What is the solution for y^{''}-4y^'+53y=0 ?

The solution for y^{''}-4y^'+53y=0 is y=e^{2t}(c_{1}cos(7t)+c_{2}sin(7t))

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y^{''}-4y^'+53y=0

y^{^{''}} - 4 y^{^{'}} + 53 y = 0

y = e^{2 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 53 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

y = e^{2 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

\frac{d}{d x} ((x^{5} + 1)^{6})

x \to 5 lim (\frac{x}{x + 1})

I nb x^{4}

\int 5 e^{x} - 3 cosh (x) d x

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

What is the solution for y^{''}-4y^'+53y=0 ?
The solution for y^{''}-4y^'+53y=0 is y=e^{2t}(c_{1}cos(7t)+c_{2}sin(7t))