What is the solution for t^2y^{''}-4ty^'+6y=0 ?

The solution for t^2y^{''}-4ty^'+6y=0 is y=c_{1}t^3+c_{2}t^2

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t^2y^{''}-4ty^'+6y=0

t^{2} y^{^{''}} - 4 t y^{^{'}} + 6 y = 0

y = c_{1} t^{3} + c_{2} t^{2}

Solution steps

t^{2} y^{''} (t) - 4 t y^{'} (t) + 6 y = 0

Solve linear ODE:

y = c_{1} t^{3} + c_{2} t^{2}

y = c_{1} t^{3} + c_{2} t^{2}

x \to 0 lim (\frac{2 x}{2 x ^{2}})

\int_{3}^{5} e^{6 x} d x

f (x) = 3 - 2 x

\int \frac{x ^{4} + 1}{x ^{2} + 1} d x

0.882 a^{0.842}

What is the solution for t^2y^{''}-4ty^'+6y=0 ?
The solution for t^2y^{''}-4ty^'+6y=0 is y=c_{1}t^3+c_{2}t^2