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JEE Mathematics 2026 Question with Solution

Let y=y(x)y = y(x) be the solution of the differential equation sec⁡xdydx−2y=2+3sin⁡x,x∈(−π2,π2)\sec x \frac{dy}{dx} - 2y = 2 + 3\sin x, x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right). If y(0)=−74y(0) = -\frac{7}{4}, then y(π6)y\left(\frac{\pi}{6}\right) is equal to :

  • A

    −52-\frac{5}{2}

  • B

    −54-\frac{5}{4}

  • C

    −32−7-3\sqrt{2} - 7

  • D

    −33−7-3\sqrt{3} - 7

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: sec⁡xdydx−2y=2+3sin⁡x\sec x \frac{dy}{dx} - 2y = 2 + 3\sin x with x∈(−π2,π2)x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) and y(0)=−74y(0) = -\frac{7}{4}.

Find: y(π6)y\left(\frac{\pi}{6}\right).

This is a first-order linear differential equation. Divide by sec⁡x\sec x to write it in standard form:

dydx−2cos⁡x⋅y=(2+3sin⁡x)cos⁡x\frac{dy}{dx} - 2\cos x \cdot y = (2 + 3\sin x)\cos x

So the integrating factor is

IF=e∫−2cos⁡x dx=e−2sin⁡xIF = e^{\int -2\cos x \, dx} = e^{-2\sin x}

Multiplying the equation by the integrating factor,

y⋅e−2sin⁡x=∫(2+3sin⁡x)cos⁡x⋅e−2sin⁡x dxy \cdot e^{-2\sin x} = \int (2 + 3\sin x)\cos x \cdot e^{-2\sin x} \, dx

Let

u=−2sin⁡xu = -2\sin x

Then

du=−2cos⁡x dxdu = -2\cos x \, dx

and the integral becomes

∫(2−3u2)eu(−12) du=−12∫(2−3u2)eu du\int \left(2 - \frac{3u}{2}\right)e^u\left(-\frac{1}{2}\right) \, du = -\frac{1}{2}\int \left(2 - \frac{3u}{2}\right)e^u \, du

Now,

−12[(2−3u2)eu−∫(−32)eu du]-\frac{1}{2}\left[\left(2 - \frac{3u}{2}\right)e^u - \int \left(-\frac{3}{2}\right)e^u \, du\right] =−12eu(2−3u2+32)= -\frac{1}{2} e^u \left(2 - \frac{3u}{2} + \frac{3}{2}\right) =eu(3u4−74)= e^u \left(\frac{3u}{4} - \frac{7}{4}\right)

Substituting back u=−2sin⁡xu = -2\sin x,

ye−2sin⁡x=e−2sin⁡x(−32sin⁡x−74)+Cy e^{-2\sin x} = e^{-2\sin x}\left(-\frac{3}{2}\sin x - \frac{7}{4}\right) + C

Therefore,

y=−32sin⁡x−74+Ce2sin⁡xy = -\frac{3}{2}\sin x - \frac{7}{4} + C e^{2\sin x}

Using y(0)=−74y(0) = -\frac{7}{4},

−74=−74+C-\frac{7}{4} = -\frac{7}{4} + C

Hence,

C=0C = 0

So,

y(x)=−32sin⁡x−74y(x) = -\frac{3}{2}\sin x - \frac{7}{4}

Now at x=π6x = \frac{\pi}{6},

y(π6)=−32(12)−74y\left(\frac{\pi}{6}\right) = -\frac{3}{2}\left(\frac{1}{2}\right) - \frac{7}{4} =−34−74=−104=−52= -\frac{3}{4} - \frac{7}{4} = -\frac{10}{4} = -\frac{5}{2}

Therefore, y(π6)=−52y\left(\frac{\pi}{6}\right) = -\frac{5}{2} and the correct option is A.

Common mistakes

  • Dividing by sec⁡x\sec x incorrectly. Since 1sec⁡x=cos⁡x\frac{1}{\sec x} = \cos x, the standard form becomes dydx−2cos⁡x⋅y=(2+3sin⁡x)cos⁡x\frac{dy}{dx} - 2\cos x \cdot y = (2 + 3\sin x)\cos x, not dydx−2sec⁡x⋅y=2+3sin⁡x\frac{dy}{dx} - 2\sec x \cdot y = 2 + 3\sin x. Always convert carefully before finding the integrating factor.

  • Using the wrong integrating factor. For a linear equation dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x), the integrating factor is e∫P(x) dxe^{\int P(x) \, dx}. Here P(x)=−2cos⁡xP(x) = -2\cos x, so the integrating factor is e−2sin⁡xe^{-2\sin x}. Missing the negative sign changes the whole solution.

  • Making an error in the substitution step. If u=−2sin⁡xu = -2\sin x, then du=−2cos⁡x dxdu = -2\cos x \, dx. Forgetting the factor of −2-2 or replacing sin⁡x\sin x incorrectly in terms of uu leads to a wrong integral. Write both substitutions explicitly before integrating.

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