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Bernoulli Differential Equation with Greatest Integer | JEE

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

Question
Let y=y(x)y = y(x) and x dy+y(1−xy) dx=0x\,dy + y(1 - xy)\,dx = 0. If y(1)=1y(1) = 1, then the value of [ y(1e)]\big[\,y\left(\tfrac{1}{e}\right)\big] is (where [ ⋅ ][\,\cdot\,] is the greatest integer function):
A00
B−1-1
C11correct
D−2-2
Solution
Step 1: Write in Bernoulli form:
xdydx+y−xy2=0⇒dydx+yx=y2.x\frac{dy}{dx} + y - xy^2 = 0 \Rightarrow \frac{dy}{dx} + \frac{y}{x} = y^2.
 Step 2: Divide by y2y^2 and put t=1yt = \dfrac{1}{y} (so dtdx=−1y2dydx\dfrac{dt}{dx} = -\dfrac{1}{y^2}\dfrac{dy}{dx}):
dtdx−tx=−1,IF=1x.\frac{dt}{dx} - \frac{t}{x} = -1, \qquad \text{IF} = \frac{1}{x}.
 Step 3: Solve:
tx=−∫1x dx+C=−ln⁡x+C⇒1xy=C−ln⁡x.\frac{t}{x} = -\int\frac{1}{x}\,dx + C = -\ln x + C \Rightarrow \frac{1}{xy} = C - \ln x.
 Apply y(1)=1y(1) = 1: 1=C⇒C=11 = C \Rightarrow C = 1, so 1xy=1−ln⁡x\dfrac{1}{xy} = 1 - \ln x. Step 4: At x=1ex = \dfrac{1}{e}, ln⁡x=−1\ln x = -1:
1(1/e) y=1−(−1)=2⇒y=e2≈1.36⇒[y(1e)]=1.\frac{1}{(1/e)\,y} = 1 - (-1) = 2 \Rightarrow y = \frac{e}{2} \approx 1.36 \Rightarrow \big[y(\tfrac1e)\big] = 1.
 Correct answer: (3)
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