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MCQMediumJEE 2024Separation of Variables

JEE Mathematics 2024 Question with Solution

If y=y(x)y = y(x) is the solution curve of the differential equation (x2−4) dy−(y2−3y) dx=0(x^2 - 4) \, dy - (y^2 - 3y) \, dx = 0, x>2x > 2, y(4)=32y(4) = \frac{3}{2}, and the slope of the curve is never zero, then y(10)y(10) equals:

  • A

    31+81/4\frac{3}{1 + 8^{1/4}}

  • B

    31+22\frac{3}{1 + 2\sqrt{2}}

  • C

    31−22\frac{3}{1 - 2\sqrt{2}}

  • D

    31−81/4\frac{3}{1 - 8^{1/4}}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: (x2−4) dy−(y2−3y) dx=0(x^2 - 4) \, dy - (y^2 - 3y) \, dx = 0, x>2x > 2, and y(4)=32y(4) = \frac{3}{2}.

Find: y(10)y(10).

Rewrite the differential equation as

dydx=y2−3yx2−4\frac{dy}{dx} = \frac{y^2 - 3y}{x^2 - 4}

Since it is separable,

dyy2−3y=dxx2−4\frac{dy}{y^2 - 3y} = \frac{dx}{x^2 - 4}

Using partial fractions,

1y(y−3)=13(1y−3−1y)\frac{1}{y(y-3)} = \frac{1}{3}\left(\frac{1}{y-3} - \frac{1}{y}\right)

So,

13(1y−3−1y)dy=dxx2−4\frac{1}{3}\left(\frac{1}{y-3} - \frac{1}{y}\right)dy = \frac{dx}{x^2 - 4}

Integrating both sides,

13(ln⁡∣y−3∣−ln⁡∣y∣)=14ln⁡∣x−2x+2∣+C\frac{1}{3}\left(\ln|y-3| - \ln|y|\right) = \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C

Hence,

13ln⁡∣y−3y∣=14ln⁡∣x−2x+2∣+C\frac{1}{3}\ln\left|\frac{y-3}{y}\right| = \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C

Now use x=4x=4 and y=32y=\frac{3}{2}:

13ln⁡∣32−332∣=14ln⁡∣4−24+2∣+C\frac{1}{3}\ln\left|\frac{\frac{3}{2}-3}{\frac{3}{2}}\right| = \frac{1}{4}\ln\left|\frac{4-2}{4+2}\right| + C

This gives

13ln⁡1=14ln⁡(13)+C\frac{1}{3}\ln 1 = \frac{1}{4}\ln\left(\frac{1}{3}\right) + C

So,

C=14ln⁡3C = \frac{1}{4}\ln 3

At x=10x=10,

13ln⁡∣y−3y∣=14ln⁡(10−210+2)+14ln⁡3\frac{1}{3}\ln\left|\frac{y-3}{y}\right| = \frac{1}{4}\ln\left(\frac{10-2}{10+2}\right) + \frac{1}{4}\ln 3

Therefore,

13ln⁡∣y−3y∣=14ln⁡(23)+14ln⁡3=14ln⁡2\frac{1}{3}\ln\left|\frac{y-3}{y}\right| = \frac{1}{4}\ln\left(\frac{2}{3}\right) + \frac{1}{4}\ln 3 = \frac{1}{4}\ln 2

Hence,

ln⁡∣y−3y∣=34ln⁡2=ln⁡(23/4)\ln\left|\frac{y-3}{y}\right| = \frac{3}{4}\ln 2 = \ln\left(2^{3/4}\right)

So,

∣y−3y∣=23/4=81/4\left|\frac{y-3}{y}\right| = 2^{3/4} = 8^{1/4}

From the condition that the slope is never zero, the relevant branch is 00

Using the sign of the solution branch

After integration, one reaches

∣y−3y∣=81/4\left|\frac{y-3}{y}\right| = 8^{1/4}

At first glance this leaves a sign ambiguity.

Since y(4)=32y(4)=\frac{3}{2}, initially 0<y<30<y<3. Then

y2−3y=y(y−3)<0y^2-3y = y(y-3) < 0

For x>2x>2,

x2−4>0x^2-4 > 0

Hence,

dydx=y(y−3)x2−4<0\frac{dy}{dx} = \frac{y(y-3)}{x^2-4} < 0

So the slope is never zero and the curve stays in the interval $$0

Therefore y−3<0y-3<0 while y>0y>0, so

y−3y<0\frac{y-3}{y} < 0

This fixes the sign as

y−3y=−81/4\frac{y-3}{y} = -8^{1/4}

which gives

y(10)=31+81/4y(10) = \frac{3}{1 + 8^{1/4}}

So the correct option is A.

Common mistakes

  • Taking the partial fraction decomposition of 1y(y−3)\frac{1}{y(y-3)} incorrectly. This changes the logarithmic relation after integration. Use 1y(y−3)=13(1y−3−1y)\frac{1}{y(y-3)} = \frac{1}{3}\left(\frac{1}{y-3} - \frac{1}{y}\right) carefully.

  • Ignoring the sign after removing the modulus from ∣y−3y∣\left|\frac{y-3}{y}\right|. This is wrong because the branch must be chosen using the condition y(4)=32y(4)=\frac{3}{2} and the nonzero slope condition. Since 00

  • Using the raw listed answer instead of the solution working. The solution's shows a disagreement between the option number in the answer key and the actual computed value. The solution working concludes y(10)=31+81/4y(10)=\frac{3}{1+8^{1/4}}, so the correct option is A.

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