Question

If a curve $y = f\left( x \right)$ passes through the point $\left( {1, - 1} \right)$and satisfies the differential equation, \[y\left( {1 + xy} \right)dx = xdy\], then $f\left( {\dfrac{{ - 1}}{2}} \right)$ is equal to:
(A) $\dfrac{{ - 2}}{5}$
(B) $\dfrac{{ - 4}}{5}$
(C) $\dfrac{2}{5}$
(D) $\dfrac{4}{5}$

Answer 1

Hint:A first order differential equation is Homogeneous which can be written as: $\dfrac{{dy}}{{dx}} = F\left( {\dfrac{y}{x}} \right)$The given differential equation is \[y\left( {1 + xy} \right)dx = xdy\] $ \Rightarrow \dfrac{y}{x}\left( {1 + xy} \right) = \dfrac{{dy}}{{dx}}$ , which is of the form $\dfrac{{dy}}{{dx}} = F\left( {\dfrac{y}{x}} \right)$. It means the given differential equation is a homogeneous differential equation. The Homogeneous differential equations can be solved by using $y = vx$ and $\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$.Firstly solve the given differential equation and then integrate it. Also find the value of constant of integration with the help of given boundary condition i.e., $\left( {1, - 1} \right)$ and get the value of $f\left( {\dfrac{{ - 1}}{2}} \right)$.


Complete step-by-step answer:
Given differential equation is \[y\left( {1 + xy} \right)dx = xdy\]
On rearranging,
$ \Rightarrow \dfrac{y}{x}\left( {1 + xy} \right) = \dfrac{{dy}}{{dx}}$ …. (1)
This differential equation is of the form $\dfrac{{dy}}{{dx}} = F\left( {\dfrac{y}{x}} \right)$. Hence it is a homogeneous differential equation, so it can be solved by assuming $y = vx$$ \Rightarrow \dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$.
Now equation (1) becomes,
$v\left( {1 + v{x^2}} \right) = v + x\dfrac{{dv}}{{dx}}$
On solving further, we get-
\[v + {v^2}{x^2} = v + x\dfrac{{dv}}{{dx}}\]
${v^2}{x^2} = x\dfrac{{dv}}{{dx}}$
$ \Rightarrow {v^2}x = \dfrac{{dv}}{{dx}}$
$ \Rightarrow xdx = \dfrac{1}{{{v^2}}}dv$
On taking integration both sides,
$ \Rightarrow \int {xdx} = \int {\dfrac{1}{{{v^2}}}dv} $
By using the formula $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $, we get-
$ \Rightarrow \dfrac{{{x^{1 + 1}}}}{{1 + 1}} = \dfrac{{{v^{ - 2 + 1}}}}{{ - 2 + 1}} + C$
$ \Rightarrow $$\dfrac{{{x^2}}}{2} = \dfrac{{ - 1}}{v} + C$ ….(2)
Where $C$ is the constant of integration.
We have,
$y = vx \Rightarrow v = \dfrac{y}{x}$
Put the value of v in equation (2),
$ \Rightarrow $$\dfrac{{{x^2}}}{2} = \dfrac{{ - x}}{y} + C$ ….(3)
The curve passes through $\left( {1, - 1} \right)$.So it satisfies the above equation. Put $x = 1$ and $y = - 1$ in above equation (3) to find out the value of $C$,
$\dfrac{{{1^2}}}{2} = \dfrac{{ - 1}}{{ - 1}} + C$
$ \Rightarrow \dfrac{1}{2} = 1 + C$
$ \Rightarrow C = \dfrac{{ - 1}}{2}$
Put the value of $C$in equation (3),
$\dfrac{{{x^2}}}{2} = \dfrac{{ - x}}{y} - \dfrac{1}{2}$ ….. (4)
Now, we have to find $f\left( {\dfrac{{ - 1}}{2}} \right)$.
Therefore, substitute $x = \dfrac{{ - 1}}{2}$ in above equation (4),
$\dfrac{{{{\left( {\dfrac{{ - 1}}{2}} \right)}^2}}}{2} = \dfrac{{ - \left( {\dfrac{{ - 1}}{2}} \right)}}{y} - \dfrac{1}{2}$
$ \Rightarrow $$\dfrac{1}{8} = \dfrac{1}{{2y}} - \dfrac{1}{2}$
$ \Rightarrow $$\dfrac{1}{8} + \dfrac{1}{2} = \dfrac{1}{{2y}}$
$ \Rightarrow $ $\dfrac{{1 + 4}}{8}$$ = \dfrac{1}{{2y}}$
$ \Rightarrow $$\dfrac{5}{8} = \dfrac{1}{{2y}}$
$ \Rightarrow 10y = 8$
$ \Rightarrow y = \dfrac{4}{5}$

So, the correct answer is “Option D”.

Note:In this question , the curve $y = f\left( x \right)$ passes through the point $\left( {1, - 1} \right)$, so it must satisfy the given differential equation. So, we can Put $x = 1$ and $y = - 1$ in the equation to find out the value of the constant of integration $\left( C \right)$.Students should remember the integration formulas and definitions of homogeneous differential equations for solving these types of problems.