Question

b) Find singular solution of y = px + p2. ​

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

here is your answer please mark as me brainlist please