Medium MCQ +4 / -1 PYQ · JEE Mains 2022

Let $y=y(x)$ 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 $(0,1)$. Then $y(1)$ is equal to :

  1. A $\frac{1}{2}$
  2. B $\frac{3}{2}$ Correct answer
  3. C $\frac{5}{2}$
  4. D $\frac{7}{2}$

Solution

$\frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$, <br/><br/>Integrating factor I.F. $=e^{\int \frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6} d x}$ <br/><br/>$$ \begin{aligned} & \text { Let } \frac{2 x^{2}+11 x+13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3} \\\\ & A=2, B=1, C=-1 \\\\ & \text { I.F. }=e^{(2 \ln |x+1|+\ln |x+2|-\ln |x+3|)} \\\\ & =\frac{(x+1)^{2}(x+2)}{x+3} \end{aligned} $$ <br/><br/>Solution of differential equation <br/><br/>$$ \begin{aligned} &y \cdot \frac{(x+1)^{2}(x+2)}{x+3}=\int(x+1)(x+2) d x \\\\ &y \frac{(x+1)^{2}(x+2)}{x+3}=\frac{x^{3}}{3}+\frac{3 x^{2}}{2}+2 x+c \end{aligned} $$ <br/><br/>Curve passes through $(0,1)$ <br/><br/>$$ \begin{gathered} 1 \times \frac{1 \times 2}{3}=0+c \Rightarrow c=\frac{2}{3} \\\\ \text { So, } y(1)=\frac{\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}}{\frac{\left(2^{2} \times 3\right)}{4}}=\frac{3}{2} \end{gathered} $$

About this question

Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree

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