Question

y=x(p+✓1+p^2) solvable for y
​

Answer 1

Step-by-step explanation:

The differential equation

\sf{y = xp + \sqrt{1 + {p}^{2} } }y=xp+

1+p

2

EVALUATION

Here the given differential equation is

\sf{y = xp + \sqrt{1 + {p}^{2} } }y=xp+

1+p

2

This equation is of the form

\sf{y = xp + f(p) }y=xp+f(p)

Which is Clairaut's equation

Now for Clairaut's equation the general solution is obtained replacing p by c

Hence the required solution is

\sf{y = cx+ \sqrt{1 + {c}^{2} } }y=cx+

1+c

2

Where C is constant

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1. M+N(dy/dx)=0 where M and N are function of

(A) x only

(B) y only

(C) constant

(D) all of these

2. This type of equation is of the form dy/dx=f1(x,y)/f2(x,y)

(A) variable seprable

(B) homogeneous

(C) exact