What is the solution for (dy)/(dt)= y/(t+1)+4t^2+4t ?

The solution for (dy)/(dt)= y/(t+1)+4t^2+4t is y=2t^2(t+1)-2(t+1)+c_{1}(t+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dt)= y/(t+1)+4t^2+4t

\frac{d y}{d t} = \frac{y}{t + 1} + 4 t^{2} + 4 t

y = 2 t^{2} (t + 1) - 2 (t + 1) + c_{1} (t + 1)

Solution steps

\frac{d y}{d t} = \frac{y}{t + 1} + 4 t^{2} + 4 t

Solve linear ODE:

y = 2 t^{2} (t + 1) - 2 (t + 1) + c_{1} (t + 1)

y = 2 t^{2} (t + 1) - 2 (t + 1) + c_{1} (t + 1)

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

x \to 2 lim (\frac{x - 3}{x ^{2} + 1})

\int sin^{2} (x co) s^{5} x d x

\int a x (x - 4) d x

\int x ln (2) d x

What is the solution for (dy)/(dt)= y/(t+1)+4t^2+4t ?
The solution for (dy)/(dt)= y/(t+1)+4t^2+4t is y=2t^2(t+1)-2(t+1)+c_{1}(t+1)