Question

solve the given initial - value problem.
$\frac{dy}{dt}+2(t + 1)y^{2}=0$, $y(0)=-\frac{1}{3}$
$y(t)=$

Explanation:

Step1: Rearrange the differential equation

Separate variables: $\frac{dy}{y^{2}}=-2(t + 1)dt$.

Step2: Integrate both sides

Integrate $\int y^{-2}dy=\int-2(t + 1)dt$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $-\frac{1}{y}=-2(\frac{t^{2}}{2}+t)+C=-t^{2}-2t + C$.

Step3: Use the initial condition

Given $y(0)=-\frac{1}{3}$, substitute $t = 0$ and $y=-\frac{1}{3}$ into $-\frac{1}{y}=-t^{2}-2t + C$. Then $- \frac{1}{-\frac{1}{3}}=C$, so $C = 3$.

Step4: Solve for $y$

We have $-\frac{1}{y}=-t^{2}-2t + 3$, then $y=\frac{1}{t^{2}+2t - 3}$.

Answer:

$y=\frac{1}{t^{2}+2t - 3}$