What is the solution for y^{''}-4y^'+10y=0 ?

The solution for y^{''}-4y^'+10y=0 is y=e^{2t}(c_{1}cos(sqrt(6)t)+c_{2}sin(sqrt(6)t))

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y^{''}-4y^'+10y=0

y^{^{''}} - 4 y^{^{'}} + 10 y = 0

y = e^{2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

y = e^{2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

\frac{\partial}{\partial x} (x^{\frac{4}{7}} y^{\frac{1}{7}} z^{\frac{2}{7}})

\int \frac{1}{324 + x ^{2}} d x

\frac{\partial}{\partial y} (e^{x} sin (y))

\int \frac{ln ( 5 x )}{x ^{3}} d x

a re a y = 4 x, y = x^{2}

What is the solution for y^{''}-4y^'+10y=0 ?
The solution for y^{''}-4y^'+10y=0 is y=e^{2t}(c_{1}cos(sqrt(6)t)+c_{2}sin(sqrt(6)t))