Question

The differential equation $2xy\dfrac{{dy}}{{dx}} = {x^2} + {y^2} + 1$ determines.
$\left( a \right)$ A family of circles with center on x-axis
$\left( b \right)$ A family of circles with center on y-axis
$\left( c \right)$ A family of rectangular hyperbola with center on x-axis
$\left( d \right)$ A family of rectangular hyperbola with center on y-axis.

Answer 1

In this particular question use the concept that first convert the given differential equation into linear differential equation by using substitution method in this equation first divide by x throughout then substitute ${y^2} = v$, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given differential equation
$2xy\dfrac{{dy}}{{dx}} = {x^2} + {y^2} + 1$
Divide the above equation by x throughout we have,
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = x + \dfrac{{{y^2}}}{x} + \dfrac{1}{x}$.............. (1)
Now let, ${y^2} = v$................ (2)
Differentiate equation (2) w.r.t, x we have,
$\dfrac{d}{{dx}}{y^2} = \dfrac{d}{{dx}}v$
Now as we know that $\dfrac{d}{{dx}}{y^n} = ny\dfrac{d}{{dx}}{y^{n - 1}}$, so use this property in the above equation we have,
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{dx}}$................ (3)
Now substitutes the values from equation (2) and (3) in equation (1) we have,
$\Rightarrow \dfrac{{dv}}{{dx}} = x + \dfrac{v}{x} + \dfrac{1}{x}$
$\Rightarrow \dfrac{{dv}}{{dx}} - \dfrac{v}{x} = x + \dfrac{1}{x}$.............. (4)
Now the solution of standard differential equation $\dfrac{{dv}}{{dx}} + px = q$, where q is free from v is
$v.\left( {I.F} \right) = \int {q\left( {I.F} \right)dx}$................... (5), where I.F = integration factor, $I.F = {e^{\int {pdx} }}$
So on comparing with equation (4), $p = \dfrac{{ - 1}}{x},q = x + \dfrac{1}{x}$.
Now first find out integrating factor
$\Rightarrow I.F = {e^{\int {\dfrac{{ - 1}}{x}dx} }} = {e^{ - \int {\dfrac{1}{x}dx} }}$
Now as we know that integration of $\int {\dfrac{1}{x}dx} = \ln x + C$, where C is some integration constant so we have,
$\Rightarrow I.F = {e^{ - \int {\dfrac{1}{x}dx} }} = {e^{ - \ln x}}$
Now as we know that $- \ln a = \ln \dfrac{1}{a},{e^{\ln \dfrac{1}{a}}} = \dfrac{1}{a}$, so use this in the above equation we have,
$\Rightarrow I.F = {e^{ - \ln x}} = {e^{\ln \dfrac{1}{x}}} = \dfrac{1}{x}$
Now from equation (5) we have,
$\Rightarrow v.\left( {\dfrac{1}{x}} \right) = \int {\left( {x + \dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)dx}$
Now simplifying we have,
$\Rightarrow v.\left( {\dfrac{1}{x}} \right) = \int {\left( {1 + \dfrac{1}{{{x^2}}}} \right)dx}$
Now integrate it we have,
$\Rightarrow v.\left( {\dfrac{1}{x}} \right) = x - \dfrac{1}{x} + C$
Now substitute the value of v we have,
$\Rightarrow {y^2}.\left( {\dfrac{1}{x}} \right) = x - \dfrac{1}{x} + C$
$\Rightarrow {y^2} = {x^2} - 1 + Cx$
Now add and subtract by half of square of coefficient of x to make the complete square in x so we have,
$\Rightarrow {y^2} = {x^2} - 1 + Cx + \dfrac{{{C^2}}}{4} - \dfrac{{{C^2}}}{4}$
$\Rightarrow {y^2} = {\left( {x + \dfrac{C}{2}} \right)^2} - 1 - \dfrac{{{C^2}}}{4}$
$\Rightarrow {\left( {x + \dfrac{C}{2}} \right)^2} - {y^2} = 1 + \dfrac{{{C^2}}}{4}$
So this is the required solution of the given differential equation, which represents the equation of hyperbola with center $\left( {\dfrac{{ - C}}{2},0} \right)$ on its center lying on x-axis.
So this is the required answer.
Hence option (c) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that the solution of the linear differential equation $\dfrac{{dv}}{{dx}} + px = q$, where q is free from v is,$v.\left( {I.F} \right) = \int {q\left( {I.F} \right)dx}$, where I.F = $I.F = {e^{\int {pdx} }}$, so simply substitute the values of p and q in the above defined equations and simplify we will get the required answer.