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.
The example discussed below is the initial value problem for separable differential equations.
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 initila 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.
Read More:
Separable Differential Equations
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.
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.
Table of Content
- Standard form of Separable Differential Equation
- What is Separable Differential Equation?
- How to Solve Separable Differential Equations
- Initial Value Problem on Separable Differential Equations
- Examples on Separable Differential Equations
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