An equation involving derivatives or a derived function is called a differential equation.
Differential equations are solved using integration.
When a differential equation is solved, there are two types of solution :
A general solution which will involve arbitrary constants. This can lead to solutions which are families of curves.
or
A particular solution where information is given allowing the values of the arbitrary constants to be determined.
Two types of differential equations are studied: first order or second order:
Type of differential equation
|
Notation
|
Example
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First order
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contain f '(x), y ' or dy⁄dx | dy⁄dx = 3x − 2 |
Second order
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contain f ''(x), y '' or d2y⁄dx2 | y '' + y ' = 2x |
In this and the next topic all three types of notation will be used.
Verifying a Given Solution
Sometimes the solution to a differential equation is provided and has be be verified. It is important to set out this type of problem in an orderly and mathematically correct way.
Example
Given that y '' + y ' = 2x, verify that y = (x − 1)2 is a particular solution.
Answer (LHS means left hand side and RHS means right hand side.)
To show that y = (x − 1)2 is a solution of y '' + y ' = 2x Let y = (x − 1)2
⇒ y ' = 2x − 2
⇒ y '' = 2
Thus, LHS= y '' + y ' = 2 + 2x − 2 = 2x = RHS Hence y = (x − 1)2 is a particular solution.
General Solutions
Differential equations are solved by integrating both sides of the equation.
Example
Find the function g(x) given that g '(x) = 4 − 2x
Solution
Integrating both sides g(x) = 4x − x2 + c
This general solution produces a family of curves.
Particular Solutions
A method of solving differential equations is known as separating the variables. The derivative dy⁄dx is not a fraction but can be treated in a similar manner when solving differential equations.
Once solved the value of the arbitrary constant can be found if enough information is given.
Example
Solve the differential equation dy⁄dx = 2x given that y = 26 when x = 5
dy = 2x. dx (separating the variables)
∫ dy = ∫ 2x.dx (integrating both sides)
y = x2 + c
When x = 5, y = 26
26 = 25 + c
c = 1
The solution is y = x2 + 1
Second Order Differential Equations
Second order differential equations are equations containing terms such as f ''(x), y '' or d2y⁄dx2
Solving this type of differential equation involves integrating twice.
Example
Find the general solution to the differential equation s ''(t) = -9.8
Integrating both sides:
∫ s ''(t) dt = ∫ -9.8 dt
s '(t) = -9.8t + A
Integrating both sides again:
∫ s '(t) dt = ∫ (-9.8t + A)dt
s(t) = -4.9t2 + At + B