Special Types Of First_Order Equations Part_3

Special Types Of First_Order Equations Part_3 :

Standard Form _ 3 :

             Separable equations i.e. equations of the type

               f₁(x,p) = f₂(y,q)      ............(1)

   The Charpit's auxiliary equation for equation(1) are

       dx/[∂f₁/∂p] = dy/-[∂f₂/∂q] 

                      = dz/[p∂f₁/∂p - q∂f₂/∂q]

                     = dp/-[∂f₁/∂x] = dq/-[∂f₂/∂y]

∴        dx/[∂f₁/∂p] = dp/-[∂f₁/∂x]

⇒ ∂f₁/∂x dx +∂f₁/∂p dp = 0

⇒df₁(x,p) = 0

Integrating , f₁(x,p) = a (say) .........(2)

  Hence , equation(1) gives 

                  f₂(y,q) = a       ..........(3)

Solving equations (2) and (3) for p and q , we obtain 

           p = g₁(a,x) and q = g₂(a,y)

Now         dz = pdx + qdy 

gives      dz = g₁(a,x) dx + g₂(a,x) dy 

Integrating , z = ∫g₁(a,x) dx +∫g₂(a,y) dy +c

                                            ...............(4)

Where c is an arbitrary constant .

Examples Related To Standard Form :

 Example _ 1 :

        Find the complete integral of 

                        q = xyp²

Solution : 

          Given that 

                      q = xyp² 

  The given equation can be written as 

                  xp² = q/y

Put       xp² = q/y = a , so that 

              p = √(a/x ) and q = ay 

∴       dz = pdx + qdy 

              = √(a/x) dx + aydy 

Integrating ,

             ∫dz = ∫[√(a/x )dx + aydy]

     ⇒      z = 2√(ax) + ay²/2 + c

    ⇒       (2z - ay² - 2c)² = 16ax , 

Which is the required complete integral .

Example _ 2 :

     Find the complete integral of the equation 

              z(p² - q²) = x - y 

Solution : 

            Given that 

                         z(p² - q²) = x - y ........(1)

  Let us put  Z = 2/3 z^3/2 , then 

            P = ∂Z/∂x = ∂Z/∂z .∂z/∂x = √(z)p

           Q = ∂Z/∂y = ∂Z/∂z. ∂z/∂y = √(z)q

Substituting in equation  (1) , we have 

    P² - Q² = x - y 

⇒P² - x = Q² - y = a 

∴    P = √(x+a) and Q = √(y+a)

Hence , dZ = Pdx + Qdy 

becomes 

          dZ = √(x+a)dx + √(y+a)dy

Integrating , 

   Z = 2/3 (x+a)^3/2 + 2/3 (y+a)^3/2 + const.

⇒ z^3/2 = (x+a)^3/2 + (y+a)^3/2 + c 

Which is the required solution .

 

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