Verified Solution • 2025–26 Session moderate

QJEEM23Ma55e7: JEE Main 2023 Mathematics - Class 12 JEEM23Ma55e7 Solution: Let be the solution of the differential equation . Then y (1)…

Let y=y(x)y = y(x) be the solution of the differential equation {x^3}dy + (xy - 1)dx = 0,x > 0,y\left( {{1 \over 2}} \right) = 3 - \mathrm{e} . Then y (1) is equal to

Multiple Choice Options

  • A 2 −- e
  • B 3
  • C 1
  • D e

Step-by-Step Verified Solution

x3dy+xydx−dx=0x^{3} d y+x y d x-d x=0

⇒dydx=1−xyx3\Rightarrow \frac{d y}{d x}=\frac{1-x y}{x^{3}}

⇒dydx+yx2=1x3\Rightarrow \frac{d y}{d x}+\frac{y}{x^{2}}=\frac{1}{x^{3}}

I.F. =e∫dxx2=e−1x=e^{\int \frac{d x}{x^{2}}}=e^{-\frac{1}{x}}

∴ye−1x=∫e−1xx3dx\therefore \quad y e^{-\frac{1}{x}}=\int \frac{e^{-\frac{1}{x}}}{x^{3}} d x

For RHS put −1x=t⇒dxx2=dt-\frac{1}{x}=t \Rightarrow \frac{d x}{x^{2}}=d t

∴ye−1x=−∫tetdt\therefore y e^{-\frac{1}{x}}=-\int t e^{t} d t

⇒ye−1x=−[tet−et]+c\Rightarrow y e^{-\frac{1}{x}}=-\left[t e^{t}-e^{t}\right]+c

⇒ye−1x=e−1xx+e−1x+c\Rightarrow y e^{-\frac{1}{x}}=\frac{e^{-\frac{1}{x}}}{x}+e^{-\frac{1}{x}}+c

↓y(12)=3−e\downarrow y\left(\frac{1}{2}\right)=3-e

⇒(3−e)e−2=2e−2+e−2+c\Rightarrow(3-e) e^{-2}=2 e^{-2}+e^{-2}+c

⇒c=−1e\Rightarrow \quad c=-\frac{1}{e}

For y(1)y(1) put x=1,c=−e−1x=1, c=-e^{-1} in equation (i) we get ye−1=e−1+e−1−e−1y e^{-1}=e^{-1}+e^{-1}-e^{-1}

⇒y=1\Rightarrow y=1

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring