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.
$$\frac{dy}{dx}=5xy$$
$$\frac{dy}{ydx}=5x$$
$$\frac{dy}{y}=5xdx$$
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Separation of variables can be used when:
$$\frac{dy}{dx}=5xy$$
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All the y terms (including dy) can be moved to one side of the equation, and
$$\frac{dy}{ydx}=5x$$
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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:
$$y=k(1+x^2)$$