Question

solve the given initial - value problem. $\frac{dy}{dt}+2(t + 1)y^{2}=0$, $y(0)=-\frac{1}{3}$. give the largest interval $i$ on which the solution is defined. (enter your answer using interval notation.)

Explanation:

Step1: Rewrite the differential equation

The given differential equation $\frac{dy}{dt}+2(t + 1)y^{2}=0$ can be rewritten as $\frac{dy}{y^{2}}=-2(t + 1)dt$ (assuming $y
eq0$).

Step2: Integrate both sides

Integrating $\int y^{-2}dy=\int-2(t + 1)dt$. The left - hand side integral is $\int y^{-2}dy=-\frac{1}{y}+C_1$, and the right - hand side integral: $\int-2(t + 1)dt=-2\int(t + 1)dt=-2(\frac{t^{2}}{2}+t)+C_2=-t^{2}-2t + C$. So, $-\frac{1}{y}=-t^{2}-2t + C$.

Step3: Use the initial condition

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

Step4: Solve for y

From $-\frac{1}{y}=-t^{2}-2t + 3$, we can solve for $y$ as $y=\frac{1}{t^{2}+2t - 3}=\frac{1}{(t + 3)(t - 1)}$.

Step5: Find the domain

The function $y(t)$ is undefined when the denominator is zero, i.e., when $(t + 3)(t - 1)=0$, so $t=-3$ or $t = 1$. Since $y(0)=-\frac{1}{3}$, the largest interval on which the solution is defined is $(-3,1)$.

Answer:

$(-3,1)$