Let x = x(y) be the solution of the differential equation: y = ( x - y dxdy) ( xy), y 0 and x(1) = 2. Then (x(2)) is equal to:

Given: y = ( x - y dxdy) ( xy), y 0, x(1) = 2 Find: (x(2)) Use the substitution v = xy so that x = vy and dxdy = v + y dvdy. Substitute into the differential equation: y = ( vy - y (v + y dvdy)) v which gives y = ( -y^2 dvdy) v. Since y 0, divide by y: 1 = -y dvdy v. Hence dvdy = - 1y v. Separate the variables: v dv = - dyy. Integrating, v dv = - dyy so - v = - y + C. Therefore, v = y + C1. Now apply the initial condition x(1) = 2. Then v(1) = x(1)1 = 2. So ( 2) = 1 + C1 which gives 0 = 0 + C1 C1 = 0. Hence ( xy) = y. At y = 2, ( x(2)2) = 2. Using the identity (2) = 2 ^2 - 1, with = x(2)2, we get (x(2)) = 2 ^2 ( x(2)2) - 1 = 2( 2)^2 - 1. Therefore, (x(2)) = 2( 2)^2 - 1, so the correct option is B. A tempting but incorrect step is to divide the original equation by ( xy) and conclude y = x - y dxdy. This is wrong because the equation is y = ( x - y dxdy) ( xy), so dividing by ( xy) gives y ( xy) = x - y dxdy, not y = x - y dxdy. The quantity xy appears inside the sine function, so the natural substitution is v = xy. That converts the equation into a separable differential equation in v and y, which leads directly to the required expression for (x(2)).

Let x = x(y) be the solution of the differential equation: y = (… — Solution | JEE Mathematics

MCQMediumJEE 2025Homogeneous Differential Equations

Let $x = x(y)$ be the solution of the differential equation:

y = \left( x - y \frac{dx}{dy} \right) \sin\left( \frac{x}{y} \right), \, y > 0 \, \text{and} \, x(1) = \frac{\pi}{2}.

Then $\cos(x(2))$ is equal to:

A
$1 - 2(\log 2)^2$
B
$2(\log 2)^2 - 1$
C
$2(\log 2) - 1$
D
$1 - 2(\log 2)$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

y = \left( x - y \frac{dx}{dy} \right) \sin\left( \frac{x}{y} \right), \quad y > 0, \quad x(1) = \frac{\pi}{2}

Find: $\cos(x(2))$

Use the substitution

v = \frac{x}{y}

so that

x = vy \quad \text{and} \quad \frac{dx}{dy} = v + y\frac{dv}{dy}.

Substitute into the differential equation:

y = \left( vy - y\left(v + y\frac{dv}{dy}\right) \right) \sin v

which gives

y = \left( -y^2\frac{dv}{dy} \right) \sin v.

Since $y > 0$ , divide by $y$ :

1 = -y\frac{dv}{dy} \sin v.

Hence

\frac{dv}{dy} = -\frac{1}{y\sin v}.

Separate the variables:

\sin v \, dv = -\frac{dy}{y}.

Integrating,

\int \sin v \, dv = -\int \frac{dy}{y}

-\cos v = -\log y + C.

Therefore,

\cos v = \log y + C_1.

Now apply the initial condition $x(1) = \frac{\pi}{2}$ . Then

v(1) = \frac{x(1)}{1} = \frac{\pi}{2}.

\cos\left(\frac{\pi}{2}\right) = \log 1 + C_1

which gives

0 = 0 + C_1 \implies C_1 = 0.

Hence

\cos\left(\frac{x}{y}\right) = \log y.

At $y = 2$ ,

\cos\left(\frac{x(2)}{2}\right) = \log 2.

Using the identity

\cos(2\theta) = 2\cos^2\theta - 1,

with $\theta = \frac{x(2)}{2}$ , we get

\cos(x(2)) = 2\cos^2\left(\frac{x(2)}{2}\right) - 1 = 2(\log 2)^2 - 1.

Therefore, $\cos(x(2)) = 2(\log 2)^2 - 1$ , so the correct option is B.

Why the direct rearrangement fails

A tempting but incorrect step is to divide the original equation by $\sin\left(\frac{x}{y}\right)$ and conclude

y = x - y\frac{dx}{dy}.

This is wrong because the equation is

y = \left( x - y\frac{dx}{dy} \right) \sin\left( \frac{x}{y} \right),

so dividing by $\sin\left(\frac{x}{y}\right)$ gives

\frac{y}{\sin\left(\frac{x}{y}\right)} = x - y\frac{dx}{dy},

not $y = x - y\frac{dx}{dy}$ .

The quantity $\frac{x}{y}$ appears inside the sine function, so the natural substitution is

v = \frac{x}{y}.

That converts the equation into a separable differential equation in $v$ and $y$ , which leads directly to the required expression for $\cos(x(2))$ .

Common mistakes

Dividing by $\sin\left(\frac{x}{y}\right)$ incorrectly. From
$y = \left( x - y\frac{dx}{dy} \right) \sin\left( \frac{x}{y} \right)$
you must get
$\frac{y}{\sin\left(\frac{x}{y}\right)} = x - y\frac{dx}{dy},$
not $y = x - y\frac{dx}{dy}$ . Keep the sine factor in the denominator.
Not recognizing the homogeneous form. Because the trigonometric term contains $\frac{x}{y}$ , the correct substitution is $v = \frac{x}{y}$ . Treating it as a linear or directly separable equation in $x$ and $y$ leads to an incorrect integral.
Differentiating $x = vy$ incorrectly. The product rule gives
$\frac{dx}{dy} = v + y\frac{dv}{dy}.$
If the extra term $v$ is missed, the transformed differential equation becomes wrong.
Using the wrong trigonometric identity at the end. From
$\cos\left(\frac{x(2)}{2}\right) = \log 2,$
you must use
$\cos(x(2)) = 2\cos^2\left(\frac{x(2)}{2}\right) - 1.$
Do not replace $\cos(x(2))$ directly by $\log 2$ .

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Answer

Step-by-step solution

Common mistakes

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