Animated Solution for Mathematics - Differential Equations: Let $y=y(x)$ be the solution of the differential equation $(x+1)y^{\prime }-y=e^{3x}(x+1{)}^2$, with $y(0)=\frac{1}{3}$. Then, the point $x=-\frac{4}{3}$ for the curve $y=y(x)$ is:

We begin with the given differential equation:

This equation is a classic example of a pattern waiting to be recognized. Observe the left-hand side: $(x+1)y^{\prime }-y$. This structure is reminiscent of the quotient rule for differentiation, where the derivative of $\frac{u}{v}$ is $\frac{vdu-udv}{v^2}$.

If we set $u=y$ and $v=x+1$, then $du=dy$ and $dv=dx$. The numerator of the quotient rule is exactly $(x+1)dy-ydx$.

To complete this pattern, we divide both sides of the original equation by $(x+1{)}^2$:

This simplifies the left side into the derivative of a quotient, yielding:

We now integrate both sides with respect to $x$:

Performing the integration, we obtain:

We are given the initial condition $y(0)=\frac{1}{3}$. Substituting $x=0$ and $y=\frac{1}{3}$ into our general solution:

This simplifies to $\frac{1}{3}=\frac{1}{3}+C$, which clearly indicates that $C=0$. Therefore, our specific solution is:

To determine the nature of the point $x=-\frac{4}{3}$, we first find the derivative $y^{\prime }(x)$ using the product rule:

Setting $y^{\prime }(x)=0$ confirms the critical point at $3x+4=0$, or $x=-\frac{4}{3}$. To classify this point, we compute the second derivative:

Evaluating the second derivative at $x=-\frac{4}{3}$:

Since $e^{-4}>0$, the function is concave upwards at this point. Thus, the function has a local minimum at $x=-\frac{4}{3}$.