## Finding the Differential Equation from a General Solution

To find the differential equation when the general solution is given, differentiate the general solution, differentiate the derived solution etc. until the number of derived equation is equal to the number of independent arbitrary constant finally eliminate the constants from the derived equations.

Finding the Differential Equation from a General Solution Examples

Example # 1:

Find the differential equation of X^2 + Y^2 = CX

Example # 2:

Find the Differential Equation of ( X – C1 )^2 + C2y = C3

Example # 3:

Find the differential Equation of all lines through the origin is

y = mx

m = y/x

Example # 4:

Find the differential equation of all circles through (0,0) and (2,0)

Solution:

The standard equation of a circle is (x-h)^2 + (y-k)^2  =  1-2

## Linear First and Second Order Differential Equation

Linear First and Second Order Differential Equation

Linear First Oder Differential Equation

The equation in the form dy/dx + Py = Q where P and Q are functions of x only is called Linear Differential Equations since y and its derivatives are of the first degree.

The solution for dy/dx + Py = Q is obtained by multiplying throughout by an Integrating factor   to become

Example: Solve the equation dy +4xy dx = 2xdx

Solution:

Rearranging:  dy/dx + 4xy = 2x

then P = 4x and Q = 2x

Linear Second Order Differential Equation

Equation in the form a (d^2y/dx^2) + b(dy/dx) + cy = 0 where a, b, and c are constants, is called a linear second order differential equation with constant coefficient

Setting D = d/dx and D^2 = d^2/dx. The following procedures may be followed.

1. write the equation in D – operator form (aD^2 + bD + C) y=0, substitute m for D and solve the auxiliary equation am^2 + bm + c=0 for m

A. If the roots are real and different (b^2 > 4ac) say .

Then the general solution is

B. If the roots are real and equal  twice the general solution is

C. If the roots are imaginary (b^2 – 4ac) Say

the general solution is

Example:  Solve the equation 2(d^2y/dx^2) + 5(dy/dx) – 3y = 0

Solution:

Writing D – operator form: (2D^2  + 5D – 3) y = 0

Substituting m for D gives the auxiliary equation 2m^2 + 5m – 3 = 0 which can be factored as (2m – 1) (m + 3) and the roots are m = ½ and m = -3

Since the roots are real and different the general solution is

then the general solution is

## How to Solve Differential Equation with Example

How to Solve Differential Equation with Example

Differential Equation

– are equation that contain differential coefficients.

Example:

– classified according to the highest derivative that occurs in Them the differential equation dy/dx = 12x is a first order differential equation d^2y/dx^2 + 4dy/dx – 3y = 0 is a second order differential equation. A solution to a differential equation that contains one or more arbitrary constant of integration is called general solution. When additional information is Given so that these constants may be calculated the particular solution of differential equation is obtained.

Variable Separable

A differential equations can be of type dy/dx = f(x)solved by direct integration by writing it in the form dy = f(x) dx

Example:

Solve the differential equation dy/dx = 2x + Sin 3x

Solution: dy = (2x + Sin 3x) dx

y = x2 – 1/3 Cos 3x + c  general solution

Differential equation of type dy/dx = f(y) can be solved by direct integrating by writing it in the form.

dx = dy/f(x)

Example:

Solve the equation (y^2 – 1) dy/dx = 3y Given that y = 1 when x = 13/6

Differential Equation of type

dy/dx = f(x)  g(y) can be solve by Direct Integration by writing it in the form dy/g(y) = f(x)dx

Differential Equation of type dq/dt = KQ the general solution of an equation of the form dq/dt = KQ is Q = Ce^kt . where C is constant

Example:

solve the equation dy/dx = 3y

here we have  Q = y   dQ = dy                     then

t = x       k = 3                          y = Ce^3x

Example:

Obtain the differential equation of the family of straight lines with slope and y intercept equal

Solution:

The standard equation of the line in slope – intercept form is y = mx + b.

Since the slope and y intercept are equal m = b = c then y = cx + c

isolating constant and differentiate