Differential EquationsmediumFree

Integro-Differential Equation and Greatest Integer | JEE

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

Question
If the solution of dydx=y+∫02y dx\dfrac{dy}{dx} = y + \displaystyle\int_0^2 y\,dx is y(x)y(x) with y(0)=1y(0) = 1, then [ ∣y(2)∣ ]\big[\,|y(2)|\,\big] equals (where [ ⋅ ][\,\cdot\,] is the greatest integer function):
A22
B33
C44correct
D55
Solution
Step 1: ∫02y dx=K\displaystyle\int_0^2 y\,dx = K is a constant, so dydx=y+K\dfrac{dy}{dx} = y + K. Separate and integrate:
dyy+K=dx⇒ln⁡(y+K)=x+λ.\frac{dy}{y + K} = dx \Rightarrow \ln(y + K) = x + \lambda.
 Using y(0)=1y(0) = 1: λ=ln⁡(1+K)\lambda = \ln(1 + K), hence y+K=(1+K)exy + K = (1 + K)e^x, i.e. y=(1+K)ex−Ky = (1+K)e^x - K. Step 2: Determine KK:
K=∫02[(1+K)ex−K]dx=(1+K)(e2−1)−2K⇒3K=(1+K)(e2−1).K = \int_0^2\big[(1+K)e^x - K\big]dx = (1+K)(e^2 - 1) - 2K \Rightarrow 3K = (1+K)(e^2 - 1).
⇒K(4−e2)=e2−1⇒K=e2−14−e2.\Rightarrow K(4 - e^2) = e^2 - 1 \Rightarrow K = \frac{e^2 - 1}{4 - e^2}.
 Step 3: Evaluate y(2)y(2). Since 1+K=34−e21 + K = \dfrac{3}{4 - e^2},
y(2)=(1+K)e2−K=3e2−(e2−1)4−e2=2e2+14−e2.y(2) = (1+K)e^2 - K = \frac{3e^2 - (e^2 - 1)}{4 - e^2} = \frac{2e^2 + 1}{4 - e^2}.
 Numerically ∣y(2)∣=2e2+1e2−4≈15.783.39≈4.66|y(2)| = \dfrac{2e^2 + 1}{e^2 - 4} \approx \dfrac{15.78}{3.39} \approx 4.66, so [ ∣y(2)∣ ]=4\big[\,|y(2)|\,\big] = 4. Correct answer: (3)
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