3.2: Separable Equations

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A first-order ode is separable if it can be written in the form $$\label{eq:1}g(y)\frac{dy}{dx}=f(x),\quad y(x_0)=y_0,$$ where the function $g(y)$ is independent of $x$ and $f(x)$ is independent of $y$. Integration from $x_0$ to $x$ results in $$\int_{x_0}^xg(y(x))y'(x)dx=\int_{x_0}^xf(x)dx.\nonumber$$

The integral on the left can be transformed by substituting $u = y(x),\: du = y'(x)dx$, and changing the lower and upper limits of integration to $y(x_0) = y_0$ and $y(x) = y$. Therefore, $$\int_{y_0}^yg(u)du=\int_{x_0}^xf(x)dx,\nonumber$$ and since $u$ is a dummy variable of integration, we can write this in the equivalent form $$\label{eq:2}\int_{y_0}^yg(y)dy=\int_{x_0}^xf(x)dx.$$

A simpler procedure that also yields $\eqref{eq:2}$ is to treat $dy/dx$ in $\eqref{eq:1}$ like a fraction. Multiplying $\eqref{eq:1}$ by $dx$ results in $$g(y)dy=f(x)dx,\nonumber$$ which is a separated equation with all the dependent variables on the left-side, and all the independent variables on the right-side. Equation $\eqref{eq:2}$ then results directly upon integration.

Example $\PageIndex{1}$

Solve $\frac{dy}{dx}+\frac{1}{2}y=\frac{3}{2}$, with $y(0)=2$.

Solution

We first manipulate the differential equation to the form $$\label{eq:3}\frac{dy}{dx}=\frac{1}{2}(3-y),$$ and then treat $dy/dx$ as if it was a fraction to separate variables:

$$\frac{dy}{3-y}=\frac{1}{2}dx.\nonumber$$

We integrate the right-side from the initial condition $x = 0$ to $x$ and the left-side from the initial condition $y(0) = 2$ to $y$. Accordingly, $$\label{eq:4}\int_2^y\frac{dy}{3-y}=\frac{1}{2}\int_0^xdx.$$

The integrals in $\eqref{eq:4}$ need to be done. Note that $y(x) < 3$ for finite $x$ or the integral on the left-side diverges. Therefore, $3 − y > 0$ and integration yields $$\begin{aligned} -\ln (3-y)]_2^y&=\frac{1}{2}x]_0^x, \\ \ln(3-y)&=-\frac{1}{2}x, \\ 3-y&=e^{-\frac{1}{2}x}, \\ y&=3-e^{-\frac{1}{2}x}.\end{aligned}$$

Since this is our first nontrivial analytical solution, it is prudent to check our result. We do this by differentiating our solution:

$$\begin{aligned}\frac{dy}{dx}&=\frac{1}{2}e^{-\frac{1}{2}x} \\ &=\frac{1}{2}(3-y);\end{aligned}$$ and checking the initial conditions, $y(0) = 3 − e^0 = 2$. Therefore, our solution satisfies both the original ode and the initial condition.

Example $\PageIndex{2}$

Solve $\frac{dy}{dx}+\frac{1}{2}y=\frac{3}{2},$ with $y(0)=4$.

Solution

This is the identical differential equation as before, but with different initial conditions. We will jump directly to the integration step:

$$\int_4^y\frac{dy}{3-y}=\frac{1}{2}\int_0^xdx.\nonumber$$

Now $y(x) > 3$, so that $y − 3 > 0$ and integration yields $$\begin{aligned}-\ln (y-3)]_4^y&=\frac{1}{2}x]_0^x, \\ \ln(y-3)&=-\frac{1}{2}x, \\ y-3&=e^{-\frac{1}{2}x}, \\ y&=3+e^{-\frac{1}{2}x}.\end{aligned}$$

Figure $\PageIndex{1}$: Solution of the following ODE: $\frac{dy}{dx}+\frac{1}{2}y=\frac{3}{2}$.

The solution curves for a range of initial conditions is presented in Fig. $\PageIndex{1}$. All solutions have a horizontal asymptote at $y = 3$ at which $dy/dx = 0$. For $y(0) = y_0$, the general solution can be shown to be $y(x) = 3 + (y_0 − 3) \text{exp}(−x/2)$.

Example $\PageIndex{3}$

Solve $\frac{dy}{dx}=\frac{2\cos 2x}{3+2y}$, with $y(0)=-1$.

1. For what values of $x>0$ does the solution exist?
2. For what value of $x>0$ is $y(x)$ maximum?

Solution

Notice that the derivative of $y$ diverges when $y = −3/2$, and that this may cause some problems with a solution.

We solve the ode by separating variables and integrating from initial conditions:

$$\begin{aligned} (3+2y)dy&=2\cos 2xdx \\ \int_{-1}^y(3+2y)dy&=2\int_0^x\cos 2xdx \\ 3y+y^2]_{-1}^y&=\sin 2x]_0^x \\ y^2+3y+2-\sin 2x&=0 \\ y_{\pm} &=\frac{1}{2}[-3\pm\sqrt{1+4\sin 2x}].\end{aligned}$$

Solving the quadratic equation for $y$ has introduced a spurious solution that does not satisfy the initial conditions. We test:

$$y_{\pm}(0)=\frac{1}{2}[-3\pm 1]=\left\{\begin{array}{c}-1; \\ -2.\end{array}\right.\nonumber$$

Only the $+$ root satisfies the initial condition, so that the unique solution to the ode and initial condition is $$\label{eq:5} y=\frac{1}{2}[-3+\sqrt{1+4\sin 2x}].$$

To determine (i) the values of $x > 0$ for which the solution exists, we require $$1+4\sin 2x\geq 0,\nonumber$$ or $$\label{eq:6}\sin 2x\geq -\frac{1}{4}.$$

Notice that at $x = 0$, we have $\sin 2x = 0$; at $x = π/4$, we have $\sin 2x = 1;$ at $x = π/2$, we have $\sin 2x = 0$; and at $x = 3π/4$, we have $\sin 2x = −1$. We therefore need to determine the value of $x$ such that $\sin 2x = −1/4$, with $x$ in the range $π/2 < x < 3π/4$. The solution to the ode will then exist for all $x$ between zero and this value.

To solve $\sin 2x = −1/4$ for $x$ in the interval $π/2 < x < 3π/4$, one needs to recall the definition of $\text{arcsin}$, or $\sin^{−1}$, as found on a typical scientific calculator. The inverse of the function $$f(x)=\sin x,\quad -\pi /2\leq x\leq \pi /2\nonumber$$ is denoted by $\text{arcsin}$. The first solution with $x > 0$ of the equation $\sin 2x = −1/4$ places $2x$ in the interval $(π, 3π/2)$, so to invert this equation using the $\text{arcsine}$ we need to apply the identity $\sin (π − x) = \sin x$, and rewrite $\sin 2x = −1/4$ as $\sin (π − 2x) = −1/4$. The solution of this equation may then be found by taking the $\text{arcsine}$, and is $$\pi -2x=\text{arcsin}(-1/4),\nonumber$$ or $$x=\frac{1}{2}\left(\pi +\text{arcsin}\frac{1}{4}\right).\nonumber$$

Therefore the solution exists for $0 ≤ x ≤ (π + \text{arcsin} (1/4)) /2 = 1.6971\ldots$, where we have used a calculator value (computing in radians) to find $\text{arcsin}(0.25) = 0.2527\ldots$. At the value $(x, y) = (1.6971\ldots , −3/2)$, the solution curve ends and $dy/dx$ becomes infinite.

To determine (ii) the value of $x$ at which $y = y(x)$ is maximum, we examine $\eqref{eq:5}$ directly. The value of $y$ will be maximum when $\sin 2x$ takes its maximum value over the interval where the solution exists. This will be when $2x = π/2$, or $x = π/4 = 0.7854\ldots$.

The graph of $y=y(x)$ is shown in Fig. 3.3.1.