Exercises — Separable ODEs — technique, implicit solutions

• (a) Separable. $g(x)=x^3,h(y)=1/y$. The RHS is a pure-$x$ piece times a pure-$y$ piece.
• (b) NOT separable. A sum $x-y$ cannot be written as (function of $x$)$\times$(function of $y$). There is no factoring of a subtraction.
• (c) Separable. $g(x)=\cos x,h(y)=1+y^2$.
• (d) Separable. The exponent is a sum, and a sum in an exponent splits into a product: $e^{x-y}=e^x{e}^{-y}$. So $g(x)=e^x,h(y)=e^{-y}$.

Here $g(x)=1$ and $h(y)=y(3-y)$. Set $h(y)=0$: either $y=0$ or $3-y=0\Rightarrow y=3$. So the constant functions $y\equiv 0$ and $y\equiv 3$ are equilibrium solutions — they'll be erased the moment we divide by $h(y)$, so we record them now.

Check for lost solutions first. Here $g(x)=x^3,h(y)=1/y$. There is no $y$ making $h(y)=0$ (since $1/y$ is never zero), and $y=0$ makes the RHS $x^3/y$ undefined — so $y\equiv 0$ is not a solution and solution curves never touch the $x$-axis. We flag this because multiplying by $y$ (next step) would otherwise silently swallow a would-be constant solution. Separate (multiply both sides by $y$, i.e. move all $y$ left, all $x$ right): $ydy=x^{3}dx.$ Integrate both sides. Each side gains its own constant, but we collect them into a single arbitrary constant $C$ on the right: $\frac{y^2}{2}=\frac{x^4}{4}+C.$ Single constant: one $C$ absorbs both. Multiply by 2; since $2C$ is still just "some unknown constant," rename $2C\to C$ (a fresh arbitrary constant): $\boxed{y^2=\frac{x^4}{2}+C.}$ Leaving it implicit is cleanest; solving for $y$ would force a $\pm \sqrt{}$ (and the sign is pinned by the sign of $y$ at your starting point, since curves cannot cross $y=0$).

Separate: divide by $1+y^2$ (never zero, so no lost solutions): $\frac{dy}{1+y^2}=\cos xdx.$ Integrate. Why $\arctan$ on the left? Because $\int \frac{dy}{1+y^2}=\arctan y$ — that integral is precisely the one whose answer is arctan. We place the single arbitrary constant $C$ on the right: $\arctan y=\sin x+C.$ Solve for $y$ (arctan is invertible everywhere): $\boxed{y=\tan (\sin x+C).}$ Where is this valid? $\tan$ blows up whenever its argument hits $\frac{\pi }{2}+k\pi$. So the solution exists only while $-\frac{\pi }{2}<\sin x+C<\frac{\pi }{2}(\text{on the branch containing the start point}).$ For the initial condition $y(0)=0$: $\arctan 0=\sin 0+C\Rightarrow C=0$, giving $y=\tan (\sin x)$. Since $|\sin x|\le 1<\frac{\pi }{2}\approx 1.571$ for all real $x$, the argument never reaches $\pm \frac{\pi }{2}$, and this particular solution is defined on all of $(-\infty ,\infty )$. (For a large $C$, e.g. $C=1$, the argument $\sin x+1$ can exceed $\frac{\pi }{2}$ and the solution then only lives on the $x$-intervals where $\sin x+1<\frac{\pi }{2}$.)

Split the exponent: $e^{x-y}=e^x{e}^{-y}$, so $g=e^x,h=e^{-y}$. Separate: divide by $e^{-y}$ = multiply by $e^y$: $e^{y}dy=e^{x}dx.$ Integrate. Why does $\int e^{y}dy$ give exactly $e^y$? Because $e^y$ is the one function that is its own derivative: $\frac{d}{dy}{e}^y=e^y$, so it is its own antiderivative. Likewise $\int e^{x}dx=e^x$. Each integral would carry a constant, but — as always — we merge the two into a single arbitrary constant $C$ and park it on the right side only: $e^y+({\text{const}}_L)=e^x+({\text{const}}_R)⟹e^y=e^x+C,C={\text{const}}_R-{\text{const}}_L.$ So the two integration constants collapse into the one $C$ shown. Solve for $y$: take logs of both sides: $\boxed{y=\ln (e^x+C).}$ Explicit domain. A log is only defined for a positive argument, so we need $e^x+C>0$, i.e. $e^x>-C$.

• If $C\ge 0$: $e^x+C>0$ automatically, so the solution is defined for all $x\in (-\infty ,\infty )$.
• If $C<0$: we need $e^x>-C\Rightarrow x>\ln (-C)$. The solution lives only on $(\ln (-C),\infty )$, blowing down to $-\infty$ as $x\to \ln (-C{)}^{+}$.
• For $y(0)=0$: $e^0=e^0+C\Rightarrow 1=1+C\Rightarrow C=0$, giving $y=\ln (e^x)=x$. Since $C=0\ge 0$, this is valid on all of $(-\infty ,\infty )$. (Indeed $y=x$ trivially checks: $y^{\prime }=1=e^{x-x}=e^0=1$.)

Separate & integrate: $\frac{dy}{y}=x^{2}dx\Rightarrow \ln |y|=\frac{x^3}{3}+C$, with $C$ the single arbitrary constant of integration. Exponentiate: $|y|=e^C{e}^{x^3/3}$, so $y=A{e}^{x^3/3}$ with $A=\pm e^C$ (a new nonzero arbitrary constant, defined as $\pm e^C$). The $|\cdot |$ is exactly what lets $A$ absorb the $\pm$ sign. Apply $y(0)=3$: $3=A{e}^0=A$, so $A=3$: $\boxed{y=3e^{x^3/3}.}$ Domain: $e^{x^3/3}$ is finite and positive for all real $x$, so the solution is defined on $(-\infty ,\infty )$.

In the figure below the horizontal axis is $x$ and the vertical axis is $y$; the blue curve is the solution $y=3e^{x^3/3}$. The yellow dot marks the initial point $(0,3)$ where the curve passes through $y=3$; notice how flat the curve is for $x<0$ (it decays toward $0$ but never reaches it) and how the red arrow highlights the explosive, faster-than-exponential rise once $x>1$ (the exponent $x^3/3$ itself grows like a cube).

Equilibria first (from 1.2): $y\equiv 0$ and $y\equiv 3$. Separate: $\frac{dy}{y(3-y)}=dx$. Partial fractions. Why? So each piece integrates to a log. Write $\frac{1}{y(3-y)}=\frac{A}{y}+\frac{B}{3-y}.$ (Here $A,B$ are decomposition constants, not integration constants — they get pinned by algebra, not by initial data.) Clearing: $1=A(3-y)+By$. At $y=0$: $1=3A\Rightarrow A=\frac{1}{3}$. At $y=3$: $1=3B\Rightarrow B=\frac{1}{3}$. So $\frac{1}{y(3-y)}=\frac{1}{3}\left(\frac{1}{y}+\frac{1}{3-y}\right)$. Integrate (one arbitrary constant $C$ on the right): $\frac{1}{3}(\ln |y|-\ln |3-y|)=x+C.$ Clear the $\frac{1}{3}$ by multiplying every term by $3$: $\ln |y|-\ln |3-y|=3x+3C.$ Now $3C$ is still just "some unknown constant," so we rename it $C^{\prime }=3C$ — a fresh arbitrary constant. Combining the two logs with $\ln p-\ln q=\ln \frac{p}{q}$: $\ln \left|\frac{y}{3-y}\right|=3x+C^{\prime }.$ Solve for $y$: exponentiate both sides, $\left|\frac{y}{3-y}\right|=e^{C^{\prime }}{e}^{3x}$; folding $\pm e^{C^{\prime }}\to A$ (yet another renamed arbitrary constant) drops the modulus: $\frac{y}{3-y}=A{e}^{3x}.$ Cross-multiply and isolate $y$: $y=A{e}^{3x}(3-y)\Rightarrow y(1+A{e}^{3x})=3A{e}^{3x}\Rightarrow y=\frac{3A{e}^{3x}}{1+A{e}^{3x}}.$ Apply $y(0)=1$: $1=\frac{3A}{1+A}\Rightarrow 1+A=3A\Rightarrow A=\frac{1}{2}$. $\boxed{y=\frac{\frac{3}{2}{e}^{3x}}{1+\frac{1}{2}{e}^{3x}}=\frac{3e^{3x}}{2+e^{3x}}.}$ Sanity check the limit: as $x\to \infty$, $y\to 3$ (approaches the upper equilibrium); as $x\to -\infty$, $y\to 0$.

In the figure below the horizontal axis is $x$ and the vertical axis is $y$; the green curve is our solution $y=\frac{3e^{3x}}{2+e^{3x}}$. The yellow dashed line at $y=3$ and the red dashed line at $y=0$ are the two equilibrium solutions: the green curve starts at the blue dot $(0,1)$, is squeezed between those two flat lines, rises as an S-shape, and levels off just below $y=3$ — visually showing why $y=0$ and $y=3$ act as a "floor" and "ceiling" the solution can approach but never cross.

Equilibrium: $h(y)=y-2=0\Rightarrow y\equiv 2$ is a solution (its slope is $0=(0)\cos x$). Record it. Generic curves: divide by $y-2$: $\frac{dy}{y-2}=\cos xdx\Rightarrow \ln |y-2|=\sin x+C,$ with $C$ the single arbitrary constant of integration. Exponentiate: $y-2=A{e}^{\sin x}$ (where $A=\pm e^C$, a new nonzero arbitrary constant): $\boxed{y=2+A{e}^{\sin x}(A\ne 0),\text{plus the equilibrium }y\equiv 2.}$ Note $A=0$ recovers $y\equiv 2$, so we may allow any real $A$ and the single formula covers everything — but only because we consciously checked.

Equilibrium: $y^2=0\Rightarrow y\equiv 0$ is a solution (won't fit $y(0)=1$, but record). Separate: $\frac{dy}{y^2}=dx\Rightarrow -\frac{1}{y}=x+C$, with $C$ the arbitrary constant of integration. Apply $y(0)=1$: $-1=0+C\Rightarrow C=-1$. So $-\frac{1}{y}=x-1\Rightarrow y=\frac{1}{1-x}$. $\boxed{y=\frac{1}{1-x}.}$ Interval of validity: the denominator $1-x=0$ at $x=1$, where $y\to \infty$ (finite-time blow-up!). The solution through $(0,1)$ lives only on $(-\infty ,1)$ — the largest interval containing $x=0$ that avoids the singularity. This is the classic lesson: a perfectly smooth-looking ODE can produce a solution that escapes to infinity in finite $x$.

Coax into separable form. Factor the RHS: $y+xy=y(1+x)$. Divide by $x$: $\frac{dy}{dx}=\frac{y(1+x)}{x}=y\cdot \frac{1+x}{x}.$ Now $g(x)=\frac{1+x}{x}=\frac{1}{x}+1$ and $h(y)=y$ — separable! Equilibrium: $y\equiv 0$. Separate & integrate (single arbitrary constant $C$): $\frac{dy}{y}=\left(\frac{1}{x}+1\right)dx\Rightarrow \ln |y|=\ln |x|+x+C.$ Solve for $y$: exponentiate, $|y|=e^C|x|e^x$, and since $x>0$ (so $|x|=x$), fold $\pm e^C\to A$: $\boxed{y=Ax{e}^x(A\text{ any constant, }A=0\text{ gives }y\equiv 0).}$

Watch the singularity you're about to divide by. The RHS is undefined when $2y-1=0$, i.e. $y=\frac{1}{2}$. There the slope $y^{\prime }$ would be infinite (a vertical tangent), so a genuine solution curve can touch $y=\frac{1}{2}$ only at an isolated point with a vertical tangent, and can never cross it as a function of $x$. Keep this in mind for the domain. Separate: $(2y-1)dy=(3x^2+1)dx$. Integrate (single arbitrary constant $C$): $y^2-y=x^3+x+C$. $\boxed{y^2-y-x^3-x=C.}$ Why stop? Isolating $y$ needs the quadratic formula with a sign ambiguity — the implicit form is the honest general solution. Completing the square, ${\left(y-\frac{1}{2}\right)}^2=x^3+x+C+\frac{1}{4}$, so $y=\frac{1}{2}\pm \sqrt{x^3+x+C+\frac{1}{4}}.$ Which sign / where valid? The initial condition picks the sign: if $y_0>\frac{1}{2}$ take "$+$" (the whole branch stays above $\frac{1}{2}$); if $y_0<\frac{1}{2}$ take "$-$" (stays below). The solution exists only where the radicand $x^3+x+C+\frac{1}{4}\ge 0$; at the point where it hits $0$ the curve reaches $y=\frac{1}{2}$ with a vertical tangent and the function-solution stops. So the interval of validity is the $x$-interval (containing $x_0$) on which $x^3+x+C+\frac{1}{4}>0$. Verify by implicit differentiation: $2y{y}^{\prime }-y^{\prime }-3x^2-1=0\Rightarrow y^{\prime }(2y-1)=3x^2+1\Rightarrow y^{\prime }=\frac{3x^2+1}{2y-1}.$ ✓

No equilibrium in the usual sense, but note $y=0$ makes the RHS $x/y$ undefined — solutions never cross the $x$-axis. Separate: $ydy=xdx$. Integrate (single arbitrary constant): $\frac{y^2}{2}=\frac{x^2}{2}+K$. Multiply through by $2$ and rename $2K\to C$ (a fresh arbitrary constant): $\boxed{y^2-x^2=C.}$ These are hyperbolas (or their degenerate limit). We must handle all three signs of $C$ and, within each, the sign of $y_0$:

Case $C>0$. Then $y^2=x^2+C\ge C>0$, so $y$ is never zero. Solving, $y=\pm \sqrt{x^2+C}$ — two separate branches:

• the upper branch $y=+\sqrt{x^2+C}$ lives entirely in $y>0$;
• the lower branch $y=-\sqrt{x^2+C}$ lives entirely in $y<0$. A solution stays on the branch matching the sign of $y_0$ (it can never jump across $y=0$). Both branches are defined for all $x\in (-\infty ,\infty )$ (the radicand is always positive).

Case $C<0$. Now $y^2=x^2+C$ needs $x^2\ge -C$, i.e. $|x|\ge \sqrt{-C}$. The hyperbola opens left/right: two pieces, one on $x\le -\sqrt{-C}$ and one on $x\ge +\sqrt{-C}$, with a gap in between where no real $y$ exists. Again $y=\pm \sqrt{x^2+C}$, and the sign of $y_0$ selects upper ($+$) vs lower ($-$). Each solution is defined only on its own $x$-interval (one of the two arms), and at the endpoint $|x|=\sqrt{-C}$ the curve reaches $y=0$ with a vertical tangent, so as a function of $x$ it stops there.

Case $C=0$ (degenerate). Then $y^2=x^2\Rightarrow y=\pm x$ — the two straight asymptote lines. Because solutions cannot pass through the origin (RHS undefined at $y=0$), each is valid only on $x>0$ or $x<0$, giving four half-line solutions: $y=x(x>0)$, $y=x(x<0)$, $y=-x(x>0)$, $y=-x(x<0)$. The sign of $y_0$ together with the sign of $x_0$ selects which half-line you are on.

Summary of "all signs of $y_0$": the sign of $y_0$ decides upper (+) vs lower (−) branch in every case; the value $C=y_0^2-x_0^2$ decides the shape (open up/down if $C>0$, open left/right if $C<0$, degenerate lines if $C=0$); and no solution ever crosses $y=0$.

The three cases are drawn together below.

Equilibrium: $y\equiv 0$ (won't meet $y(1)=1$, but note it). Separate: $\frac{dy}{y^2}=\frac{dx}{x(x+1)}$. Partial fractions on the right. The denominator $x(x+1)$ has two distinct linear factors, so we split into one term per factor: $\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}(\text{check: }\frac{1}{x}-\frac{1}{x+1}=\frac{(x+1)-x}{x(x+1)}=\frac{1}{x(x+1)}✓).$ Integrate (single arbitrary constant $C$): $-\frac{1}{y}=\ln |x|-\ln |x+1|+C=\ln \left|\frac{x}{x+1}\right|+C.$ Apply $y(1)=1$: LHS $=-1$; RHS $=\ln \frac{1}{2}+C$. So $C=-1-\ln \frac{1}{2}=-1+\ln 2$. Thus $-\frac{1}{y}=\ln \left|\frac{x}{x+1}\right|-1+\ln 2=\ln \left|\frac{2x}{x+1}\right|-1$. Solve for $y$: $\boxed{y=\frac{1}{1-\ln \left|\frac{2x}{x+1}\right|}.}$ Explicit interval of validity around $x=1$. Three things must hold near the start point $x=1$:

1. Stay off the ODE's own singularities $x=0$ and $x=-1$ (where $x(x+1)=0$). Since $1>0$, we require $x>0$, ruling out everything at or left of $0$.
2. Keep the log's argument positive: for $x>0$ we have $\frac{2x}{x+1}>0$ automatically, so no extra constraint here.
3. Keep the denominator of $y$ nonzero: $1-\ln \frac{2x}{x+1}=0\Rightarrow \ln \frac{2x}{x+1}=1\Rightarrow \frac{2x}{x+1}=e$. Solving: $2x=e(x+1)\Rightarrow x(2-e)=e\Rightarrow x=\frac{e}{2-e}\approx \frac{2.718}{-0.718}\approx -3.79$. This root is negative, hence not in our region $x>0$, so the denominator never vanishes for $x>0$.

Combining, the largest interval containing $x=1$ on which all conditions hold is $\boxed{(0,\infty ).}$ The point is the method: two-distinct-factor partial fractions untangled $x(x+1)$, then singularity-plus-denominator analysis pinned the explicit interval $(0,\infty )$.

$x+y$, $e^{x+y}$, $x-y$, $\cos x(1+y^2)$ are separable? $e^{x+y}=e^x{e}^y$ ✓ and $\cos x(1+y^2)$ ✓ are separable; the bare sums $x+y$ and $x-y$ are not (a $\pm$ between $x$ and $y$, outside an exponent, blocks factoring).

(1) Did dividing by $h(y)$ delete an equilibrium $y\equiv y_0$ with $h(y_0)=0$? (2) What is the interval of validity — where does a denominator vanish, a log go non-positive, or the solution blow up?

$C$ (or $C^{\prime }$, $A$) stand for on this page? An arbitrary constant of integration — an as-yet-unknown fixed number that appears because antiderivatives are only determined up to a constant. Reshaping it (multiplying, exponentiating) just renames it; there is always exactly one such constant for a first-order ODE, pinned by an initial condition.