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)y=y(x) satisfy (tan⁡x)1/2 dy=(sec⁡3x−(tan⁡x)3/2y)dx(\tan x)^{1/2}\,dy=\left(\sec^3x-(\tan x)^{3/2}y\right)dx on (0,π2)\left(0,\frac{\pi}{2}\right), with y ⁣(π4)=625y\!\left(\frac{\pi}{4}\right)=\frac{6\sqrt{2}}{5}. If y ⁣(π3)=45αy\!\left(\frac{\pi}{3}\right)=\frac{4}{5}\alpha, then α4\alpha^4 equals
Solution
Answer: 48
Step 1: Rewrite in standard linear form Dividing by (tan⁡x)1/2(\tan x)^{1/2}:
dydx+ytan⁡x=sec⁡3xtan⁡x\frac{dy}{dx}+y\tan x = \frac{\sec^3x}{\sqrt{\tan x}}
 Step 2: Determine the integrating factor
I.F.=e∫tan⁡x dx=eln⁡sec⁡x=sec⁡x\text{I.F.} = e^{\int\tan x\,dx} = e^{\ln\sec x} = \sec x
 Step 3: Solve the resulting equation
ddx(ysec⁡x)=sec⁡4xtan⁡x\frac{d}{dx}(y\sec x) = \frac{\sec^4x}{\sqrt{\tan x}}
 Substituting t=tan⁡xt=\tan x, dt=sec⁡2x dxdt=\sec^2x\,dx:
ysec⁡x=∫(1+t2)sec⁡2xt dtsec⁡2x=∫(t−1/2+t3/2)dt=2t+25t5/2+Cy\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
ysec⁡x=2tan⁡x+25(tan⁡x)5/2+Cy\sec x = 2\sqrt{\tan x}+\frac{2}{5}(\tan x)^{5/2}+C
 Step 4: Apply the initial condition y ⁣(π4)=625y\!\left(\frac{\pi}{4}\right)=\frac{6\sqrt{2}}{5} At x=π4x=\frac{\pi}{4}: tan⁡=1\tan=1, sec⁡=2\sec=\sqrt{2}:
625⋅2=2+25+C  ⟹  125=125+C  ⟹  C=0\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=π3x=\frac{\pi}{3} and compute α4\alpha^4 At x=π3x=\frac{\pi}{3}: tan⁡=3\tan=\sqrt{3}, sec⁡=2\sec=2:
2y=2⋅31/4+25⋅35/4=2⋅31/4 ⁣(1+35)=165⋅31/42y = 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=85⋅31/4=45α  ⟹  α=2⋅31/4y = \frac{8}{5}\cdot3^{1/4} = \frac{4}{5}\alpha \implies \alpha = 2\cdot3^{1/4}
α4=24⋅3=16×3=48\alpha^4 = 2^4\cdot3 = 16\times3 = 48
 Answer: 48
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Differential Equations · medium
Let f:[1,∞)→Rf:[1,\infty)\to\mathbb{R} be defined as f(x)=∫1xf(t) dt+(1−x)(ln⁡x−1)+ef(x)=\displaystyle\int_1^x f(t)\,dt+(1-x)(\ln x-1)+e. Then the value of f(f(1))f(f(1)) is:
Differential Equations · easy
The order of the differential equation corresponding to y=c1cos⁡2x+c2cos⁡2x+c3sin⁡2x+c4y = c_1\cos 2x + c_2\cos^2 x + c_3\sin^2 x + c_4 is:
Differential Equations · medium
The degree of the differential equation satisfying the relation 1+x2+1+y2=λ(x1+y2−y1+x2)\sqrt{1+x^2} + \sqrt{1+y^2} = \lambda\big(x\sqrt{1+y^2} - y\sqrt{1+x^2}\big) is:
Differential Equations · medium
If pp and qq are the order and degree of the differential equation y2(d2ydx2)2+3x(dydx)1/3+x2y2=sin⁡xy^2\left(\dfrac{d^2y}{dx^2}\right)^2 + 3x\left(\dfrac{dy}{dx}\right)^{1/3} + x^2y^2 = \sin x, then:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath