What is the solution for t^2y^{''}+5ty^'-45y=0 ?

The solution for t^2y^{''}+5ty^'-45y=0 is y=c_{1}t^5+(c_{2})/(t^9)

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t^2y^{''}+5ty^'-45y=0

t^{2} y^{^{''}} + 5 t y^{^{'}} - 45 y = 0

y = c_{1} t^{5} + \frac{c _{2}}{t ^{9}}

Solution steps

t^{2} y^{''} (t) + 5 t y^{'} (t) - 45 y = 0

Solve linear ODE:

y = c_{1} t^{5} + \frac{c _{2}}{t ^{9}}

y = c_{1} t^{5} + \frac{c _{2}}{t ^{9}}

\frac{( x d y - y d x )}{x ^{2} + y ^{2}} = 0

\int - 2 e^{8 x} d x

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

\frac{1}{x y}

f^{^{'}} (θ) = θ cos (θ) sin (θ)

What is the solution for t^2y^{''}+5ty^'-45y=0 ?
The solution for t^2y^{''}+5ty^'-45y=0 is y=c_{1}t^5+(c_{2})/(t^9)