Preparation for differential equations:

Calculus is one of important and higher level of topic in math. Calculus is divided into two types one is differentiation and integration. Differentiation deals with the rate of change of variables.Differential equation involves variables and numbers.

  •  In mathematics, a differential equation is defined as the derivatives of a function that appear as variables.
  •  A differential equation deals with unknown function of one or more variables that relates the values derivatives of different orders and its functions.

 

EXAMPLES FOR DIFFERENTIAL EQUATION PREPARATION:

 

Differential Equation Preparation Example 1:

Solve the differential equation dy / dx – 3xy = x

Solution:

Here P(x) = -3x and Q(x) = x

u(x) = e ∫ P(x)  dx     

       = e ∫ -3x dx    

       =   e – x^3  

Substitute the value of u(x) and Q(x) in the equation

       = e –x^3 y = ∫x e –x^3 dx         

       =     e –x^3 = - (1/2) e –x^3 + c, where c is the constant of integration                                      

    y = c e –x^3 – ½

 

Differential Equation Preparation Example 2:

 X3-3x2+14x+7.

Solution: 

=3x2-6x+14.

 

Differential Equation Preparation Example 3:

 Differentiate the given equation:

6x5+3x3+4.

Solution:

                  = 30x4+9x2.

 

Differential Equation Preparation Example 4:

 Find the derivative of xsin x , with respect to x

Solution:

    Let y = xsin x

    Taking log of the both sides, we get

    Log y = sin x. log x.

    Differentiating with respect to x, we get

    1/y. dy/dx = sin x. d/dx (log x) + log x. d/dx(sin x)

    1/y.dy/dx = sin x. 1/x + log x.cos x

     dy/dx = y [ sin x/x +cos x. log x]

     = xsin x [sin x/x + cosx log x]

    In both the examples, the alphabets ‘x’, and ‘y’ denote the differential equation constants.

These can be other alphabets as well but in math generally X and Y, are the most commonly used alphabets. 

 

 

Differential Equation Preparation Example 5:

 Solve the differential equation dy / dx – 4xy = x

Solution:

Here P(x) = -4x and Q(x) = x

u(x) = e ∫ P(x)  dx     

       = e ∫ -4x dx    

       =   e –x^4    

Substitute the value of u(x) and Q(x) in the equation,

       = e –x^4 y = ∫x e –x^4 dx         

       =     e –x^4 = - (1/2) e –x^4 + c, where c is the constant of integration                                      

    y = c e –x^4 – 1/2