A separable differential equation is a type of first-order ordinary differential equation (ODE) in which the variables x and y can be separated and placed on opposite sides of the equation. This makes the equation easier to solve by integrating both sides separately.
The general form of a separable differential equation is:
dy/dx = f(x) × g(y)
By rearranging the equation, we get:
\frac{1}{g(y)}dy = f(x) dx,
After separating the variables, both sides are integrated to find the solution. The method of separation of variables is one of the simplest and most widely used techniques for solving first-order differential equations.
Examples of Separable Differential Equations
Various examples of separable differential equations are:
- dy/dx = (2x3 + 6)(y2 - 7)
- dy/dx = sec x cosec y
- dy/dx = eyex
- dy/dx = sin y cos x
Identify Separable Equation
The first-order differential equation is a separable differential equation if and only if it can be written as
\frac{dy}{dx} = f(x).g(y)
where,
- f(x) is a function of x that does not contain y
- g(y) is a function of y that does not contain x
Note: In order to solve this type of differential equation we have to separate all y's on one side and x's on another side of the equal sign.
How to Solve Separable Differential Equation
- Step 1: Arrange the given differential equation, in the form, dy/dx = f(x) g(y).
- Step 2: Separate the dependent and the independent variable on either side of the equal sign. As, dy/g(y) = d(x)f(x).
- Step 3: Integrating both sides individually to get the required solution ∫dy/g(y) = ∫ f(x) dx.
Example: Solve dy/dx = x3/y2.
Solution:
Given DE, dy/dx = x3/y2
⇒ dy(y2) = x3(dx)
Integrating both sides
∫dy(y2) = ∫x3(dx)
⇒ y2+1/(2+1) = x3+1/(3+1) + c
⇒ y3/3 = x4/4 + c
This is the solution to the given differential equation.
Initial Value Problem on Separable Differential Equation
We know how to solve the differential equation given in the separable form and this can also be achieved if the initial condition is given. In the differential equation if the initial condition is given we can easily find the value of the integration constant and this helps us to find the exact solution to the differential equation.
Example: Solve dy/dx = (x3 + 2)/(y2 - 2) and when y(0) = 0.
Solution:
Given DE, dy/dx = (x3 + 2)/(y2 - 2)
⇒ dy(y2 - 2) = (x3 + 2)(dx)
Integrating both sides
∫dy(y2 - 2) = ∫x3(x3 + 2)
⇒ y2+1/(2+1) -2y = x3+1/(3+1) + 2x + c
⇒ y3/3 - 2y = x4/4 + 2x + c...(i)
Using the initial condition y(0) = 0
x = 0 and then y = 0
⇒ 0 - 0 = 0 + 0 + c
⇒ c = 0
Thus from eq (i)
y3/3 - 2y = x4/4 + 2x + c
⇒ y3/3 - 2y = x4/4 + 2x
This is the solution to the given differential equation.
Solved Examples
Example 1: Find whether the differential equation dy/dx = f(x) + g(y) is variable separable or not.
Solution:
Given DE,
dy/dx = f(x) + g(y)
We know that in the above equation we can not separate the variables.
dy/dx = f(x) + g(y)
Thus, the given equation is not variable separable.
Example 2: Find the general solution of the differential equation dy/dx = -4xy2.
Solution:
Given DE, dy/dx = -4xy2
⇒ dy/y2 = -4x dx
Integrating both side
∫dy/y2 = ∫-4x dx
⇒ -(1/y) = -4(x2/2) + C
⇒ -(1/y) = -2x+C
⇒ -2x + (1/y) + C =0
This is the general solution of the differential equation.
Example 3: Find the solution of the differential equation dy/dx = (x+1)/(2-y), (y≠2).
Solution:
Given DE,
dy/dx = (x+1)/(2-y)
⇒ (2-y) dy = (x+1) dx
Integrating both side
∫(2-y) dy = ∫(x+1) dx
⇒ ∫2 dy- ∫y dy = ∫x dx + ∫1 dx
⇒ 2y − (y2/2) = (x2/2) + x + C
⇒ 2y − (y2/2) − (x2/2) − x − C = 0
This is the general solution of the given differential equation.
Example 4: Find the general solution of the differential equation dy/dx = (1+y2)/(1+x2).
Solution:
Given DE, dy/dx = (1+y2)/(1+x2)
⇒ dy/(1+y2) = dx/(1+x2)
Inytegrating both side
∫dy/(1+y2) = ∫dx/(1+x2)
⇒ tan-1y = tan-1x + C
⇒ tan-1x - tan-1y + C = 0
This is the general solution of the given differential equation.
Example 5: Solve the following differential equation dy/dx = 6xy2, y(1) = 1/25
Solution:
Given DE, dy/dx = 6xy2
⇒ dy/y2 = 6x dx
Integrating both sides
∫dy/y2 = ∫6x dx
⇒ ∫dy/y2 = 6∫x dx
⇒ -(1/y) = 6(x2/2) + C
⇒ -(1/y) = 3x2 + C...(i)
This is the general solution of the differential equation.
Using the Intial condition,
x = 1 then y = 1/25
⇒ -25 = 3 + C
⇒ C = -28
Substituting the value of C in eq (i)
-(1/y) = 3x2-28
⇒ y(3x2-28) + 1 = 0
