Separable Differential Equations

A separable differential equation is a type of first-order ordinary differential equation (ODE) that can be written so that all terms involving x are on one side and all terms involving y are on the other.

The general form of the separable differential equation is dy/dx = f(x) g(y).

It is separable because we can rewrite it as: \frac{1}{g(y)}dy = f(x) dx, which separates the variables x and y.

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 = e^ye^x
• 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.

Related Articles:

• How to solve Differential Equations?
• Exact Differential Equations
• Partial Differential Equations

Solved Examples on Separable Differential Equation

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

Example 6: Solve the following differential equation dy/dx = (3x² + 4x - 4)/(2y - 4), y(1) = 3.

Solution:

Given DE, dy/dx = (3x² + 4x - 4)/(2y - 4)

⇒ (2y-4)dy = (3x²+4x-4)dx

Integrating both side,

∫(2y-4)dy = ∫(3x²+4x-4)dx

⇒ ∫2y dy - ∫4 dy = ∫3x² dx + ∫4x dx - ∫4 dx

⇒ y² - 4y = x³ + 2x² - 4x + C...(i)

Using the intial condition,

x = 1 then y = 3

⇒ 3² - 4×3 = 1³ + 2×1² - 4×1 + C

⇒ C = -2

From eq (i)

y²-4y = x³+2x²-4x-2

Which is the required solution.