Separable Equations — Question 4

Question 4 – Solution

Strategy. Treat the left side of the integrated equation as an invertible function rather than forcing an elementary inverse.

Step 1: Integrate. Multiplication and integration give $$(1+e^y)\,dy=x\,dx,\qquad \boxed{y+e^y=1+\frac{x^2}{2}}.$$ The constant is $1$ because $y(0)=0$.

Step 2: Prove global existence and uniqueness of the implicit value. Let $F(u)=u+e^u$. It is continuous and strictly increasing because $F'(u)=1+e^u>0$. Also $F(u)\to-\infty$ as $u\to-\infty$ and $F(u)\to\infty$ as $u\to\infty$. It therefore has an inverse on all of $\mathbb R$. Since its derivative never vanishes, the inverse is differentiable. Thus $$y(x)=F^{-1}(1+x^2/2)$$ defines a unique differentiable solution for every real $x$. Any solution through $(0,0)$ must satisfy the same integrated identity, establishing uniqueness for the initial-value problem.

Step 3: Extract shape and an exact bound. Uniqueness of the implicit value and the dependence on $x^2$ show $y(-x)=y(x)$. Since $F(0)=1$, $y\ge 0$, with equality only at $x=0$. The differential equation makes $y$ decreasing for $x<0$ and increasing for $x>0$, so its unique minimum is $0$.

Using $e^y\ge 1+y$, with equality only at $y=0$, gives $$1+\frac{x^2}{2}=y+e^y\ge 1+2y, \qquad \boxed{0\le y\le\frac{x^2}{4}}.$$ Equality in either bound occurs only at $x=0$.

Step 4: Verify and interpret. Implicit differentiation yields $(1+e^y)y'=x$, exactly the given equation. At $x=0$, strict monotonicity of $F$ forces $y=0$. The implicit formula uniquely specifies every solution value, its full domain, and its derivative; an elementary inverse is unnecessary.