Bernoulli Differential Equations — Question 8

Question 8 – Solution

Strategy. Keep the forcing integral exact, then use its monotonicity to determine the entire admissible denominator interval.

Step 1: Linearize and integrate. For $a\ne 0$, set $v=1/y$. Then $v'-xv=-e^{x^2}$, with integrating factor $e^{-x^2/2}$. Thus $$(e^{-x^2/2}v)'=-e^{x^2/2},\qquad v=e^{x^2/2}\left(\frac 1a-J(x)\right).$$ Inverting gives $$\boxed{y(x)=\frac{a e^{-x^2/2}}{1-aJ(x)}}.$$ For $a=0$, $y\equiv 0$ is the unique solution, since the original right-hand side is smooth in $y$.

Step 2: Locate the unique obstruction. By the fundamental theorem, $J'=e^{x^2/2}>0$. The integrand is even, so $J$ is odd. Also $J(x)\ge x$ for $x\ge 0$, so its limits at the two infinities are $\pm\infty$. Therefore it is one-to-one and onto, and there is exactly one $T$ with $J(T)=1/a$.

Step 3: Select the interval and sign. If $a>0$, then $T>0$, and the maximal interval is $\boxed{(-\infty,T)}$. The denominator approaches zero from above as $x\uparrow T$, so $y\to+\infty$. If $a<0$, then $T<0$, and the maximal interval is $\boxed{(T,\infty)}$; as $x\downarrow T$, the denominator approaches zero from above and $y\to-\infty$. A pole cannot be crossed by a finite differentiable continuation.

Step 4: Verify and interpret. With $D=1-aJ$, differentiation using $J'=e^{x^2/2}$ gives $$y'=-xy+\frac{a^2}{D^2}=-xy+e^{x^2}y^2.$$ Also $J(0)=0$ gives $y(0)=a$. The definite integral specifies an exact differentiable function and uniquely determines the endpoint; elementary notation is unnecessary for either the solution or its interval.