Differential EquationsmediumFree

Curve with Concavity Proportional to log x | JEE DE

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

Question
A curve, whose concavity is directly proportional to the logarithm of its xx-coordinate at any point of the curve, is given by:
AC1x2(2log⁡x−3)+C2x+C3C_1 x^2(2\log x - 3) + C_2 x + C_3correct
BC1x2(2log⁡x+3)+C2x+C3C_1 x^2(2\log x + 3) + C_2 x + C_3
CC1x2(2log⁡x)+C2C_1 x^2(2\log x) + C_2
DNone of these
Solution
Concavity means the second derivative, so the condition is
d2ydx2=Kln⁡x.\frac{d^2y}{dx^2} = K\ln x.
 Step 1: Integrate once (by parts):
dydx=K∫ln⁡x dx=K(xln⁡x−x)+c.\frac{dy}{dx} = K\int\ln x\,dx = K(x\ln x - x) + c.
 Step 2: Integrate again (by parts on ∫xln⁡x dx=x22ln⁡x−x24\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4}):
y=K(x22ln⁡x−x24−x22)+cx+d=K(x22ln⁡x−3x24)+cx+d.y = K\left(\frac{x^2}{2}\ln x - \frac{x^2}{4} - \frac{x^2}{2}\right) + cx + d = K\left(\frac{x^2}{2}\ln x - \frac{3x^2}{4}\right) + cx + d.
 Step 3: Factor:
y=K4x2(2ln⁡x−3)+cx+d=C1x2(2log⁡x−3)+C2x+C3.y = \frac{K}{4}x^2(2\ln x - 3) + cx + d = C_1 x^2(2\log x - 3) + C_2 x + C_3.
 Correct answer: (1)
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 · medium
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
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:

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