Separation of variables in ODE

Separation of variables in ODE. Dear friends, today I will talk about the separation of variables in ordinary differential equations. This is one of the methods of solving first order ODE.

Separation of variables

As the name says, in this method, the variables are separated first.

Then both sides are integrated to get the solution to the equation.

Here I will show you two examples of ‘separation of variables’ method. Have a look!!

If interested, you can also check out more posts in first order ODE such as

Examples of separation of variables

Note: None of these examples are mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here is the first example.

Example 1

According to Stroud and Booth (2013)* “Find the general solution of $(1+x^2)\cfrac{dy}{dx}=x(1+y^2).$ ”

Solution

Here the given equation is

$(1+x^2)\frac{dy}{dx}=x(1+y^2).$

This equation does not have any term like $m \pm (xy)^n$ . Here $m, n$ can be real number except zero.

This means in this example it will be possible to separate $x$ and $y$ variables.

In case if it would have any term like $(1+xy)$ or $(1-xy)$ , then the ‘separation of variables’ method would not work.

Step 1

I will now try to separate $x$ and $y$ components of this equation.

For this, I will divide the equation with $(1+x^2)(1+y^2)$ to get

$\begin{eqnarray*} (1+x^2)\frac{dy}{dx}&=&x(1+y^2)\\ \frac{(1+x^2)}{(1+x^2)(1+y^2)}\frac{dy}{dx}&=&\frac{x(1+y^2)}{(1+x^2)(1+y^2)}\\ \frac{1}{1+y^2}\frac{dy}{dx}&=&\frac{x}{1+x^2}~~~(1+x^2) \neq 0, (1+y^2) \neq 0\\ \frac{1}{1+y^2} dy &=& \frac{x}{1+x^2} dx. \end{eqnarray*}$

Now I will integrate both sides to the solution of the given equation.

Step 2

Thus the equation will look like

(1) $\begin{equation*} \int\frac{1}{1+y^2} dy =\int \frac{x}{1+x^2} dx. \end{equation*}$

From the standard formula of integration, I already know that

$\int \frac{1}{1+x^2}= \tan^{-1}x.$

Thus the left hand side of the equation (1) will be

(2) $\begin{equation*} \int\frac{1}{1+y^2} dy = \tan^{-1}y. \end{equation*}$

Now I will work on the right hand side of the equation.

Suppose

$I =\int \frac{x}{1+x^2} dx.$

From the standard rule of differentiation, I already know that if $t = 1+x^2$ , $\cfrac{dt}{dx} = 2x.$

In other words, I can also say that $x dx = \cfrac{1}{2}dt.$

Thus I can rewrite $I$ as

$I = \int \frac{1}{2}\frac{1}{t} dt.$

Now I will try to get the value of $I$ . Therefore

$\begin{eqnarray*} I&=& \int \frac{1}{2}\frac{1}{t} dt\\ &=&\frac{1}{2}\int \frac{1}{t} dt\\ &=&\frac{1}{2} \ln t + C. \end{eqnarray*}$

Here $C$ is the integration constant.

Now I substitute back $t = 1+x^2$ in $I$ to get

(3) $\begin{equation*} I = \frac{1}{2} \ln |1+x^2| + C. \end{equation*}$

At the end, I will put the values of equations (2) and (3) in equation (1). So it becomes

$\begin{eqnarray*} \tan^{-1}y &=&\frac{1}{2} \ln |1+x^2| + C\\2\tan^{-1}y &=&\ln |1+x^2| + 2C\\2\tan^{-1}y &=&\ln |1+x^2| + A ~(A = 2C). \end{eqnarray*}$

So the conclusion is $2\tan^{-1}y &=&\ln |1+x^2| + A$ is the solution of the equation.

Now I will go to the second example.

Example 2

According to Stroud and Booth (2013)* “Solve $\cfrac{dy}{dx}+x+xy^2 = 0,$ given $y = 0$ when $x = 1.$ ”

Solution

Here the given differential equation is:

$\frac{dy}{dx}+x+xy^2 = 0.$

Like my first example, here also I will separate $x$ and $y$ components.

Step 1

First of all, I will rewrite the equation as

$\begin{eqnarray*} \frac{dy}{dx}+x+xy^2 &=& 0\\ \frac{dy}{dx} &=& -(x+xy^2)\\ \frac{dy}{dx} &=& -x(1+y^2). \end{eqnarray*}$

Now I will divide the equation through out with $(1+y^2)$ to get

$\begin{eqnarray*} \frac{dy}{dx} &=& -x(1+y^2)\\ \frac{1}{1+y^2}\frac{dy}{dx} &=& \frac{-x(1+y^2)}{1+y^2}\\ \frac{1}{1+y^2}\frac{dy}{dx} &=& -x~ (1+y^2) \neq 0 \\ \frac{1}{1+y^2} dy &=& - x dx. \end{eqnarray*}$

Next I will integrate both sides of the equation to get the solution.

Step 2

Thus the equation will look like

(4) $\begin{equation*} \int \frac{1}{1+y^2} dy = \int - x dx. \end{equation*}$

From the standard formula of integration, I already know that

$\int \frac{1}{1+x^2}= \tan^{-1}x.$

Thus the left hand side of the equation (4) will be

(5) $\begin{equation*} \int\frac{1}{1+y^2} dy = \tan^{-1}y. \end{equation*}$

For the right hand side of the equation (4), I already know that from the standard formula of integration. Thus it will be

(6) $\begin{equation*} \int - x dx = - \frac{x^2}{2}+C. \end{equation*}$

Now I will put the values from (5) and (6) to (4) to get the solution of the given differential equation as

(7) $\begin{equation*} \tan^{-1}y = - \frac{x^2}{2}+C. \end{equation*}$

Well, this is the general solution of the equation.

Next I have to get the particular solution of the equation for $y = 0, x = 1.$

Step 3

Now I put $y = 0$ and $x = 1$ in the equation (7) to get

$\begin{eqnarray*} \tan^{-1}(0) &=& - \frac{1^2}{2}+C\\ 2\tan^{-1}(0) &=& - 1 + 2C\\ 2 (0) &=& -1+2C\\ 2C&=&1\\C &=&\frac{1}{2}. \end{eqnarray*}$

Next I substitute this value of $C= \cfrac{1}{2}$ in the equation (7) to get the particular solution of the given equation as

$\begin{eqnarray*} \tan^{-1}y &=& - \frac{x^2}{2}+\frac{1}{2}\\ 2\tan^{-1}y &=& - x^2+1\\ 2\tan^{-1}y &=&1 - x^2. \end{eqnarray*}$

Finally I can say that $2\tan^{-1}y = 1 - x^2$ is the solution of the given equation for $y = 0, x = 1.$

Dear friends, this is the end of my today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

*Reference: K. A. Stroud and Dexter J. Booth (2013): Engineering mathematics, Industrial Press, Inc. ; 7th Edition (March 8, 2013), Chapter: First-order Differential Equations, Further problems 25, p. 1000, Q. No.s 33 (Example 1), 43 (Example 2).

Separation of variables in ODE