What is the solution for t^2y^{''}-6ty^'+10y=0 ?

The solution for t^2y^{''}-6ty^'+10y=0 is y=c_{1}t^5+c_{2}t^2

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t^2y^{''}-6ty^'+10y=0

t^{2} y^{^{''}} - 6 t y^{^{'}} + 10 y = 0

y = c_{1} t^{5} + c_{2} t^{2}

Solution steps

t^{2} y^{''} (t) - 6 t y^{'} (t) + 10 y = 0

Solve linear ODE:

y = c_{1} t^{5} + c_{2} t^{2}

y = c_{1} t^{5} + c_{2} t^{2}

\frac{d}{d x} (ln (x^{5} (x + 4)^{7} (x^{2} + 4)^{5}))

\int_{0}^{4} x + 2 - x d x

\int \frac{y ^{2} + 16}{y} d y

\frac{d}{d x} (ln (\frac{x ^{3} - 2 x}{4 + x}))

f (x) = \frac{2 x + 1}{x + 2}

What is the solution for t^2y^{''}-6ty^'+10y=0 ?
The solution for t^2y^{''}-6ty^'+10y=0 is y=c_{1}t^5+c_{2}t^2