<s>[INST] Domain of solution of differential equation $y’=\frac{1}{(y+7)(t-3)}$ If we are given the initial-value problem $$\frac{dy}{dt}=\frac{1}{(y+7)(t-3)}, y(0)=0$$
I want to solve said initial value problem & state the domain of the solution. I also want to observe what happens when $t$ approaches the limits of the solution’s domain.
Via separation of variables, one obtains: $$\implies(y+7) \space dy = \frac{dt}{t-3}\implies\frac{1}{2}y^2+7y=\ln|t-3|+c_1$$ $$\iff y^2+14y+49=2\ln|t-3|+(2c_1+49)$$
Now call $C=2c_1+49$, then $$(y+7)^2=2\ln|t-3|+C\implies y(t) = \pm\sqrt{2\ln|t-3|+C}-7$$
Substituting the initial condition:
$$0=\pm\sqrt{2\ln|0-3|+C}-7\iff 7 = \pm\sqrt{2\ln(3)+C}$$
This shows we must choose the positive square root in order for our solution $y(t)$ to pass through the initial condition & solve the IVP. Then $$C=49-2\ln(3)$$ so that
$$y(t)=\sqrt{2\ln\left|\frac{t-3}{3}\right|+49}-7$$
We know from the original differential equation that $y(t)\neq-7$ & $t\ne3$. Thus: $$-7\neq \sqrt{2\ln\left|\frac{t-3}{3}\right|+49}-7 \iff\ln\left|\frac{t-3}{3}\right|\neq-\frac{49}{2}\iff t\neq\pm 3\exp\left(-\frac{49}{2}\right)+3$$
Does this tell us that the domain of the solution has to be $(-\infty,-3\exp\left(-\frac{49}{2}\right)+3)$ in order for $t=0$ to be on the domain of this solution (so the solution passes through the initial condition)?
Would we just say $y(t)\to0$ as $t\to-3\exp\left(-\frac{49}{2}\right)+3$?
Then finally $y(t)=\sqrt{2\ln\left(1-\frac{t}{3}\right)+49}-7$, for $t\in(-\infty, -3\exp\left(-\frac{49}{2}\right)+3)$.
 [/INST] The equation in question is $$y'(t)=\frac1{[y(t)+7](t-3)}.$$ Thus, there is a singularity at $t=3,$ which suggests that the solutions will have a domain that is a subset of $(-\infty,3)\cup(3,\infty).$ To solve the equation, we consider $$[y(t)+7]y'(t)=\frac1{t-3}.$$ Notice that $$\left(\frac{y^2}2+7y\right)'(t)=[y(t)+7]y'(t).$$ Therefore, we have that $$\frac{y(t)^2}2+7y(t)=\begin{cases}\ln(3-t)+A&t\lt3\\\ln(t-3)+B&t\gt3\end{cases}.$$ Given the initial condition $y(0)=0,$ one has that $$\frac{y(0)^2}2+7y(0)=0=\ln(3)+A,$$ implying $A=-\ln(3).$ Therefore, $$\frac{y(t)^2}2+7y(t)=\begin{cases}\ln(3-t)-\ln(3)&t\lt3\\\ln(t-3)+B&t\gt3\end{cases},$$ which is equivalent to $$y(t)^2+14y(t)=\begin{cases}2\ln(3-t)-2\ln(3)&t\lt3\\2\ln(t-3)+2B&t\gt3\end{cases},$$ which is equivalent to $$[y(t)+7]^2=\begin{cases}2\ln(3-t)-2\ln(3)+49&t\lt3\\2\ln(t-3)+2B+49&t\gt3\end{cases}.$$ Here, it gets complicated. It is required that $$2\ln(3-t)-2\ln(3)+49\geq0$$ and $$2\ln(t-3)+2B+49\geq0.$$ This implies, respectively, that $$\ln(3-t)\geq\ln(3)-\frac{49}2$$ and $$\ln(t-3)\geq{B}-\frac{49}2.$$ Therefore, $$3-t\geq3\exp\left(-\frac{49}2\right)$$ and $$t-3\geq\exp\left(B-\frac{49}2\right)=C\exp\left(-\frac{49}2\right)$$ with $C\gt0.$ Therefore, $$t\leq3-3\exp\left(-\frac{49}2\right)$$ and $$t\geq3+C\exp\left(-\frac{49}2\right).$$ However, at the endpoints, the function is not differentiable, so the domain of the function is $\left(-\infty,3-3\exp\left(-\frac{49}2\right)\right)\cup\left(3+C\exp\left(-\frac{49}2\right),\infty\right).$
With this in mind, the solutions to the equation are given by the two families $$y(t)=-7-\sqrt{f(t)}$$ and $$y(t)=-7+\sqrt{f(t)},$$ where $f$ is the family of functions $$f(t)=\begin{cases}2\ln(3-t)-2\ln(3)+49&t\lt3\\2\ln(t-3)+2B+49&t\gt3\end{cases}.$$
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