Special Types Of First_Order Equations Part_1

Special Types Of First_Order Equations  Part_1 :

Standard Form 1 :

                Equations involving only p and q ,The equations of this form are 

               f(p,q) = 0 .......(1)

Charpit's equations take the forms 

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

                     = dp/0 = dq/0

∴       dp = 0

⇒p = constant = a (say) .......(2)

Substituting in equation(1) , we get

                f(a,q) = 0 ......(3)

which gives q = φ(a) = constant 

Hence dz = pdx + qdy 

                  = a dx + φ(a) dy 

Which on integration yields

           z = ax + φ(a)y + c .......(4)

Example _ 1 :

         Find the complete integral of the following equation .

                          pq = 1

Solution : 

                  pq = 1

        F(p,q) = pq - 1 ..........(1)

∴       ∂F/∂p = q , ∂F/∂q = p

Now , 

       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)

 ⇒ dx/ q = dy/p = dz / pq + pq = dp/0 = dq/0

 ⇒dp = 0 ⇒ p = a (say)

From equation (1) we get ,

                                      aq - 1 = 0

                               ⇒    aq = 1

                               ⇒     q = 1/a

   ∴       dz = pdx + qdy 

       ⇒ dz  = a dx + 1/a dy 

So  z = ax + 1/a y + c 

Which is the required solution.

Example _ 2 :

           Find the complete integral of the equation 

                            x²p²+ y²q² = z²

  Solution :

             Given that 

                             x²p² + y²q² = z² .........(1)

     The equation can be written as

         (x/z ∂z/∂x)² + (y/z ∂z/∂y)² = 1...........(2)

 Let us put dz /z = dZ ⇒ Z = logz

                    dx/x = dX ⇒ X = logx

                    dy/y = dY ⇒ Y = logy

∴    Equation (1) becomes 

             (∂Z/∂X)² + (∂Z/∂Y)² = 1

or               P²  +  Q²  = 1,

 Where P = ∂Z/∂X   and Q = ∂Z/∂Y

The complete integral is 

      Z = aX + √(1-a² )Y +c 

⇒ log z = a log x + √(1-a² )log y +log k ,

                              where c = log k

⇒ z = kxᵃ y^√(1-a² )

Which is the required solution .

  This is a different type of problem of the same standard form . So it has a different type of solution .

          

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