What is the solution for t^2y^{''}+9ty^'+16y=0 ?

The solution for t^2y^{''}+9ty^'+16y=0 is y=(c_{1})/(t^4)+(c_{2}ln(t))/(t^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^2y^{''}+9ty^'+16y=0

t^{2} y^{^{''}} + 9 t y^{^{'}} + 16 y = 0

y = \frac{c _{1}}{t ^{4}} + \frac{c _{2} ln ( t )}{t ^{4}}

Solution steps

t^{2} y^{''} (t) + 9 t y^{'} (t) + 16 y = 0

Solve linear ODE:

y = \frac{c _{1}}{t ^{4}} + \frac{c _{2} ln ( t )}{t ^{4}}

y = \frac{c _{1}}{t ^{4}} + \frac{c _{2} ln ( t )}{t ^{4}}

\int_{1}^{\infty} \frac{1}{x ^{2} + 1} d x

(3 - x)^{^{'}}

2 + 8 x^{\frac{3}{2}}

(t^{2} - 2)^{2}

25 y^{^{''}} - 20 y^{^{'}} + 4 y = 0

What is the solution for t^2y^{''}+9ty^'+16y=0 ?
The solution for t^2y^{''}+9ty^'+16y=0 is y=(c_{1})/(t^4)+(c_{2}ln(t))/(t^4)