Separation of variables in ODE Engineering math blog

(Last Updated On: August 14, 2019)

Separation of variables in ODE. Dear friends, today I will show how to use the ‘separation of variables’ method in ordinary differential equations.

If you are looking for more in the first-order ODE, do check out:

First order linear ODE

First order homogeneous ODE

Solve first order ODE using ‘transformations of variables’

Solve Bernoulli’s equations

Separation of variables in ODE_compressed

Separation of variables in ODE

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

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

Now I will give you three examples of ‘separation of variables’ method. Have a look!!

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 $(1-x y)$ , then the ‘separation of variables’ method would not work.

Step 1

Now I will separate $x$ and $y$ components of this equation. So I will divide the given differential equation with $(1+x^2)(1+y^2)$ .

Thus I get

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

Next, I’ll simplify it to get

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

Now I will integrate both sides of $\cfrac{1}{1+y^2} dy = \cfrac{x}{1+x^2} dx$ .

Step 2

Thus it will be

(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.

Let’s say

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

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 get the value of $I$ . So it will be

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

And here $C$ is the integration constant.

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

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

Hence I can rewrite equation (3) as

$I = \frac{1}{2} \ln |1+x^2| + C.$

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

$\tan^{-1}y &=&\frac{1}{2} \ln |1+x^2| + C.$

Now I’ll simplify it a bit. So this gives

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

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

$\frac{dy}{dx} = -(x+xy^2).$

Next, I’ll take out $x$ as a common factor like

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

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

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

So this gives

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

And this is only because $(1+y^2)$ is not equal to $0$ . Now I’ll separate the $x$ and $y$ variables. So this gives

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

Next, I will integrate both sides of the equation.

Step 2

Thus the equation will look like

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

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

Also I’ll use the standard formula of integration for the right hand side of the equation (4). Thus it will be

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

Next, 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.

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

Step 3

So I put $y = 0$ and $x = 1$ in the equation (7). And this gives

$\tan^{-1}(0) = - \frac{1^2}{2}+C.$

Then I’ll simplify it to get

$2\tan^{-1}(0) = - 1 + 2C.$

Also, I already know that $\tan^{-1}(0) = 0$ . Thus it gives

$2 (0) = -1+2C.$

Then I’ll simplify it to get the value of $C$ as

$2C = 1.$

So this means

$C = \frac{1}{2}.$

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

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

Then I’ll simplify it to get

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

But I can also rewrite it as

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

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

Now I’ll give another example.

Example 3

According to Stroud and Booth (2013)* “Solve $\cfrac{dy}{dx} = e^{3x-2y}$ , given that $y = 0$ when $x = 0$ .”

Solution

Here the given differential equation is:

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

Like the first examples, here also I will separate $x$ and $y$ components.

Step 1

First of all, I will rewrite the equation as

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

Next, I’ll rewrite $e^{3x-2y}$ as $e^{3x}/e^{2y}$ . And this is because, as per the law of indices $x^{m-n} = x^m / x^n$ .

So the differential equation becomes

$\frac{dy}{dx} = \frac{e^{3x}}{e^{2y}}.$

Next, I’ll separate the $x-$ and $y-$ component of the equation. So this gives

$e^{2y}dy = e^{3x}dx.$

Now I’ll solve this equation. And for that, I’ll integrate this equation.

Step 2

Thus it gives

$\int e^{2y}dy = \int e^{3x}dx.$

As per the standard laws of integration, $\int e^{mx} dx = e^{mx}/m + C$ .

Hence this equation becomes

$\frac{e^{2y}}{2} = \frac{e^{3x}}{3} + C.$

And here $C$ is the integration constant.

Thus I can say that $\cfrac{e^{2y}}{2} = \cfrac{e^{3x}}{3} + C$ is the general solution of the equation. Now I have to get the particular solution of the equation for $x = 0, y = 0$ .

So this means I have to get the value of $C$ for $x = 0, y = 0$ .

Therefore I’ll substitute $x = 0, y = 0$ to the general solution of the given differential equation. So this gives

$\begin{eqnarray*}\frac{e^{0}}{2} &=& \frac{e^{0}}{3} + C\\\frac{1}{2} &=& \frac{1}{3} + C.\end{eqnarray*}$

Next, I’ll simplify it to get the value of $C$ as

$C = \frac{1}{2} - \frac{1}{3} \Rightarrow C = \frac{1}{6}.$

Thus the particular solution to the given differential equation is

$\frac{e^{2y}}{2} = \frac{e^{3x}}{3} + \frac{1}{6}.$

As I can see, it is still possible to simplify it a bit. So this gives

$3 e^{2y} = 2 e^{3x} + 1.$

Hence I can conclude that this is the answer to the given example.

Dear friends, this is the end of 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. 33 (Example 1), Q. No. 43 (Example 2).

Separation of variables in ODE

Examples of separation of variables

Example 1

Solution

Step 1

Step 2

Example 2

Solution

Step 1

Step 2

Step 3

Example 3

Solution

Step 1

Step 2

*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. 33 (Example 1), Q. No. 43 (Example 2).

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