What is the solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=6 ?

The solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=6 is y=2t^2(t+1)+t+1

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(dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=6

\frac{d y}{d t} = \frac{y}{t + 1} + 4 t^{2} + 4 t, y (1) = 6

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

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

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x \to 3 lim (5 x^{2} - 7 x - 3)

(10 x^{2} - 20 x + 20) e^{- 6 x}

\int \frac{( sin ^{3} ( x ) )}{cos ( x )} d x

\frac{d}{d x} (4 x^{3} + 3 x)

What is the solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=6 ?
The solution for (dy)/(dt)= y/(t+1)+4t^2+4t,y(1)=6 is y=2t^2(t+1)+t+1