Section 10.2: Separation of variables

The method of separation of variables applies to differential equations of the form

$$y' = p(t) q(y)$$

where $p(t)$ and $q(x)$ are functions of a single variable.

Example

Find the general solution to the differential equation

$$y' = t y^2$$

Solution

Any constant solution to this equation would have $0 \equiv t y^2$ so that $y \equiv 0$.

Avoiding the constant solution, we may divide both sides of the equation by $y^2$ and then we solve:

$$\begin{eqnarray*} \frac{T^2}{2} & = & \int_0^T t dt \\ & = & \int_0^T \frac{y'... ...} \vert_{y(0)}^{y(T)} \\ & = & \frac{1}{y(0)} - \frac{1}{y(T)} \end{eqnarray*}$$

So, if we set $C := y(0)$, we have $y = \frac{2 C}{2 - C t^2}$.

General procedure

To solve the differential equation $y' = p(t) q(y)$:

• Find the constant solutions by solving for $q(c) = 0$.
• Find $P(t)$ an antiderivative of $p(t)$ and $Q(x)$ an antiderivative of $\frac{1}{q(x)}$.
• Write $c = y(0)$. We find $Q(y) - Q(c) = \int_0^y \frac{dy}{q(y)} = \int_{0}^T p(t) dt = P(T) - P(0)$.
• Solve for $y$.

Example

Find the general solution of $y' = y\sin(t) - \sin(t)$

Solution

We begin by rewriting the equation at $y' = (y - 1) \sin(t)$.

The only constant solution is $y \equiv 1$.

Integrating, we find that $\ln(\vert y - 1\vert)$ is an antiderivative of $\frac{1}{y -1}$ while $-\cos(t)$ is an antiderivative of $\sin(t)$.

Let $C = y(0)$. Then we have $\ln(\vert y - 1\vert) - \ln(\vert C - 1\vert) = 1 -\cos(t)$.

Adding $\ln(\vert C - 1\vert)$ to both sides and applying the exponential function, we conclude that $\vert y - 1\vert = \vert C - 1\vert e^{1 - \cos(t)}$.

As the solution $y$ must be continuous, the signs of $y - 1$ and $C - 1$ agree. Thus, $y = 1 + (C -1)e^{1 - \cos(t)}$.

Note: In this case the constant solution has the same form.

Yet another example

Find the general solution to the differential equation $y' = ty + 1$

No elementary solution!

The method of separation of variables does not apply as the function $ty + 1$ cannot be written as the product of a function of $y$ by a function of $t$.

Scholium: Using Taylor series expansions (a topic which we shall discuss next month), one can compute an expression for solutions to the equation $y' = ty + 1$.

Another Example

Find the general solution to the equation

$$y' = \frac{\sec^2{t}}{y + 1}$$

Solution

There are no constant solutions as $\frac{1}{x + 1}$ is never zero. Note, however, that we cannot have $y(t) = -1$ as the differential equation would require $y$ to be nondifferentiable at such a point.

As before, we set $C = y(0)$. Multiplying by $y + 1$ and integrating, we find

$$\begin{eqnarray*} \tan(T) & = & \int_0^T \sec^(t) dt \\ & = & \int_0^T (y(t) +... ...+ 1) dy \\ & = & \frac{1}{2}y(T)^2 + y(T) - \frac{1}{2}C^2 - C \end{eqnarray*}$$

Solution, continued

So, $y$ satisfies the equation

$$y^2 + 2y - C^2 - 2C - 2 \tan(t) = 0$$

From the quadratic formula, we compute that

$$\begin{eqnarray*} y & = & \frac{-2 \pm \sqrt{2^2 - 4(-C^2 - 2C - 2 \tan(t))}}{2}... ... + 2C + 2 \tan(t)} \\ & = & -1 \pm \sqrt{(C+1)^2 + 2 \tan(t)} \end{eqnarray*}$$

Undefined solutions, multiple solutions

• For each choice of $C \neq -1$, we found two solutions to the initial value problem $y' = \frac{\sec^2(t)}{y + 1}$, namely $y = -1 + \sqrt{(C+1)^2 + 2 \tan(t)}$ and $y = -1 - \sqrt{(C+1)^2 + 2 \tan(t)}$. However, only $y = -1 + \sqrt{(C+1)^2 + 2 \tan(t)}$ satisfies $y(0) = C$.
• Strictly speaking, there is no solution with $y(0) = -1$, but there is a solution having $\lim_{t \to O^+} y(t) = -1$.
• No solution to the differential equation is defined for all values of $t$. If $t \ll 0$ so that $\tan(t) < \frac{-(C+1)^2}{2}$, then $\sqrt{(C+1)^2 + 2 \tan(t)}$ is not a real number.

Another Example

Find a function $y$ satisfying $y(0) = 5$ and $y' = \frac{t}{e^{y}}$.

Solution

As the exponential function never attains the value zero, there are no constant solutions to this differential equation. Multiplying both sides of the equation by $e^y$ and integrating, we obtain:

$$\begin{eqnarray*} \frac{1}{2}T^2 & = & \int_0^T t dt \\ & = & \int_0^T e^{y(t)... ...(t) dy \\ & = & \int_5^{y(T)} e^y dy \\ & = & e^{y(T)} - e^5 \end{eqnarray*}$$

Solution, continued

Addding $e^5$ to both sides of this equation and taking the natural logarithm, we compute

$$\begin{eqnarray*} y & = & \ln (e^y) \\ & = & \ln (e^5 + \frac{1}{2}t^2) \end{eqnarray*}$$