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

Popular >

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}}}

h \to 0 lim (\frac{( 1 + h ) ^{9} - 1}{h})

\frac{d}{d x} (x^{3} + 6 x^{2})

5 x y (\frac{d y}{d x}) = 3

\int_{0}^{2} x 1 + 2 x^{2} d x

\int 3 + tan^{2} (x) d x

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)}}