Question bank — Separable ODEs — technique, implicit solutions

False — $\frac{dy}{dx}=x+y$ has a sum on the right that never factors into a pure-$x$ times pure-$y$ product; you need integrating factors instead.

True — the sum in the exponent splits: $e^{x+y}=e^x{e}^y$, so $g(x)=e^x$, $h(y)=e^y$; an exponential of a sum is a product.

False — it is a fully valid solution; if isolating $y$ needs a $\pm$ quadratic formula it is often cleaner and more honest to stop at the implicit form.

True — one integration on each side gives two constants, but they merge into a single $C$; the order of the ODE (here 1) fixes the count.

False — $dy/dx$ is a limit, not a fraction; the split is valid shorthand for the substitution/chain-rule step $\int \frac{1}{h(y)}\frac{dy}{dx}dx=\int \frac{dy}{h(y)}$.

False — if $h(y_0)=0$, the constant $y\equiv y_0$ solves the ODE but division by $h(y)$ erases it; those equilibrium solutions must be restored by hand.

False — it is $\ln |y|+C$; the modulus is exactly what later lets the constant absorb a $\pm$ sign when you exponentiate.

True — here $h(y)=y$ and $h(0)=0$, but $y\equiv 0$ is recovered by taking $A=0$, so nothing is truly lost in this case (unlike the logistic).

True — it is $g(x)h(y)$ with $g=\sin x$ and $h=1/y^2$; a quotient with pure-$x$ on top and pure-$y$ on the bottom still factors.

False — that double-counts; the difference of two arbitrary constants is one arbitrary constant, so a single $C$ on one side is complete.

The error is stopping too early: dividing by $y(1-y)$ deleted the constants $y\equiv 0$ and $y\equiv 1$, which are genuine solutions and must be reported alongside the family.

Sign error in the partial fractions: the correct split is $\frac{1}{y}+\frac{1}{1-y}$ (both plus), since $\frac{1}{y}+\frac{1}{1-y}=\frac{(1-y)+y}{y(1-y)}=\frac{1}{y(1-y)}$.

Missing minus sign from the inner derivative: $\frac{d}{dy}(1-y)=-1$, so $\int \frac{dy}{1-y}=-\ln |1-y|+C$.

Illegal move — you cannot pull the $x$ off a sum $x+y$; only a product factors, and $x+y$ is not a product.

Exponentiating a sum turns it into a product, not a sum: $y=e^{x^3/3}\cdot e^C=A{e}^{x^3/3}$, not $e^{x^3/3}+e^C$.

Even implicit solutions carry the constant: it is $y+y^2=x^2+C$, and $C$ is what an initial condition pins down.

One $C$ suffices; writing it on both sides is redundant because the two constants merge into one.

Because separation moves every $y$ to the left and every $x$ to the right; that clean split is only possible when the two variables were multiplied (or divided), never added.

Here it is a faithful shorthand for one specific chain-rule substitution that is provably correct; blindly manipulating $dy$ and $dx$ elsewhere has no such guarantee.

The antiderivative of $1/y$ must be valid for negative $y$ too; the modulus keeps the formula true on both sides of $y=0$, and its sign later gets absorbed into the arbitrary constant.

If $h(y_0)=0$ then a constant $y\equiv y_0$ gives $y^{\prime }=0=g(x)\cdot h(y_0)$, so it satisfies the ODE — but that is exactly the value we divided by, so the algebra hid it.

Solving for $y$ can introduce a quadratic $\pm$ branch or a transcendental tangle; the implicit relation $F(x,y)=C$ states the full solution without picking a branch or losing information.

A solution is defined as a relation whose derivative reproduces the original ODE; recovering $y^{\prime }=g(x)h(y)$ confirms exactly that, independent of how you found it.

$h(y)=y(1-y)$ has two roots $y=0,1$ that the fraction $\frac{A{e}^x}{1+A{e}^x}$ can never reach; $h(y)=y$ has the single root $y=0$, which the family recovers at $A=0$.

The unique solution through that point is the constant equilibrium $y\equiv y_0$; you never separate, because dividing by $h(y_0)=0$ is forbidden and unnecessary.

$y^{\prime }=0$, so $y\equiv$ constant — every horizontal line is a solution; separation degenerates but the answer is immediate.

No — uniqueness forbids two solutions sharing a point, so non-equilibrium curves approach $y=1$ asymptotically but never touch or cross it.

Yes, the exponential is defined and smooth for every real $x$, so this solution is global; not all separable solutions are — implicit ones can restrict the domain.

Not globally — solving gives $y=\frac{-1\pm \sqrt{1+4(x^2+C)}}{2}$, two branches; the initial condition selects which branch, and the curve may only be a function on part of its extent.

All of them — here both $y\equiv 1$ and $y\equiv 2$ are separate constant solutions, each a root of $h$, and each must be reported beyond the separated family.

Yes, it separates to $ydy=xdx$; but $y=0$ makes the original right side blow up, so no solution passes through the $x$-axis and $h(y)=1/y$ is simply undefined there.