What is the solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=12 ?

The solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=12 is y=2t^2(t+1)+4(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,y(1)=12

\frac{d y}{d t} = \frac{y}{t + 1} + 4 t^{2} + 4 t, y (1) = 12

y = 2 t^{2} (t + 1) + 4 (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) + 4 (t + 1)

y = 2 t^{2} (t + 1) + 4 (t + 1)

y = 5 x^{8} - 3 x^{5} + 2, (0, 4)

\frac{d}{d x} (θ (x)), tan (θ (x)) = \frac{x}{35}

(x) \frac{d y}{d x} + y = \frac{1}{y ^{2}}

\frac{x ^{2} - 9 x + 20}{( x + 2 ) ( x - 2 )}

\int 20 x^{\frac{3}{2}} d x

What is the solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=12 ?
The solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=12 is y=2t^2(t+1)+4(t+1)