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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

ty^'+t^2+ty-y=0

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

y = - t + \frac{c _{1} t}{e ^{t + c_{1}}}

Solution steps

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

Solve linear ODE:

y = - t + \frac{c _{1} t}{e ^{t + c_{1}}}

y = - t + \frac{c _{1} t}{e ^{t + c_{1}}}

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