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 seperate the variables.

dy/dx = f(x) + g(y)

Thus, the given equation is not variable seperable.

Example 2: 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 3: 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 4: 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 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.

Separable differential equations are a special type of ordinary differential equation (ODE) that can be solved by separating the variables and integrating each side separately. Any differential equation that can be written in form of y’ = f(x).g(y), is called a separable differential equation.

Basic form of the Separable differential equations is dy/dx = f(x) g(y), where x is the independent variable and y is the dependent variable.