Differential EquationseasyFree

Exact Differential Equation with Initial Condition | JEE

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

Question
The differential equation ey dx+(eyx+2y) dy=0e^y\,dx + (e^y x + 2y)\,dy = 0 has the particular solution y(0)=1y(0) = 1. The value of xx when y=0y = 0 is:
A11correct
B00
C22
D−1-1
Solution
Group as exact differentials:
ey dx+xey dy+2y dy=0⇒d(xey)+d(y2)=0⇒xey+y2=C.e^y\,dx + x e^y\,dy + 2y\,dy = 0 \Rightarrow d(xe^y) + d(y^2) = 0 \Rightarrow xe^y + y^2 = C.
 Apply y(0)=1y(0) = 1 (i.e. x=0x = 0 at y=1y = 1): 0⋅e+1=C⇒C=10\cdot e + 1 = C \Rightarrow C = 1, so xey+y2=1xe^y + y^2 = 1. When y=0y = 0: x⋅e0+0=1⇒x=1x\cdot e^0 + 0 = 1 \Rightarrow x = 1. 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