Separable equations: six questions from setup to interpretation

Separation requires a product of a function of $x$ and a function of $y$. Distinguish a zero of a numerator from a value where the original equation is undefined.

(a) is separable. The equilibrium is $y=2$. Away from it,

$$\frac{dy}{y-2}=3x\,dx.$$

(b) gives $y'=x-2y$, which is linear but not separable in these variables on an open region. Writing $dy/y=(x-2)\,dx$ would incorrectly replace $x/y$ with $x$. Use integrating factors.

(c) is separable on intervals excluding $x=0$. The constant solution $y=0$ is valid on such an interval. For nonzero $y$,

$$\begin{gathered}\frac{dy}{y}=\frac{dx}{x}, \\ x\ne0.\end{gathered}$$

This equation is also linear. Dividing by $y$ does not remove the original restriction $x\ne0$.

It is separable, with $y=-1$ an equilibrium. For $y\ne-1$, use $dy/(y+1)=x\,dx$.

The integrated relation is $\ln|y-2|=3x^2/2+C$. At the initial point, $y-2$ is positive.

The constant solution $y=2$ works in the original equation but does not meet this initial value. Substituting $x=0$, $y=4$ into the integrated relation gives $C=\ln2$. Thus

$$y=2+2e^{3x^2/2}.$$

Its derivative is $6xe^{3x^2/2}$; the original right side is $3x(2e^{3x^2/2})$, the same expression. At zero it gives 4. It is differentiable on $\mathbb R$, and $y-2>0$ everywhere, so the division is consistent with this solution.

The absolute value is needed before the initial condition selects a sign. A negative branch of $y-2$ would be allowed for other data.

(a) gives $y=2-e^{3x^2/2}$ on $\mathbb R$. Its derivative is $-3xe^{3x^2/2}=3x(y-2)$, and its value at zero is 1. (b) gives the equilibrium $y=2$. The logarithmic equation cannot accept $y-2=0$; check that case directly.

“Proportional to the current membership” means $N'=0.08N$. Keep the coefficient’s units and compare $N(1)/N(0)$ with $1.08$.

The model is $N'=0.08N$, with equilibrium $N=0$. For the positive initial value,

$$\begin{gathered}\ln\left(\frac{N(t)}{250}\right)=0.08t, \\ N(t)=250e^{0.08t}.\end{gathered}$$

Differentiating gives $N'=20e^{0.08t}=0.08N$, and $N(0)=250$. Its mathematical interval is $\mathbb R$; the forward model uses $t\ge0$.

Doubling requires $e^{0.08t}=2$, so $t=\ln2/0.08\approx8.664$ years. The first-year multiplier is $e^{0.08}\approx1.083287$, an increase of about 8.329%. The coefficient 0.08 specifies the instantaneous relative rate, not a discrete one-year percentage change. Real membership is integer-valued; this smooth model is an approximation.

$N(t)=250e^{-0.08t}$. Its derivative is $-0.08N$, and its value at zero is 250. Setting it equal to 125 gives $t=\ln2/0.08\approx8.664$ years. It approaches zero without reaching it at finite time.

The equation requires $y\ne0$. Multiplying by $y$ gives $y\,dy=x\,dx$. The initial value chooses one square-root sign.

Integration gives $y^2/2=x^2/2+C$. At the initial point, $C=2$, so $y^2=x^2+4$. The initial value is negative, which selects

$$y=-\sqrt{x^2+4}.$$

The derivative is $-x/\sqrt{x^2+4}$, equal to $x/y$. The function equals $-2$ at zero and never reaches zero, so its maximal interval is $\mathbb R$. Writing both signs as the final initial-value answer would include a branch with the wrong initial value.

The function $y=0$ is not an equilibrium: the original right side is undefined there. The equation obtained by multiplication does not expand the original domain.

$y^2=x^2+9$, and the positive initial value selects $y=\sqrt{x^2+9}$ on $\mathbb R$. Its derivative is $x/\sqrt{x^2+9}=x/y$, and $y(0)=3$.

Record $y=0$ first. For nonzero $y$, integrate $y^{-2}\,dy=2x\,dx$. Look for the nearest singularity on either side of zero.

The equilibrium $y=0$ solves the equation but not this initial value. Separation gives $-1/y=x^2+C$, and $y(0)=1$ gives $C=-1$. Therefore

$$y=\frac1{1-x^2}.$$

Its derivative is $2x/(1-x^2)^2=2xy^2$, and its value at zero is 1. The poles are $x=-1$ and $x=1$. The maximal interval containing the initial input is $(-1,1)$. The two outer intervals also support branches of this formula, but they are disconnected from the prescribed point. The solution cannot pass through either blow-up.

The constant is now $C=1$, so $y=-1/(1+x^2)$. Its denominator never vanishes for real $x$; the maximal interval is $\mathbb R$. Its derivative $2x/(1+x^2)^2$ equals $2xy^2$, and its initial value is $-1$. A changed initial value can change the interval.

Use $1/[P(1-P/500)]=1/P+1/(500-P)$ for $P\ne0,500$. On the branch through 100, both logarithm arguments are positive. The general nonzero form can be written $P=500/(1+Ae^{-0.2t})$.

The equilibria are $P=0$ and $P=500$, checked in the original equation. Separating between them gives

$$\ln\left(\frac P{500-P}\right)=0.2t+C.$$

At zero the ratio is $1/4$, giving

$$P(t)=\frac{500}{1+4e^{-0.2t}}.$$

With $D=1+4e^{-0.2t}$, differentiation gives $P'=400e^{-0.2t}/D^2$. The original right side is $0.2(500/D)(4e^{-0.2t}/D)$, exactly the same. The initial value is 100, and the denominator is positive on $\mathbb R$. Use $t\ge0$ for the forward population model.

Setting $P=250$ gives $4e^{-0.2t}=1$, so $t=\ln4/0.2\approx6.931$ years. The population increases toward 500 and stays below it at every finite time. Above 500, the model’s derivative is negative, so a positive initial population above capacity decreases toward it for forward time.

The constant is $A=500/1000-1=-1/2$, so

$$P(t)=\frac{500}{1-\tfrac12e^{-0.2t}}.$$

It has a pole at $t=-\ln2/0.2$. The maximal interval containing zero is $(-\ln2/0.2,\infty)$. On $t\ge0$, the denominator lies between $1/2$ and 1, so the population stays above 500 and decreases toward it. The derivative is

$$P'=-\frac{50e^{-0.2t}}{(1-\tfrac12e^{-0.2t})^2},$$

which equals the original logistic right side. It gives $P'(0)=-200$ people per year, also obtained from $0.2(1000)(1-1000/500)$. The earlier positive-denominator argument for all real time cannot be reused for this changed initial value.