What is the solution for y^{''}-16y^'+64y=0 ?

The solution for y^{''}-16y^'+64y=0 is y=c_{1}e^{8t}+c_{2}te^{8t}

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y^{''}-16y^'+64y=0

y^{^{''}} - 16 y^{^{'}} + 64 y = 0

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

Solution steps

y^{''} (t) - 16 y^{'} (t) + 64 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (ln (\frac{( x + 2 ) e ^{3 x}}{x ^{2} + 1}))

\frac{d y}{d x} = (\frac{9 x}{9 x ^{2} - 1}) e^{- y}

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

\int \frac{1}{( a + x ) ^{2}} d x

x^{4} - 3 x^{2} + 1, 1

What is the solution for y^{''}-16y^'+64y=0 ?
The solution for y^{''}-16y^'+64y=0 is y=c_{1}e^{8t}+c_{2}te^{8t}