Special Types Of First _ Order Equations (Part _2)

Special  Types  Of First Order  Differential  Equations (Part_2) :

Standard Form _2 :

                Equations not involving the independent variables i.e. equation of the form 

                  f(z,p,q) = 0          ...........(1)

       The auxiliary equations for equation(1)  are 

 dx/[∂f/∂p] = dy/[∂f/∂q]= dz/[p∂f/∂p+q∂f/∂q]

                                 = dp/-p∂f/∂z = dq/-q∂f/∂z

      ∴  dp/p = dq/q

Integrating   q = ap              ............(2)

  Where a is a constant .

 Substituting in equation(1) , we have 

         f(z,p,ap) = 0

 Now dz = pdx + qdy            ...............(3)

                = pdx + apdy

                = pd(x+ay)

Let us take X = x + ay

So that  dz/dX = p and adz/dX = q .

Hence the given equation becomes

 f(z,dz/dX,adz/dX) = 0        ...................(4)

          The equation (4) is an ordinary differential equation of first order and can easily be  solved . The complete integral is thus obtained by replacing X by (x + ay) .

Examples Related To This Type Of Standard Form :

Example_1 :

        Find the complete integral of the equation 

                  p³ + q³ = 27 z 

Solution :

          Given that

                         p³ + q³ = 27 z ...........(1)

       Let us put p = dz/dX and q = a dz/dX in equation (1) ,

   where X = x + ay .

∴ (1+a³)(dz / dX)³ = 27 z

 ⇒ dz / dX = 3 z^1/3 / (1+a³)^1/3

 ⇒ z^-1/3 dz = 3dX / (1+a³)^1/3

Integrating , 3/2 z^2/3 (1+a³)^1/3 

                                        = 3X + constant

   ⇒z^2/3 (1+a³)^1/3 = 2(X+c)

   ⇒ z² (1+a³) = 8(x + ay + c)³

 Which is the required solution .

Example_2 :

             Find the complete integral of

                      p²x² = z(z- qy)

Solution :

          Given that ,

                    p²x² = z(z - qy) ...............(1)

   Let us put 

                   u = log x 

and            v = log y ,

Then       p = ∂z/∂x

                   = ∂z/∂u . ∂u/∂x = 1/x ∂z/∂u

and          q = ∂z/∂y 

                    = ∂z/∂v . ∂v/∂z = 1/y ∂z/∂v

Substituting in equation (1) , we get 

              (∂z/∂u)² = z(z- ∂z/∂v)

or             P² = z(z - Q) [standard form 2]

                                                  ...........(2)

where      P = ∂z/∂u and Q = ∂z/∂v

Now replacing P by dz/dX and Q by adz/dX , 

     where X = u + av , we get 

        (dz/dX)² = z² - za dz/dX

⇒     (dz/dX)² + az dz/dX - z² = 0

⇒    dz/dX = -az ± √(a²z²+4z²) / 2

                   = z/2 [- a ± √(a²+4)

⇒2/z dz = [-a + √(a²+4 ) dX

                                     [taking +ve sign].

Integrating , 

       2 log z = [-a + √(a²+4 )]X + 2 log c

                  = [-a + √(a²+4 )](u+av) + 2 log c

                  = [-a + √(a²+4 )]log x 

                    + a[-a + √(a²+4 )]log y + 2 log c

             = 2 A log x + 2(1-A²) log y + 2 log c 

where     2A = - a + √(a² + 4 )

∴               z = cxᴬy^(1 - A²)

    Which is the required solution .

         These are some problems which better explain the standard form .

               "Tough Times Don't Last"

                    Tough  People Do!