What is the solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=-5 ?

The solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=-5 is y=2t^2(t+1)-9/2 (t+1)

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(dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=-5

\frac{d y}{d t} = \frac{y}{t + 1} + 4 t^{2} + 4 t, y (1) = - 5

y = 2 t^{2} (t + 1) - \frac{9}{2} (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) - \frac{9}{2} (t + 1)

y = 2 t^{2} (t + 1) - \frac{9}{2} (t + 1)

x \to 0 lim (\frac{1}{x} - 3)

f (y) = \frac{y - 1}{y ^{2} - y + 1}

\int (x + 1) 4 x + 5 d x

\frac{\partial}{\partial x} (sin (x^{2} + y^{2}))

\int cos (15 x) d x

What is the solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=-5 ?
The solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=-5 is y=2t^2(t+1)-9/2 (t+1)