Question

solve y=px+a/p, where p=dy/dx​

Answer 1

Answer:

Given the differential equation,

y=px+p−p

2

..(1)

Now, differentiating both sides with respect to x we get,

p=p+(x+1−2p)

dx

dp

[ Since

dx

dy

=p]

or, (x+1−2p)

dx

dp

=0

or, x+1=2p and

dx

dp

=0.

or, p=

2

x+1

gives particular solution and

dx

dp

=0 leads to general solution.

Now putting p=

2

x+1

in (1) we get,

y=

2

(x+1)

2

−

4

(x+1)

2

or, y=

4

(x+1)

2

or, (x+1)

2

=4y.

Answer 2

Answer:

The given equation is y = post exchange + (a/p), ......(1)

Which is in Clairaut's kind. Thus replacement p by c in (1) the answer is

y = cx + (a/c) or c2x - yc + a = zero. ........(2)

Now, c - discriminant relation of (2) is

B^2 - 4AC = 0; i.e.,

(-y)2 - 4xa = zero or y2 = 4ax .......(3)

Now, y2 = 4ax offers 2y(dy/dx) = 4a or p = 2a/y. Putt this price of p in (1), we have a tendency to get y = (2ax)/y + (y/2) or y2 = 4ax that is true by (3) satisfies (1) thus y2 = 4ax is that the needed singular resolution.

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