What is the solution for 5t^2y^{''}+ty^'-12y=0 ?

The solution for 5t^2y^{''}+ty^'-12y=0 is y=c_{1}t^2+(c_{2})/(t^{6/5)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5t^2y^{''}+ty^'-12y=0

5 t^{2} y^{^{''}} + t y^{^{'}} - 12 y = 0

y = c_{1} t^{2} + \frac{c _{2}}{t ^{\frac{6}{5}}}

Solution steps

5 t^{2} y^{''} (t) + t y^{'} (t) - 12 y = 0

Solve linear ODE:

y = c_{1} t^{2} + \frac{c _{2}}{t ^{\frac{6}{5}}}

y = c_{1} t^{2} + \frac{c _{2}}{t ^{\frac{6}{5}}}

\int \frac{x ^{2}}{x ^{6} - 9} d x

(3, 0), (6, 2)

\frac{\partial}{\partial z} (\frac{1}{x ^{2} + y ^{2} + z ^{2}})

e^{x^{2}}

\frac{1}{s ^{2} + 16}

What is the solution for 5t^2y^{''}+ty^'-12y=0 ?
The solution for 5t^2y^{''}+ty^'-12y=0 is y=c_{1}t^2+(c_{2})/(t^{6/5)}