e^{8t}+c_{2}e^{6t}