What is the solution for ty^'+2y= 2/(1+t^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ty^'+2y= 2/(1+t^2)

t y^{^{'}} + 2 y = \frac{2}{1 + t ^{2}}

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

Solution steps

t y^{'} (t) + 2 y = \frac{2}{1 + t ^{2}}

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x} sin (2 x)

\frac{\partial}{\partial x} (arctan (2 x y))

\int \frac{( e ^{x} - e ^{- x} )}{e ^{2 x}} d x

\frac{\partial}{\partial s} (s cos (t))

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