Complete Integral Of Partial Differential Equations

Complete Solution Or Complete Integral Of Partial Differential Equations:

 Definition : 

            A solution of partial differential equation is said to be a complete solution or complete integral if it contains as many arbitrary constants as there are independent variables .

Definition Of General Solution Or Integral :

       A general solution or integral of a partial differential equation is a relation involving arbitrary functions which provides a solution to that equation .

Linear Partial Differential Equation Of The First Order :

       A partial differential equation of first order is said to be linear if it is of the first degree in P and Q otherwise it is            non linear .

Example :

  (i) Linear Partial Differential                         Equation :

             x²p + y²q = zq 

  (ii) Non _ Linear Partial Differential               Equation :

            xpq + yq² = 1 

Rule For The General Solution Of Linear Differential Equation : 

              The general solution of the linear partial differential equation

           Pp + Qq = R .......(1) is F(u,v) = 0 ...(2)

  where F is an arbitary function and 

u(x,y,z) = c₁ and v(x,y,z) = c₂ from a solution of the equations

               dx/P = dy/Q = dz/R .....(3)

Equation (1) is Known as Lagrange's Equation .

Example :

       Find the general integrals of the following partial differential equation.

          (x²+y²)p + 2xyq = (x+y) z

Solution : 

         Given 

         (x²+y²) p + 2xyq = (x+y)z

 The subsidiary equations are 

   dx/(x²+y²) = dy/2xy = dz/(x+y)z

      = [zdx + zdy - (x+y)dz]/[z(x²+y²)+2xyz-                                                          (x+y)²z]

      = z(dx+dy) - (x+y)dz / 0

⇒z(dx+dy) - (x+y) dz = 0

⇒z d(x+y) - (x+y)dz = 0

⇒zd(x+y) - (x+y)dz / z² = 0

⇒d(x+y / z) = 0 

⇒x+y / z = c₁ 

From first two ratios we get 

      dx/x²+y² = 2dy/4xy

  ⇒2(x²+y²) dy = 4xy dx

  ⇒2(x²+y²) dy - 4xy dx = 0

  ⇒2x² dy - 2y² dy + 4y² dy -4xy dx = 0

  ⇒2(x²-y²) dy - 2y(2xdx - 2ydy) = 0

  ⇒[(x²-y²) 2 dy - 2y d(x²-y²)]/(x²-y²)²

                                 = 0/(x²-y²)

  ⇒d(2y / x²-y²) = 0

Integrating , we get 

   2y/x²-y² = c₂ 

∴ The solution is x+y / z = Φ(2y / x²-y²)

  Which is required solution .

              

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