What is the solution for (d^2y)/(dt^2)-8(dy)/(dt)+16y=0 ?

The solution for (d^2y)/(dt^2)-8(dy)/(dt)+16y=0 is y=c_{1}e^{4t}+c_{2}te^{4t}

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(d^2y)/(dt^2)-8(dy)/(dt)+16y=0

\frac{d ^{2} y}{d t ^{2}} - 8 \frac{d y}{d t} + 16 y = 0

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

Solution steps

\frac{d ^{2} y}{d t ^{2}} - 8 \frac{d y}{d t} + 16 y = 0

Solve linear ODE:

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

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

f (x) = x^{3}, at x = 1

y = ln (18 x^{2} - 5 x + 11)

2^{x + 3}

\int \frac{x ^{3}}{3 x ^{2} + 3} d x

Y^{3}

What is the solution for (d^2y)/(dt^2)-8(dy)/(dt)+16y=0 ?
The solution for (d^2y)/(dt^2)-8(dy)/(dt)+16y=0 is y=c_{1}e^{4t}+c_{2}te^{4t}