ADVANCED MATHEMATICS FORM 6 – DIFFERENTIAL EQUATION

First order homogenous D.E

          In first order H.D.E all the terms are of the same dimension

Consider the table about dimensions

O shows there is no effect on the term

Which of the following equation are first order homogeneous

a) x²  = y²

b) xy  = x² + y²

c) x²
 = 1 + xy

d) (x²
 =

e) (x² – y²)  = 2xy

f) (1 + y²)  = x

Solution of 1^st order Homogeneous  differential equation

1^st order homogeneous d.e can be written in the form  = Q

Where both P and Q are function of and have same dimension.

Suppose p and Q hence the dimension

          Then divide by xⁿ and use the substitution

          Y = vx =      v =

          i.e

Example 1.

Solve  =

Solution

   Since all terms are of degree 2, i.e the equation is homogeneous

          Let y = vx

           = v +

          And

                    =

                           =  ()

                             =

                   V +  =

                              =  – v

                             =

                              =

                              =

           =  dx

             = lnx + c

But v =

   ² = ln x + c

           = lnx + c

          Y^{2 =} 2x² (lnx + c)

Example 2

Solve xy = x² + y²

          Solution

          All terms are of the order 2

          xy  = x² + y²……

          Let y = vx

           = v +

          From the equation…

           =  +

          But y = vx

          =  =  +

                   = x²

                     =

          V +  =

                    =

                         =

                    =

                    =

                    = lnx + A

                   But v =

                   ² = lnx + A

                   Y² = 2x² (lnx + A)

Example 3

Solve the following

(x² + xy)  = xy – y²

            Solution

    

          Let y = vx

           = v +

          And

           =

                   =

                   v +  =

                    = -v

                   =

                   =

           =

          [( dv +  dv)] =

          –½ ( + ln v) = lnx + c

    

           = ln (x² V) + 2c

          But v =

           = ln (x²) + 2c

           – 2c = ln (x².)

          Let ln A = -2c

           + ln A = ln xy

     
Xy = Ae^x/y

          EXERCISE

Solve the differential equation

1.  x²  = y (x + y)

2. (x² + y²)  = xy

First order exact differential equation

We know that

Now consider

           + y = e^x

Integrating both side w.r.t.x

          = xy = e² + c

                   Y =  (e^x + c)

                   Y =  (e^x + C)

2.  Find the general solution of the following exact differential equation.

          i)e^xy + e^x = 2

        ii) cos x – ysin x = x²

          Solution

          i) Given

       

                    (e^xy) = 2

          Integrating both sides with respect to x

                    (e^xy) dx =

                

          ii) Given

                   cosx  – y sin x = x²

                    (y cos x) = x²

          Integrating both side with respect to x

          =  (e^xy) dx =

                   Y cos x =  + c

                   3ycosx = x3 + 3C

                   3ycosx = x3 + 4

3.        + lny = x + 1

                   Soln

                   Given   + lny = x + 1

          Integrating both sides w.r.t x

           (x ln y) dx =

                   Xln y = ½ ^x2 + x + C

                   2xlny + x² + 2x + 2c

                   2x lny + x² + x + A

          Integrating factors

          Consider first order differential equation of the form

                    + py =Q

          Where p and Q are functions of x

          Multiplying by integrating factor F both sides will make an exact equation

          i.e F  + Fpy + FQ…….i

                    =  +

                    + …..ii

          Comparing (i) (ii)

           + Fpy =  +

                        Fpy =

                    = Fp

                    dF =  by separating of variables

F =  is the required integrating factor

Examples

1. Solve  + y = x³

            Solution

                    + y = x³

                    +  = x²

          Compare with  + py = Q

                             P =

                   Then  =  dx

                                      = ln x

                   If F = e^{ln x}

                   F       = e^{ln x}

                    (xy) = x² x

                    (xy) =

                   Xy =  x⁴ + C

2. Solve

          (x + 1)  + y + (x + 1)²

            Soln

                   (x + 1)  + y + (x + 1)²

                   =  +  = x + 1

                   =  + , y = x + 1

          P =

         

          F = e^sp

             = e^{ln (x + 1)}

              = x + 1

          Y (x + 1) =

                   =

          = x³ + x² + x + c

          =  + c

          Y =  +

3.      Solve (1 – x²)  – xy = 1

          Solution   –  y =

                   P =

                    =  dx

                             = ½ ln (1 – )

                             F = e ^ln (1 ^{–  ½}

                                     = (1 – x²) ^½

                   Y  =   dx

                                      =  dx

                   Y =

4.      tanx  + y = e^x tan x

          Solution

                    + , y = e^x

                   P =  = cot x

                    =

                             = ln

                             F = e^ln

                             =

                   Y =  dx

          Integrating by parts the R.H.S

=  = e^x
–  dx

                   = e^x sin x – (e^x cos x +  dx

                   = e^x
– e^x
–  dx

          2 sin x dx = e ^x ( –)

                    =  ()

                   Y sin x =  () + c

Bernoulli’s equation

This is a first order D.E of the form

           + p (x) y = Q (x) y ⁿ

  p (x) and Q (x) are functions of x or constant

Steps

           + p (x) y = Q (x) yⁿ……i

          Divide both sides by yⁿ gives

          Y^-n   + p (x) y^{1 – n} = Q (x)…… (ii)

          Let z = y^1-n

          (1 – n) y^-n

Multiplying (ii) by (1 – n) both sides gives

          (1 – n) y^-n   + (1 – n) p (x) y (1 – n) = (1 – n) Q (x)

           + (1 – n) p (x) y ^{(1 – n)} = (1 – n) Q (x)

           + p₁ (x) y ^{(1 – n)} = Q₁ (x)

  p₁ (x) and Q₁ (x) are functions of x or constant

          But z = y ^{(1 -n)}

          =  + p₁ (x) z = Q₁ (x)……..ii

iii) is linear the use of integrating  factor  can be used

Example 1

Solve  +  = xy²

Solution

 

Dividing both sides by  gives

 

Let Z =

 

Multiply (ii) by -1 gives

 

 

But Z =

 

 

 

               =

               =

 

 

=

Then Z.F=

 

       = -1

       =

But Z =

 

 

 

Example 2

Solve

Solution

1^st Expressing the equation in the form

 

–        

–        

Let Z =

 

 

 

But Z =

 

 

 

            =

            =

 

 

Then

 

 

 

          

But Z =

 

Example 3

Solve

Solution

Expressing the equation in the form

 

 

Then dividing by

 

let  Z =

 

Multiply (ii) by -2 gives

 

 

But Z =

 

 

 

 

 

 

Then,   

            

         

         

 But Z =

 

Exercise

Solve the following first order D.es

1.    

2.