Verified Solution • 2025–26 Session moderate

QJEEM22Ma16fe: JEE Main 2022 Mathematics - Class 12 JEEM22Ma16fe Solution: Let be the solution curve of the differential equation , which…

Let y=y(x)y=y(x) be the solution curve of the differential equation \frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1 , which passes through the point (0,1)(0,1) . Then y(1)y(1) is equal to :

Multiple Choice Options

  • A 12\frac{1}{2}
  • B 32\frac{3}{2}
  • C 52\frac{5}{2}
  • D 72\frac{7}{2}

Step-by-Step Verified Solution

\frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1 ,

Integrating factor I.F. =e∫2x2+11x+13x3+6x2+11x+6dx=e^{\int \frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6} d x}

amp; Let 2x2+11x+13(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+3amp;A=2,B=1,C=−1amp; I.F. =e(2ln⁡∣x+1∣+ln⁡∣x+2∣−ln⁡∣x+3∣)amp;=(x+1)2(x+2)x+3\begin{aligned}& \text { Let } \frac{2 x^{2}+11 x+13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3} \\\\& A=2, B=1, C=-1 \\\\& \text { I.F. }=e^{(2 \ln |x+1|+\ln |x+2|-\ln |x+3|)} \\\\& =\frac{(x+1)^{2}(x+2)}{x+3}\end{aligned}

Solution of differential equation

amp;y⋅(x+1)2(x+2)x+3=∫(x+1)(x+2)dxamp;y(x+1)2(x+2)x+3=x33+3x22+2x+c\begin{aligned}&y \cdot \frac{(x+1)^{2}(x+2)}{x+3}=\int(x+1)(x+2) d x \\\\&y \frac{(x+1)^{2}(x+2)}{x+3}=\frac{x^{3}}{3}+\frac{3 x^{2}}{2}+2 x+c\end{aligned}

Curve passes through (0,1)(0,1)

1×1×23=0+c⇒c=23 So, y(1)=13+32+2+23(22×3)4=32\begin{gathered}1 \times \frac{1 \times 2}{3}=0+c \Rightarrow c=\frac{2}{3} \\\\\text { So, } y(1)=\frac{\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}}{\frac{\left(2^{2} \times 3\right)}{4}}=\frac{3}{2}\end{gathered}

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