Separation of Variables

Separation of Variables
Written by: Jack Methew

Jack Methew knows that successful students become successful adults. This is her 15th year at Edison Elementary School and her 10th year teaching fourth grade. So far, fourth grade is her favorite grade to teach! Mrs. Carroll was the 2011 Newell Unified School District Teacher of the Year, and received her National Board Certification in 2013. She loves science and majored in biology at Arizona State University, where she also earned her teaching credential and Master of Education degree. Mrs. Carroll is excited to begin the best year ever!

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Separation of Variables

Separation of Variables is a very important and special method to solve some Differential Equations.

A Differential Equation is an equation with a function and one or more of its derivatives

$$\frac{dy}{dx}=5xy$$

Example#1

An equation with the function y and its derivatives as dy/dx.

$$\frac{dy}{dx}=5xy$$

$$\frac{dy}{ydx}=5x$$

$$\frac{dy}{y}=5xdx$$

  1. Separation of variables can be used when:

    $$\frac{dy}{dx}=5xy$$

  2. All the y terms (including dy) can be moved to one side of the equation, and

    $$\frac{dy}{ydx}=5x$$

  3. All the x terms (including dx can be moved to the other side of an equation)

    $$\frac{dy}{y}=5xdx$$

Working of Method

Three Steps:

Step 1:Shift all the y terms (including dy) to one side of the equation and all the x terms (including dx) to the other side.

Step 2:Integrate one side with respect to y and the other side with respect to x. Don’t forget “+ C” (the constant of integration).

Step 3:Simplify

Examples

Solve this (k is a constant)

$$\frac{dy}{dx}=ky$$

Step-1: Separate the variables by moving all the y terms to one side of the equation and all the x terms to the other side:

Multiply both sides by dx:

$$dy = ky dx$$

Divide both sides by y:

$$\frac{dy}{y}=k dx$$

Step-2:Integrate both sides of the equation separately:

Put the integral sign in front:

$$\int \frac{dy}{y}=\int kdx$$

Integrate left side

$$ln(y) + C=\int kdx$$

=Integrate right side:

$$ln ln (y) + C = kx+D$$

C is the constant of integration. And we use D for the other, as it is a different constant.

Step - 3:Simplify:

We can use the two constants into one(b = D - C):

$$\text{ln ln y}=kx+a$$

$$e^{(ln\;ln\;y)}=y$$

So let's take exponents on both sides:

$$y\;=\;e^{kx+a}$$

And

$$e^{kx+a}=e^{kx}e^a$$

so we get:

$$y = e^{kx}e^a$$

ea is just a constant so we replace it with c :

$$ y =Ce^{kx}$$

We have solved it:

$$y=ce^{kx}$$

This is a general type of differential equation.

Example#2

Solve this:

$$\frac{dy}{dx} = \frac{2xy}{1+x2}$$

Step#1: First we Separate the variables:

Multiply both sides by dx, divide both sides by y

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

Step#2: Integrate both sides of the equation separately

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

The left side is a simple logarithm , the right side can be integrated using substitution

Let

$$u=1+x^2$$

So

$$du=2udx$$

$$\int \frac{1}{y}dy=\int \frac{1}{u}du$$

Integrate:

$$ln (y)=ln (u) +C$$

Then we make

$$C = ln k$$

$$ln(y) = ln (u) +ln (k)$$

So we can get this

$$y=uk$$

Now put

$$u=1+x^2$$

back again:

$$y= k(1+x^2)$$

Step#3: Let us Simplify:

It is already as simple as can it is possible .we have solved it:

$$y=k(1+x^2)$$