What is the solution for t^2y^{''}+21ty^'+100y=0 ?

The solution for t^2y^{''}+21ty^'+100y=0 is y=(c_{1})/(t^{10)}+(c_{2}ln(t))/(t^{10)}

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t^2y^{''}+21ty^'+100y=0

t^{2} y^{^{''}} + 21 t y^{^{'}} + 100 y = 0

y = \frac{c _{1}}{t ^{10}} + \frac{c _{2} ln ( t )}{t ^{10}}

Solution steps

t^{2} y^{''} (t) + 21 t y^{'} (t) + 100 y = 0

Solve linear ODE:

y = \frac{c _{1}}{t ^{10}} + \frac{c _{2} ln ( t )}{t ^{10}}

y = \frac{c _{1}}{t ^{10}} + \frac{c _{2} ln ( t )}{t ^{10}}

f (x) = sin (x) + \frac{6}{7} cot (x)

x \to 2 lim (5 a + a x)

\frac{d y}{d x} - 5 y = e^{x}

x \to 0 lim (\frac{cos ( 5 x ) - 1}{sin ( 9 x )})

\frac{d y}{d x} = \frac{2 - e ^{x}}{3 + 2 y}

What is the solution for t^2y^{''}+21ty^'+100y=0 ?
The solution for t^2y^{''}+21ty^'+100y=0 is y=(c_{1})/(t^{10)}+(c_{2}ln(t))/(t^{10)}