Solved Example Problems & Exercise based on Type 3(b) f(x, p, q) = 0

Problems based on Type 3(b) f(x, p, q) = 0

Example 1.3b(25): Solve p = 2qx.to

Solution: Given: p = 2qx, this equation is of the form f (x, p, q) = 0.

equation (1) is the complete integral of the given equation.

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

Differentiating partially w.r.to b, we get 1 = 0.

Hence, there is no singular integral.

General integral can be found out in the usual way.

[We can solve this easily by using Lagrange's idea]

Example 1.3b(26): Solve q = px + p².

Solution: Given : q = px + p² …………(1)

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

Assume  q = a  (constant)

Then p² + px - a = 0

is the complete solution.

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

Singular integral does not exist

General integral can be found out in the usual way.

Example 1.3b(27) : Solve √p + √q = √x.

Solution: Given: √p + √q = √x

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

Assume q (1) ⇒ √p + √a √x

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

Example 1.3b(28): Find the complete integral of q = 2px

 Solution: Given : q = 2px

This equation is of the form f(x, p, q)

Let q = a Then P

[We can solve this easily by Lagrange's idea to get the general integral]

EXERCISE 1.3.b [Type 3 - Case (ii)] f(x, p, q) = 0

Find the complete integrals of the following:

Problems based on Type 3(c) f(x, p, q) = 0

Example 1.3b(29): Solve pq = y.

Solution: Given: pq = y…………..(1)

This equation is of the form f (v,p,q)

Assume p = a (constant)

Example 1.3b (30) : Solve q=py + p²

Solution: Given : q = py + p²

This equation of the form f(y,p,q) = 0

Assume p = a (constant)

Eliminate 'a' between (3) and (4), we get general solution.