Differential EquationsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Differential Equations: Let Satisfy Equals (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question

Let

y=y(x)

satisfy

(\tan x)^{1/2}\,dy=\left(\sec^3x-(\tan x)^{3/2}y\right)dx

\left(0,\frac{\pi}{2}\right)

, with

y\!\left(\frac{\pi}{4}\right)=\frac{6\sqrt{2}}{5}

. If

y\!\left(\frac{\pi}{3}\right)=\frac{4}{5}\alpha

, then

\alpha^4

equals

Solution

Answer: 48

Step 1: Rewrite in standard linear form Dividing by

(\tan x)^{1/2}

\frac{dy}{dx}+y\tan x = \frac{\sec^3x}{\sqrt{\tan x}}

Step 2: Determine the integrating factor

\text{I.F.} = e^{\int\tan x\,dx} = e^{\ln\sec x} = \sec x

Step 3: Solve the resulting equation

\frac{d}{dx}(y\sec x) = \frac{\sec^4x}{\sqrt{\tan x}}

Substituting

t=\tan x

dt=\sec^2x\,dx

y\sec x = \int\frac{(1+t^2)\sec^2x}{\sqrt{t}}\,\frac{dt}{\sec^2x} = \int\left(t^{-1/2}+t^{3/2}\right)dt = 2\sqrt{t}+\frac{2}{5}t^{5/2}+C

y\sec x = 2\sqrt{\tan x}+\frac{2}{5}(\tan x)^{5/2}+C

Step 4: Apply the initial condition

y\!\left(\frac{\pi}{4}\right)=\frac{6\sqrt{2}}{5}

x=\frac{\pi}{4}

\tan=1

\sec=\sqrt{2}

\frac{6\sqrt{2}}{5}\cdot\sqrt{2} = 2+\frac{2}{5}+C \implies \frac{12}{5} = \frac{12}{5}+C \implies C=0

Step 5: Evaluate at

x=\frac{\pi}{3}

and compute

\alpha^4

x=\frac{\pi}{3}

\tan=\sqrt{3}

\sec=2

2y = 2\cdot3^{1/4}+\frac{2}{5}\cdot3^{5/4} = 2\cdot3^{1/4}\!\left(1+\frac{3}{5}\right) = \frac{16}{5}\cdot3^{1/4}

y = \frac{8}{5}\cdot3^{1/4} = \frac{4}{5}\alpha \implies \alpha = 2\cdot3^{1/4}

\alpha^4 = 2^4\cdot3 = 16\times3 = 48

Answer: 48

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