Calculus 1

Separable differential equations

A differential equation gives the slope of its solutions but not the solutions. When can the equation be undone by integrating, and how is the one solution through a given point picked out?

You need

u-substitutionCalculus 1 Slope fieldsCalculus 1

Unlocks

Exponential growth and decay modelsCalculus 1

Where you are 7 in Differential equations

Calculus8

Differential equations7

▴ 2 above Slope fields Euler's method

Separable differential equations

Exponential growth and decay models Logistic growth

When you reach for it

Differential equations and their solutions checked a proposed solution by substituting it, and Slope fields drew the shape of solutions without any formula. Neither produces a formula. For one large and useful class of first-order equations, the ones whose right side splits into a factor in x times a factor in y, a formula can be found by integrating, using nothing beyond the antiderivatives already known, u-substitution and the trigonometric antiderivatives. Many equations written in Writing a differential equation from a description are of this kind.

Reach for separation when \dfrac{dy}{dx} is given as a product g(x)\,h(y), or can be rearranged into one, and you want every solution, or the one solution through a given starting point.

The procedure

Day 1: separating and integrating

A first-order equation \dfrac{dy}{dx} = f(x, y) is separable if its right side factors as

\frac{dy}{dx} = g(x)\,h(y),

a function of x alone times a function of y alone. Either factor may be constant: \dfrac{dy}{dx} = \cos x has h(y) = 1, and an autonomous equation \dfrac{dy}{dx} = h(y) has g(x) = 1. The equation \dfrac{dy}{dx} = x + y is not separable, since a sum does not split into such a product (Example 1 shows how to prove that).

Separating. Where h(y) \neq 0, divide by h(y) so that every y sits on the left with the derivative:

\frac{1}{h(y)}\,\frac{dy}{dx} = g(x).

Why integrating both sides works. Let H be an antiderivative of \dfrac{1}{h} and G an antiderivative of g. Suppose y = y(x) is a solution on an interval where h\big(y(x)\big) \neq 0. By the chain rule,

\begin{aligned} \frac{d}{dx}\Big[H\big(y(x)\big)\Big] &= H'\big(y(x)\big)\,y'(x) \\ &= \frac{y'(x)}{h\big(y(x)\big)} = g(x). \end{aligned}

So H\big(y(x)\big) and G(x) have the same derivative on the interval, and two functions with the same derivative on an interval differ by a constant. Every such solution therefore satisfies

H(y) = G(x) + C.

Conversely, a differentiable y(x) obeying H(y) = G(x) + C gives, on differentiating, \dfrac{y'}{h(y)} = g(x), which is the equation again. The shorthand for all this is the substitution u = y(x), du = y'(x)\,dx from u-substitution: multiply by dx and integrate,

\int \frac{1}{h(y)}\,dy = \int g(x)\,dx.

One constant. Each integral produces its own constant, but H(y) + C_1 = G(x) + C_2 is the same statement as H(y) = G(x) + (C_2 - C_1), so one constant, written on the x side, is enough.

Explicit or implicit. If H(y) = G(x) + C can be solved for y, the result is an explicit solution y = \dots. When it cannot, the equation H(y) = G(x) + C itself is an implicit solution: it defines y as a function of x even though no formula for that function is written. Steps that rewrite the constant (multiplying C by 2, writing e^C as A) keep it an arbitrary constant and are allowed, provided the set of values it can take is tracked (Example 3).

Checking. An explicit answer is checked by substituting it, as in Differential equations and their solutions; an implicit one by implicit differentiation.

Day 2: initial conditions, branches and intervals

An initial condition y(x_0) = y_0 picks one member of the family: substitute x = x_0 and y = y_0 into H(y) = G(x) + C and solve for C. It is usually easiest to do this before solving for y.

Choosing the branch. Solving for y often gives more than one formula. From y^2 = \dots come \displaystyle y = \sqrt{\dots} and \displaystyle y = -\sqrt{\dots}; from \ln|y - c| = \dots come y - c > 0 and y - c < 0. A solution is a single continuous function, so only the branch through (x_0, y_0) belongs to the initial value problem: the sign of y_0, or of y_0 - c, decides.

The interval where the solution exists. The solution of an initial value problem is a differentiable function on an interval containing x_0, along which the equation holds. That interval ends where the formula breaks: where y grows without bound (the solution blows up), where it reaches a value at which the equation is undefined, or where g(x) itself is undefined. A formula may keep producing values beyond such a point, but those values lie on a separate curve that the initial condition says nothing about. Blow-up can happen even when g and h are defined everywhere (Example 6).

Solutions lost by dividing. Separating divided by h(y), which is not allowed where h(y) = 0. If h(c) = 0, the constant function y = c is a solution, since both sides of the equation are then 0: these are the equilibria seen as rows of level segments in a slope field. They must be added to the family by hand. Sometimes a value of the constant recovers them (Example 3); sometimes none does (Example 6). An initial condition y(x_0) = c is answered by the equilibrium itself.

Comparing with the slope field. The formula and the field should tell the same story: where the equation’s sign makes solutions rise or fall, where they level off, and which equilibrium they stay beside. A solution curve does not cross an equilibrium line where the right side is smooth, which explains which branch a solution stays on.

Worked examples

Example 1

Day 1: Kalthum sorts four equations

Kalthum has four equations. Decide which are separable and write each separable one in the form \dfrac{1}{h(y)}\,\dfrac{dy}{dx} = g(x):

\begin{aligned} &\text{(a)}\ \frac{dy}{dx} = x^2 y - y \\ &\text{(b)}\ \frac{dy}{dx} = x - 2y \\ &\text{(c)}\ \frac{dy}{dx} = e^{x + 2y} \\ &\text{(d)}\ \frac{dy}{dx} = \frac{y}{x} + 2y \end{aligned}

Solution

(a) Factor out y: x^2 y - y = (x^2 - 1)\,y, so g(x) = x^2 - 1 and h(y) = y, and for y \neq 0 the equation is \dfrac{1}{y}\,\dfrac{dy}{dx} = x^2 - 1.

(c) A sum in an exponent is a product of powers: e^{x + 2y} = e^x e^{2y}. Dividing by e^{2y}, which is never 0, gives e^{-2y}\,\dfrac{dy}{dx} = e^x.

(d) Factor out y: \dfrac{y}{x} + 2y = \left(\dfrac{1}{x} + 2\right) y, so \dfrac{1}{y}\,\dfrac{dy}{dx} = \dfrac{1}{x} + 2, for x \neq 0 and y \neq 0.

(b) Failing to find a factorization is not a proof that none exists, so here is a test. If f(x, y) = g(x)\,h(y), then for any numbers a and b,

\begin{aligned} f(x, y)\,f(a, b) &= g(x)h(y)\,g(a)h(b) \\ &= f(x, b)\,f(a, y), \end{aligned}

because both sides are the same four factors in a different order. For f(x, y) = x - 2y, take (a, b) = (0, 0) and (x, y) = (1, 1): the left side is f(1, 1)f(0, 0) = (-1)(0) = 0 and the right side is f(1, 0)f(0, 1) = (1)(-2) = -2. They differ, so x - 2y is not a product g(x)\,h(y), and (b) is not separable.

Example 2

Day 1: Yousra’s explicit solution

Yousra wants every solution of \dfrac{dy}{dx} = e^{-y}\cos x.

Solution

Here h(y) = e^{-y} is never 0, so dividing by it loses nothing. Dividing gives e^{y}\,\dfrac{dy}{dx} = \cos x, and integrating both sides,

\begin{aligned} \int e^y\,dy &= \int \cos x\,dx \\ e^y &= \sin x + C. \end{aligned}

The exponential is positive, so a solution can only live where \sin x + C > 0. There, take the natural logarithm of both sides:

y = \ln(\sin x + C).

Check by substitution, using the chain rule: y' = \dfrac{\cos x}{\sin x + C}, while e^{-y}\cos x = \dfrac{1}{\sin x + C}\cos x. The two agree. For C > 1 the expression \sin x + C is positive for every x and the solution is defined on the whole line; for -1 < C \le 1 each solution lives on an interval between two zeros of \sin x + C, and for C \le -1 there is no solution at all.

Example 3

Day 1: a constant with a sign

Find every solution of \dfrac{dy}{dx} = \dfrac{2xy}{x^2 + 4}.

Solution

The factors are g(x) = \dfrac{2x}{x^2 + 4} and h(y) = y. First the constant solution: h(0) = 0, and y = 0 makes both sides 0, so y = 0 is a solution. For y \neq 0, separate and integrate:

\int \frac{1}{y}\,dy = \int \frac{2x}{x^2 + 4}\,dx.

On the right put u = x^2 + 4, du = 2x\,dx, which gives \ln|u| = \ln(x^2 + 4); no absolute value is needed since x^2 + 4 > 0. So

\ln|y| = \ln(x^2 + 4) + C.

Exponentiate: |y| = e^{C}(x^2 + 4). A solution is continuous and never 0 here, so it keeps one sign on its interval, and y = \pm e^C (x^2 + 4). As C runs over all real numbers, \pm e^C runs over every nonzero number, so write A = \pm e^C:

y = A(x^2 + 4), \quad A \neq 0.

The value A = 0 gives exactly the lost solution y = 0, so the whole family is y = A(x^2 + 4) with A any real number. Check: y' = 2Ax, and \dfrac{2x \cdot A(x^2 + 4)}{x^2 + 4} = 2Ax.

Example 4

Day 1: Ziad’s implicit solution

Ziad solves \dfrac{dy}{dx} = \dfrac{x}{2 + \cos y}. Find the solutions and decide whether they can be written as y = \dots.

Solution

Since \cos y \ge -1, the factor 2 + \cos y is at least 1, so multiplying through by it is always allowed and loses nothing:

\begin{aligned} \int (2 + \cos y)\,dy &= \int x\,dx \\ 2y + \sin y &= \frac{x^2}{2} + C. \end{aligned}

No combination of familiar functions solves 2y + \sin y = k for y, so this implicit equation is the answer. It still defines y as a function of x: the left side H(y) = 2y + \sin y has H'(y) = 2 + \cos y \ge 1, so H increases and takes each value exactly once, giving exactly one y for each x.

Check by implicit differentiation (Implicit differentiation): differentiating both sides with respect to x gives (2 + \cos y)\,y' = x, which is the equation multiplied out.

Example 5

Day 2: choosing the square root

Solve \dfrac{dy}{dx} = \dfrac{1 - x}{y} with y(1) = -2, find the interval where the solution exists, and compare it with the slope field.

Solution

Multiply by y (the equation already requires y \neq 0) and integrate:

\begin{aligned} \int y\,dy &= \int (1 - x)\,dx \\ \frac{y^2}{2} &= x - \frac{x^2}{2} + C. \end{aligned}

Doubling, and renaming 2C as K, gives y^2 = 2x - x^2 + K. The point (1, -2) gives 4 = 2 - 1 + K, so K = 3:

y^2 = 3 + 2x - x^2 = 4 - (x - 1)^2.

Both \displaystyle y = \sqrt{3 + 2x - x^2} and \displaystyle y = -\sqrt{3 + 2x - x^2} satisfy this, but the solution passes through y = -2 < 0, so it is the negative branch:

y = -\sqrt{3 + 2x - x^2}.

The radicand (3 - x)(1 + x) is positive for -1 < x < 3. At x = -1 and x = 3 it reaches y = 0, where the equation is undefined, so the solution exists on the open interval -1 < x < 3. Check: \displaystyle y' = -\dfrac{2 - 2x}{2\sqrt{3 + 2x - x^2}} = \dfrac{x - 1}{\sqrt{3 + 2x - x^2}}, and \displaystyle \dfrac{1 - x}{y} = \dfrac{1 - x}{-\sqrt{3 + 2x - x^2}} is the same.

The relation (x - 1)^2 + y^2 = 4 is a circle of radius 2 centered at (1, 0), and the field (Figure 1) shows segments circling that point. Each segment is tangent to such a circle, since the slope \dfrac{1 - x}{y} is the negative reciprocal of the slope \dfrac{y}{x - 1} of the radius. The solution is only the lower half: the upper half is a different solution, and the two are not joined because the curve turns vertical at y = 0.

Figure 1The field of \dfrac{dy}{dx} = \dfrac{1 - x}{y} (Example 5). The solution through (1, -2) is the solid lower half circle, defined for -1 < x < 3; the dashed upper half is the other branch, the solution through (1, 2).

Example 6

Day 2: a solution that blows up

Solve \dfrac{dy}{dx} = 2xy^2 with y(0) = 1 and find where the solution exists. Then solve it with y(0) = -1 and with y(0) = 0.

Solution

The constant y = 0 is a solution, since h(y) = y^2 vanishes there. For y \neq 0:

\begin{aligned} \int y^{-2}\,dy &= \int 2x\,dx \\ -\frac{1}{y} &= x^2 + C. \end{aligned}

With y(0) = 1: -1 = 0 + C, so C = -1 and -\dfrac{1}{y} = x^2 - 1, that is,

y = \frac{1}{1 - x^2}.

The formula is undefined at x = \pm 1, and y \to \infty as x approaches 1 from the left or -1 from the right. The solution is the piece through (0, 1), on -1 < x < 1. For |x| > 1 the same formula gives negative values, but those curves are cut off from the starting point by the blow-up (Figure 2). Nothing in g(x) = 2x or h(y) = y^2 is undefined anywhere; the blow-up comes from the solution itself. Check: y' = \dfrac{2x}{(1 - x^2)^2} = 2x\left(\dfrac{1}{1 - x^2}\right)^2.

With y(0) = -1: 1 = C, and y = -\dfrac{1}{x^2 + 1}, defined for every x. The same equation, started below the axis instead of above it, never blows up.

With y(0) = 0, no value of C works, since -\dfrac{1}{y} is never 0. The answer is the lost constant solution y = 0, which the separated family cannot produce.

Figure 2The field of \dfrac{dy}{dx} = 2xy^2 (Example 6). The solution through (0, 1) is y = \dfrac{1}{1 - x^2} on -1 < x < 1 only. The dashed gray pieces come from the same formula but are not part of that solution. The solution through (0, -1) is y = -\dfrac{1}{1 + x^2}, defined for every x.

Example 7

Day 2: the absolute value and an equilibrium

Solve \dfrac{dy}{dx} = x(y - 2) with y(0) = 1, and compare with the slope field.

Solution

The factor h(y) = y - 2 is 0 at y = 2, so y = 2 is an equilibrium solution. For y \neq 2:

\begin{aligned} \int \frac{1}{y - 2}\,dy &= \int x\,dx \\ \ln|y - 2| &= \frac{x^2}{2} + C. \end{aligned}

At the start y - 2 = -1 < 0, and a solution cannot reach the equilibrium: at a point with y = 2 the only solution is y = 2 itself, since the right side is smooth (Boundaries). So |y - 2| = 2 - y all along. The point (0, 1) gives \ln 1 = C, so C = 0, and 2 - y = e^{x^2/2}:

y = 2 - e^{x^2/2},

defined for every x. Check: y' = -x e^{x^2/2}, and x(y - 2) = x\left(-e^{x^2/2}\right).

The field (Figure 3) agrees. Below y = 2 the slope x(y - 2) is positive for x < 0 and negative for x > 0, so the curve rises to a highest point on the y-axis, the point (0, 1), and falls on both sides, faster and faster as e^{x^2/2} grows. Started at (0, 3) instead, y - 2 > 0, \ln 1 = C again, and y = 2 + e^{x^2/2}: the mirror image above the line. All three pieces fit one formula, y = 2 + Ae^{x^2/2} with A any real number, the equilibrium being A = 0.

Figure 3The field of \dfrac{dy}{dx} = x(y - 2) (Example 7): the equilibrium y = 2, lost when dividing by y - 2, and the solutions y = 2 - e^{x^2/2} and y = 2 + e^{x^2/2} through (0, 1) and (0, 3).

Boundaries

The right side must factor. \dfrac{dy}{dx} = x + y, \dfrac{dy}{dx} = \sin(xy) and most equations are not separable; the method does not apply, and rearranging terms does not make it apply (Exercise 21).
Dividing by h(y) needs h(y) \neq 0. Constant solutions y = c with h(c) = 0 are not produced by the separated equation and must be added (Examples 3, 6 and 7).
The antiderivatives have to be found. Separation turns the equation into two integrals. If one has no antiderivative in familiar functions, the solution stays in integral form; if H(y) = G(x) + C cannot be solved for y, it stays implicit (Example 4).
A solution lives on an interval. It ends where it blows up, reaches a value where the equation is undefined, or meets a point where g is undefined (Examples 5 and 6). A formula is not a solution beyond that point, even where it gives numbers.
One initial condition, one solution, needs a smooth right side. That exactly one solution passes through (x_0, y_0) is a theorem, stated without proof, valid when g(x)\,h(y) is continuous near the point and has a continuous derivative in y. For \displaystyle \dfrac{dy}{dx} = 2\sqrt{y} with y(0) = 0, both y = 0 and y = x^2 work on x \ge 0 (there \displaystyle 2\sqrt{x^2} = 2x), so the initial condition does not decide.

Where this goes next

The most used separable equation is \dfrac{dy}{dt} = ky, whose solutions are the exponential functions; Exponential growth and decay models solves it once and builds growth, decay and cooling models from it. Logistic growth is separable too. When an equation is not separable, Euler’s method approximates its solution step by step, and the formulas found here are what such approximations are measured against.

Exercises

Day 1: separable or not

Decide which are separable, and write each separable one as \dfrac{1}{h(y)}\,\dfrac{dy}{dx} = g(x): (a) \dfrac{dy}{dx} = xy + 3y; (b) \dfrac{dy}{dx} = x^2 + y^2; (c) \dfrac{dy}{dx} = \dfrac{x}{y} - x; (d) \dfrac{dy}{dx} = xy + x + y + 1.
Show that \dfrac{dy}{dx} = e^x + e^y is not separable, and solve \dfrac{dy}{dx} = e^{x + y}.

Day 1: general solutions

Find every solution of \dfrac{dy}{dx} = y^2 \sin x.
Find every solution of \dfrac{dy}{dx} = 3x^2(y + 1).
Find every solution of \dfrac{dy}{dx} = x e^{x^2 - y}.
Find every solution of \dfrac{dy}{dx} = \dfrac{\cos x}{y^2} as an explicit formula.
Find every solution of \dfrac{dy}{dx} = y \cos x and check it by substitution.

Day 1: implicit solutions

Find an implicit solution of \dfrac{dy}{dx} = \dfrac{x^2}{y + e^y} and check it by implicit differentiation.
Show that \sin y = x^2 + C defines solutions of \dfrac{dy}{dx} = \dfrac{2x}{\cos y} wherever \cos y \neq 0, and obtain it by separating.

Day 2: initial value problems

Solve \dfrac{dy}{dx} = \dfrac{x}{y} with y(2) = -1 and give the interval where the solution exists.
Solve \dfrac{dy}{dx} = y^2 \sin x with y(0) = \tfrac{1}{2} and give the interval where the solution exists.
Solve \dfrac{dy}{dx} = \dfrac{y}{x + 1} with y(0) = 3 and give the interval where the solution exists.
Solve \dfrac{dy}{dx} = 3x^2 y^2 with y(0) = 1 and give the interval where the solution exists.
Solve \dfrac{dy}{dx} = x(y - 3)^2 (a) with y(0) = 3; (b) with y(0) = 2.
Solve \dfrac{dy}{dx} = (y + 1)\cos x with y(0) = -3.
Solve \dfrac{dy}{dx} = \dfrac{\cos x}{2y} with y(0) = 1 and give the interval where the solution exists.

Day 2: solutions and slope fields

Solve \dfrac{dy}{dx} = -xy with y(0) = 2. Explain from the equation alone why the solution has its highest point at x = 0 and stays above the x-axis.
For \dfrac{dy}{dx} = x^2(y - 4) with y(0) = 1, explain without solving why the solution stays below 4 and never rises. Then solve it.

Reasoning

Explain why writing a constant on both sides after integrating, H(y) + C_1 = G(x) + C_2, gives no more solutions than one constant.
Show that if h(c) = 0, the constant function y = c solves \dfrac{dy}{dx} = g(x)\,h(y), and explain why separating the variables does not find it.
Yousra rewrites \dfrac{dy}{dx} = x + y as dy - y\,dx = x\,dx and proposes to integrate. Explain why this does not solve the equation, and show the right side is not of the form g(x)\,h(y).
In Exercise 13 the formula also gives values for x > 1. Explain why that part of its graph is not part of the solution of the initial value problem.
Let H' = \dfrac{1}{h} and G' = g. Prove that a differentiable function y(x) with H\big(y(x)\big) = G(x) + C on an interval solves \dfrac{dy}{dx} = g(x)\,h(y) there, wherever h\big(y(x)\big) \neq 0.

Answers

(a) Separable: \dfrac{1}{y}\,\dfrac{dy}{dx} = x + 3. (b) Not separable: f(1, 1)f(0, 0) = 2 \cdot 0 = 0 but f(1, 0)f(0, 1) = 1 \cdot 1 = 1. (c) Separable, since \dfrac{x}{y} - x = x \cdot \dfrac{1 - y}{y}: \dfrac{y}{1 - y}\,\dfrac{dy}{dx} = x. (d) Separable, by grouping xy + x + y + 1 = (x + 1)(y + 1): \dfrac{1}{y + 1}\,\dfrac{dy}{dx} = x + 1.
With f = e^x + e^y: f(1, 1)f(0, 0) = 2e \cdot 2 = 4e \approx 10.87 but f(1, 0)f(0, 1) = (e + 1)^2 \approx 13.83, so it is not separable. For e^{x + y} = e^x e^y: \int e^{-y}\,dy = \int e^x\,dx gives -e^{-y} = e^x + C, so e^{-y} = K - e^x and y = -\ln(K - e^x), where K - e^x > 0.
y = \dfrac{1}{\cos x + C}, and y = 0. From \int y^{-2}\,dy = \int \sin x\,dx: -\dfrac{1}{y} = -\cos x + C; rename -C as C.
y = -1 + Ae^{x^3}, A any real number. \ln|y + 1| = x^3 + C gives y + 1 = \pm e^C e^{x^3}; A = 0 is the equilibrium y = -1.
y = \ln\left(\tfrac{1}{2}e^{x^2} + C\right), where \tfrac{1}{2}e^{x^2} + C > 0. Separate: e^y\,dy = x e^{x^2}\,dx; with u = x^2, \int x e^{x^2}\,dx = \tfrac{1}{2}e^{x^2}.
\displaystyle y = \sqrt[3]{3\sin x + C}, on intervals where 3\sin x + C \neq 0. From \int y^2\,dy = \int \cos x\,dx: \dfrac{y^3}{3} = \sin x + C; triple and rename the constant.
y = Ae^{\sin x}, A any real number (A = 0 is y = 0). Check: y' = Ae^{\sin x}\cos x = y\cos x.
\dfrac{y^2}{2} + e^y = \dfrac{x^3}{3} + C. Differentiating: y\,y' + e^y y' = x^2, so (y + e^y)\,y' = x^2.
Differentiating \sin y = x^2 + C: \cos y \cdot y' = 2x, so y' = \dfrac{2x}{\cos y} where \cos y \neq 0. Separating: \int \cos y\,dy = \int 2x\,dx gives the same.
\displaystyle y = -\sqrt{x^2 - 3} for \displaystyle x > \sqrt{3}. From y^2 = x^2 + C and 1 = 4 + C, C = -3; the negative branch since y(2) < 0; y = 0 at \displaystyle x = \sqrt{3}, where the equation is undefined.
y = \dfrac{1}{1 + \cos x} for -\pi < x < \pi. From Exercise 3, \tfrac{1}{2} = \dfrac{1}{1 + C} gives C = 1; 1 + \cos x = 0 at x = \pm\pi, where y blows up.
y = 3x + 3 for x > -1. \ln|y| = \ln|x + 1| + C gives y = A(x + 1), and A = 3; the equation is undefined at x = -1.
y = \dfrac{1}{1 - x^3} for x < 1. From -\dfrac{1}{y} = x^3 + C and -1 = C; 1 - x^3 > 0 exactly when x < 1, and y \to \infty as x \to 1^-.
(a) y = 3, the equilibrium. (b) y = 3 - \dfrac{2}{x^2 + 2} for every x. From -\dfrac{1}{y - 3} = \dfrac{x^2}{2} + C and 1 = C: y - 3 = -\dfrac{1}{x^2/2 + 1}.
y = -1 - 2e^{\sin x} for every x. \ln|y + 1| = \sin x + C with |{-3} + 1| = 2 gives C = \ln 2; y + 1 < 0 at the start, so y + 1 = -2e^{\sin x}.
\displaystyle y = \sqrt{1 + \sin x} for -\dfrac{\pi}{2} < x < \dfrac{3\pi}{2}. From y^2 = \sin x + C and 1 = C; the positive branch; y = 0 at x = -\dfrac{\pi}{2} and x = \dfrac{3\pi}{2}, where the equation is undefined.
y = 2e^{-x^2/2}. The constant y = 0 is a solution, so the curve through (0, 2) never reaches the axis; above it -xy is positive for x < 0 and negative for x > 0, so the curve rises, then falls, with its top at x = 0.
y = 4 - 3e^{x^3/3}. The equilibrium y = 4 cannot be crossed, so y < 4, and then x^2(y - 4) \le 0: the curve never rises. Solving: \ln|y - 4| = \dfrac{x^3}{3} + C, \ln 3 = C, and 4 - y = 3e^{x^3/3}.
H(y) + C_1 = G(x) + C_2 is the same as H(y) = G(x) + (C_2 - C_1), and C_2 - C_1 is again an arbitrary constant, so the family is the same.
For y = c: the left side is 0, the right side g(x)\,h(c) = g(x) \cdot 0 = 0. Separating divides by h(y), which is 0 along y = c, so that step assumes y \neq c.
Integrating y\,dx needs y as a known function of x, which is what is being sought, so the equation is no closer to solved. With f = x + y: f(1, 1)f(0, 0) = 2 \cdot 0 = 0 but f(1, 0)f(0, 1) = 1 \cdot 1 = 1, so f is not g(x)\,h(y).
The solution is a differentiable function on an interval containing 0. At x = 1 the formula is undefined and y \to \infty, so no such interval reaches past 1. The values for x > 1 form a separate curve, which satisfies the equation but not the initial condition.
Differentiate H\big(y(x)\big) = G(x) + C by the chain rule: H'\big(y(x)\big)\,y'(x) = g(x), that is, \dfrac{y'(x)}{h\big(y(x)\big)} = g(x). Multiplying by h\big(y(x)\big) gives y' = g(x)\,h(y).