What is the solution for y^{''}-8y^'+25y=0 ?

The solution for y^{''}-8y^'+25y=0 is y=e^{4t}(c_{1}cos(3t)+c_{2}sin(3t))

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y^{''}-8y^'+25y=0

y^{^{''}} - 8 y^{^{'}} + 25 y = 0

y = e^{4 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) - 8 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{4 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{4 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

\frac{d}{d x} (\frac{2 x ^{2}}{x ^{2} - 1})

\frac{d}{d x} ((3 x^{2} - 5)^{3})

\frac{d}{d x} ((x + 8)^{\frac{2}{3}})

\int \frac{3 ( x + 1 ) + 4}{x ( 2 x + 7 ) + 3} d x

x \to 0 lim (\frac{( x ^{2} - 3 x )}{x})

What is the solution for y^{''}-8y^'+25y=0 ?
The solution for y^{''}-8y^'+25y=0 is y=e^{4t}(c_{1}cos(3t)+c_{2}sin(3t))