What is the solution for t^2(d^2y)/(dt^2)+2t(dy)/(dt)-6y=0 ?

The solution for t^2(d^2y)/(dt^2)+2t(dy)/(dt)-6y=0 is y=c_{1}t^2+(c_{2})/(t^3)

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t^2(d^2y)/(dt^2)+2t(dy)/(dt)-6y=0

t^{2} \frac{d ^{2} y}{d t ^{2}} + 2 t \frac{d y}{d t} - 6 y = 0

y = c_{1} t^{2} + \frac{c _{2}}{t ^{3}}

Solution steps

t^{2} \frac{d ^{2} y}{d t ^{2}} + 2 t \frac{d y}{d t} - 6 y = 0

Solve linear ODE:

y = c_{1} t^{2} + \frac{c _{2}}{t ^{3}}

y = c_{1} t^{2} + \frac{c _{2}}{t ^{3}}

t an g e n t (x + 2)^{2} + (y - 3)^{2} = 37, (- 8, 4)

\int cos^{2} (u) sin^{2} (u) d u

\frac{d}{d x} (6 x^{5} e^{6 x})

\frac{d}{d x} (\frac{1 - 2 x}{6 + x})

t an g e n t f (x) = 4 x - 3 x

What is the solution for t^2(d^2y)/(dt^2)+2t(dy)/(dt)-6y=0 ?
The solution for t^2(d^2y)/(dt^2)+2t(dy)/(dt)-6y=0 is y=c_{1}t^2+(c_{2})/(t^3)