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NVAMediumJEE 2026Linear Differential Equations

JEE Mathematics 2026 Question with Solution

If the solution curve y=f(x)y = f(x) of the differential equation (x2−4)y′−2xy+2x(4−x2)2=0,x>2(x^2 - 4) y' - 2xy + 2x(4 - x^2)^2 = 0, x > 2, passes through the point (3,15)(3, 15), then the local maximum value of ff is _____

Answer

Correct answer:16

Step-by-step solution

Standard Method

Given: The differential equation is

(x2−4)y′−2xy+2x(4−x2)2=0(x^2 - 4) y' - 2xy + 2x(4 - x^2)^2 = 0

with x>2x > 2, and the solution curve passes through (3,15)(3, 15).

Find: The local maximum value of f(x)f(x).

Rewrite the equation in linear form:

y′−2xx2−4y=−2x(x2−4)y' - \frac{2x}{x^2-4}y = -2x(x^2-4)

The integrating factor is

I.F.=e∫−2xx2−4 dx=e−ln⁡(x2−4)=1x2−4I.F. = e^{\int \frac{-2x}{x^2-4} \, dx} = e^{-\ln(x^2-4)} = \frac{1}{x^2-4}

Multiplying through by the integrating factor,

y⋅1x2−4=∫−2x dx=−x2+Cy \cdot \frac{1}{x^2-4} = \int -2x \, dx = -x^2 + C

So,

y=(x2−4)(C−x2)y = (x^2-4)(C-x^2)

Using the point (3,15)(3, 15),

15=(9−4)(C−9)15 = (9-4)(C-9) 15=5(C−9)15 = 5(C-9) 3=C−93 = C-9 C=12C=12

Hence,

y=(x2−4)(12−x2)=−x4+16x2−48y = (x^2-4)(12-x^2) = -x^4 + 16x^2 - 48

For a local maximum,

y′=−4x3+32x=0y' = -4x^3 + 32x = 0 −4x(x2−8)=0-4x(x^2-8)=0

Since x>2x > 2, we take

x2=8x^2 = 8

Now substitute into y=(x2−4)(12−x2)y = (x^2-4)(12-x^2):

y=(8−4)(12−8)=4×4=16y = (8-4)(12-8) = 4 \times 4 = 16

Therefore, the local maximum value of ff is 1616.

The solution lists correct answer as 2020, but the extracted working gives the local maximum value as 1616.

Checking the extremum point

From

y=−x4+16x2−48y = -x^4 + 16x^2 - 48

we differentiate to locate critical points:

y′=−4x3+32xy' = -4x^3 + 32x

Setting y′=0y' = 0 gives

−4x(x2−8)=0-4x(x^2-8)=0

So the critical points are x=0x=0 and x2=8x^2=8. Because the question gives x>2x>2, only x=22x=2\sqrt{2} is valid.

Evaluating the function there:

y=(x2−4)(12−x2)y = (x^2-4)(12-x^2) y=(8−4)(12−8)y = (8-4)(12-8) y=16y = 16

Thus the relevant local maximum value is 1616.

Common mistakes

  • A common mistake is using the raw listed answer 2020 without checking the solution steps. The extracted working clearly gives y=(x2−4)(12−x2)y = (x^2-4)(12-x^2) and then the maximum value as 1616. Always trust the derived mathematics over a mismatched answer key.

  • Students may make an error while finding the integrating factor by missing the negative sign in −2xx2−4-\frac{2x}{x^2-4}. That changes the complementary form and gives the wrong family of curves. Compute the integrating factor carefully from the coefficient of yy in the linear form.

  • Another mistake is forgetting the domain condition x>2x>2 and taking x=0x=0 from y′=0y' = 0. That critical point is not allowed here. After solving for stationary points, always apply the given domain restriction before evaluating the function.

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