If ( yx) = |x| + 2 is the solution of the differential equation x ( yx) dydx = y ( yx) + x, and y(1) = 3, then ^2 is equal to:

Given: x ( yx) dydx = y ( yx) + x and y(1)= 3. Find: ^2 when the solution is ( yx)= |x|+ 2. Use the substitution z= yx. Then y = zx and dydx = z + x dzdx Substitute into the differential equation: x z (z+x dzdx)=zx z + x Expanding, xz z + x^2 z dzdx=zx z + x So, x^2 z dzdx=x Hence, z dzdx= 1x Now integrate both sides: z dz = 1x dx This gives z = |x| + C Since z= yx, ( yx)= |x|+C Comparing with the given form, C= 2 Apply the initial condition y(1)= 3: ( 3)= |1|+ 2 So, 32=0+ 2 Therefore, = 3 and hence ^2=3 Therefore, the correct option is A. Given: ( yx)= |x|+ 2 and y(1)= 3. Find: ^2. Substitute x=1 and y= 3 directly into the given solution form: ( 3)= |1|+ 2 Now evaluate the terms: ( 3)= 32, |1|=0 So, 32= 2 which gives = 3 Hence, ^2=3 Therefore, the correct option is A.

If ( yx) = |x| + 2 is the solution of the differential equation x… — Solution | JEE Mathematics

MCQMediumJEE 2024Homogeneous Differential Equations

If $\sin\left(\frac{y}{x}\right) = \ln|x| + \frac{\alpha}{2}$ is the solution of the differential equation $x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x$ , and $y(1) = \frac{\pi}{3}$ , then $\alpha^2$ is equal to:

A
$3$
B
$12$
C
$4$
D
$9$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x$ and $y(1)=\frac{\pi}{3}$ .

Find: $\alpha^2$ when the solution is $\sin\left(\frac{y}{x}\right)=\ln|x|+\frac{\alpha}{2}$ .

Use the substitution $z=\frac{y}{x}$ . Then

y = zx

and

\frac{dy}{dx} = z + x\frac{dz}{dx}

Substitute into the differential equation:

x\cos z\left(z+x\frac{dz}{dx}\right)=zx\cos z + x

Expanding,

xz\cos z + x^2\cos z\frac{dz}{dx}=zx\cos z + x

So,

x^2\cos z\frac{dz}{dx}=x

Hence,

\cos z\frac{dz}{dx}=\frac{1}{x}

Now integrate both sides:

\int \cos z \, dz = \int \frac{1}{x} \, dx

This gives

\sin z = \ln|x| + C

Since $z=\frac{y}{x}$ ,

\sin\left(\frac{y}{x}\right)=\ln|x|+C

Comparing with the given form,

C=\frac{\alpha}{2}

Apply the initial condition $y(1)=\frac{\pi}{3}$ :

\sin\left(\frac{\pi}{3}\right)=\ln|1|+\frac{\alpha}{2}

So,

\frac{\sqrt{3}}{2}=0+\frac{\alpha}{2}

Therefore,

\alpha=\sqrt{3}

and hence

\alpha^2=3

Therefore, the correct option is A.

Using the given solution form directly

Given: $\sin\left(\frac{y}{x}\right)=\ln|x|+\frac{\alpha}{2}$ and $y(1)=\frac{\pi}{3}$ .

Find: $\alpha^2$ .

Substitute $x=1$ and $y=\frac{\pi}{3}$ directly into the given solution form:

\sin\left(\frac{\pi}{3}\right)=\ln|1|+\frac{\alpha}{2}

Now evaluate the terms:

\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}, \qquad \ln|1|=0

So,

\frac{\sqrt{3}}{2}=\frac{\alpha}{2}

which gives

\alpha=\sqrt{3}

Hence,

\alpha^2=3

Therefore, the correct option is A.

Common mistakes

Taking the substitution incorrectly as $y=z$ instead of $z=\frac{y}{x}$ . This breaks the homogeneous structure of the differential equation. Use $z=\frac{y}{x}$ so that $y=zx$ and differentiate correctly.
Differentiating $y=zx$ as $\frac{dy}{dx}=\frac{dz}{dx}$ . This is wrong because both $z$ and $x$ vary. Apply the product rule: $\frac{dy}{dx}=z+x\frac{dz}{dx}$ .
Using the initial condition incorrectly by substituting $y(1)=\frac{\sqrt{3}}{2}$ instead of $y(1)=\frac{\pi}{3}$ . The value $\frac{\sqrt{3}}{2}$ is $\sin\left(\frac{\pi}{3}\right)$ , not $y(1)$ itself. First substitute $y=\frac{\pi}{3}$ , then evaluate the sine.

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Answer

Step-by-step solution

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