What is the solution for t*ty^{''}+5ty^'+4y=0 ?

The solution for t*ty^{''}+5ty^'+4y=0 is y=(c_{1})/(t^2)+(c_{2}ln(t))/(t^2)

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t*ty^{''}+5ty^'+4y=0

t \cdot t y^{^{''}} + 5 t y^{^{'}} + 4 y = 0

y = \frac{c _{1}}{t ^{2}} + \frac{c _{2} ln ( t )}{t ^{2}}

Solution steps

t \cdot t y^{''} (t) + 5 t y^{'} (t) + 4 y = 0

Solve linear ODE:

y = \frac{c _{1}}{t ^{2}} + \frac{c _{2} ln ( t )}{t ^{2}}

y = \frac{c _{1}}{t ^{2}} + \frac{c _{2} ln ( t )}{t ^{2}}

(\frac{x ^{2} + 25}{2} + \frac{9 - x}{3})^{^{'}}

\int x^{7} - 4 x^{- 3} + 1 d x

\int 4 x e^{- x^{2}} d x

y^{^{'}} + y = - 1

\frac{d}{d x} (tan (arccos (x)))

What is the solution for t*ty^{''}+5ty^'+4y=0 ?
The solution for t*ty^{''}+5ty^'+4y=0 is y=(c_{1})/(t^2)+(c_{2}ln(t))/(t^2)