Solved Example Problems based on First order Partial Differential Equations (p.d.e)

III (b) Problems based on First order p.d.e

Problems based on Type 1 f (p, q) = 0

Example 1.3(b)(1): [Type 1]

Solve: Vp + √q = 1

Solution: Given: √p + √q……………………….(1)

This equation is of the form f (p, q) = 0………….(2)

Hence, the trial solution is z = ax + by + c………….(3)

To get the complete integral (solution)

Since, number of a.c. = number of I.V.

To find the singular integral :

Differentiating (4) p.w.r.to c we get,

0 = 1 [absurd]

Hence, there is no singular integral.

To get the general integral (solution)

Eliminating 'a' between (5) and (6), we get the general solution of the given p.d.e.

Example 1.3b(2) : [Type 1]

Solve p+q=pq

Solution: Given: p + q = pq ………..(1)

This equation is of the form f (p,q) = 0………..(2)

Hence, the trial solution is z = ax + by + c ………..(3)

To get the complete integral (solution)

Since, number of a.c. = number of I.V

To find the singular integral :

Differentiating (4) p.w.r.to c we get,

(0=1 [absurd]

Hence, there is no singular integral.

To get the general integral (solution)

Eliminating 'a' between (5) & (6), we get the general solution of the given p.d.e.

Example 1.3b(3) : [Type 1] Solve: pq = k

Solution: Given: pq = k …..(1)

This equation is of the form f (p, q) = 0…..(2)

Hence, the trial solution is z = ax + by + c…..(3)

To get the complete integral :

To find the singular integral

Differentiating (4) p.w.r.to c we get,

0 = 1 [absurd]

There is no singular integral.

To get the general integral (solution)

Put c = f (a) in (4), we get

Eliminating 'a' between (5) & (6), we get the general solution of the given p.d.e.

Example 1.3b(4): [Type 1] Solve: p² + q² = npq

Solution: Given:  p² + q2 = npq ... (1)

This equation is of the form f (p, q) = 0  ... (2)

Hence, the trial solution is z = ax + by + c... (3)

To get the complete integral (solution) :

Since, number of a.c. = number of I.V

To find the singular integral :

Differentiating (4) p.w.r.to c we get,

0 = 1 [absurd]

There is no singular integral.

To get the general integral (solution)

Put c = (a) in (4), we get

Eliminating ‘a' between (5) & (6), we get the general solution of the given p.d.e.

Note: Solve p² + q² − 4pq = 0

In the above problem put n = 4 we get the solution.

Example 1.3b(5): [Type 1] Find the complete integral of p-q=0

Solution: Given P-q = 0….(1)

This equation is of the form  f(p, q) = 0…..(2)

Hence, the trial solution is  z = ax + by + c…..(3)

To get the complete integral (solution)

Hence, the complete integral is z = ax + ay + c.

Hence, number of a.c.= number of I.V.

 [We can get general solution easily by using Lagrange's idea]