Non_ Linear Partial Differential Equation Of First Order

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Non_ Linear Partial Differential Equation Of First Order :


         Let us consider a differential equation of the form F(x,y,p,q) = 0 in which the function F is not necessarily linear in p and q .

 Singular solution / Singular Integral :


              The equation of the envelop of the surface represented by the complete integral of a partial differential equation is called its singular solution or singular integral .

      The envelop of the surface F(x,y,a,b)=0  is obtained by eliminating a and b from the equations F = 0 , ∂F/∂a = 0,
 ∂F/∂b = 0 

Charpit's Method :


         Let us consider the equation f(x,y,p,q) = 0 then Charpit's subsidiary equations are given by 

 dx/ [∂f/∂p] = dy/ [∂f/∂q] 

                    = dz/[p(∂f/∂p) + q(∂f/∂q)]
   
                    = dp/-(∂f/∂x + p∂f/∂z)

                    = dq / -(∂f/∂y + q∂f/∂z)

Application Of Charpit's Method :


Example :


            Find the complete integrals of the following equation :

             px + qy = pq

Solution :


            Given that 

                      px + qy = pq ......(1)

  Let f(x,y,z,p,q) = px + qy - pq

   ∂f/∂x = p , ∂f/∂y = q ,∂f/∂z = 0, ∂f/∂p = x-q

    ∂f/∂q = y-p

  According to Charpit's method , corresponding subsidiary equations are 
given by 

       dx/[∂f/∂p] = dy/[∂f/∂q] 

                         = dz/[p(∂z/∂q) + q(∂f/∂q)]

                         = dp/-(∂f/∂x + p ∂f/∂z)

                         = dq/-(∂f/∂y + q∂f/∂z)

⇒dx/ x-q = dy/y-q = dz/p(x-q)+q(y-p)

                                 = dp/-(p-0) 

                                 = dq / -(q-0)

⇒dx/x-q = dy/y-p = dz/px+qy = dp/-p=dq/-q

From last two ratios we get 

⇒dp/-p = dq/-q ⇒dp/p = dq/q

Integrating both sides we get 

    ⇒∫dp/p = ∫dq/q

    ⇒log p = log q + c

    ⇒log p - log q = log a , where c = log a

    ⇒log(p/q) = log a

    ⇒p/q = a

     ⇒q = p/a ........(2)

Putting the value of q in the given equation(1) , we get 

    px + p/a y = p. p/a 

⇒ p[x+y/a] = p. p/a 

⇒x + y/a = p/a

⇒ax + y /a = p/a 

⇒ax + y = p

Putting the value of p in equation (2) we get 

        q = ax+y / a = x + y/a

Hence 

             dz = pdx + qdy 

      ⇒dz = (ax+y ) dx + (x+ y/a) dy 

      ⇒dz = (ax + y) dx + (ax+y /a) dy

      ⇒dz = 1/a [ a(ax+y)dx + (ax+y) dy]

      ⇒a dz = a(ax+y)dx + (ax+y)dy

      ⇒a dz = 2[a(ax+y)dx + (ax+y)dy] / 2

      ⇒a dz = d[(ax+y)² / 2]

Integrating , we get 

  az = (ax+y)²/2 + b (Ans)

   Which is required solution .

   

     " Don't Change The Thing , Change  "      The Way You Look At The Thing                

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