Question

The solution of $v=u$ $\frac{dv}{du}+{\left(\frac{dv}{du}\right)}^2$ , where $u=y$ and $v=xy$ are-

• A. $y=0$
• B. $y=-4x$
• C. $xy=cy+c^2$
• D. $x^{2}y=cy+c^2$

Answer 1

Answer: (A), (B), (C)

$y=0$
$y=-4x$
$xy=cy+c^2$

Let,

$\frac{dv}{du}=q$.

Then the given differential equation takes the form

$v=uq+q^2$. [Which is Clairut's equation].

So the general solution will be

$v=uc+c^2$ [$c$ being integrating constant.]

or, $xy=cy+c^2$ [Using values of $u$ and $v$].......(1)

Now the singular solution of the differential equation be the envelope of the family of straight lines represented by (1) and it is given by the equation

$y^2+4xy=0$ [(discriminant of (1)$)=0$ is the envelope.]

or, $y(y+4x)=0$.

$y=0$ or $y=-4x$ represent the singular solution of the given differential equation.

So options (A), (B) and (C) are correct.