Question

what is the solution of the differential equation y=px+logp (a) y=cx+logc (b) y=xy + logx (c) cx+ logp (d) all of these​

Answer 1

Answer:

If ylogx=x-y prove that

dy

dx

=

logx

(1+logx)2

Answer 2

Answer: y=cx+logc

Step-by-step explanation: The solution of the differential equation y=px+logp,

Here the given differential equation is y = px + logp

Which is of the form y = px + f(p)

So this is a Clairaut’s equation

Clairaut’s equation, in mathematics, is a differential equation of the form

y = x ($\frac{dy}{dx}$) + f($\frac{dy}{dx}$) where f($\frac{dy}{dx}$) is a function of $\frac{dy}{dx}$ only.

So the required solution is obtained, replacing p by c

Hence, the required solution is y = cx + log c

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