Question

singular solutions of p^3-4pxy+y^2​

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.

Step-by-step explanation:

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