Q109 / 180 · KCET 2022 · Mathematics

If $\frac{d y}{d x}+\frac{y}{x}=x^2$, then $2 y(2)-y(1)=$

  • A
    $\frac{11}{4}$
  • Correct answer
    B
    $\frac{15}{4}$
  • C
    $\frac{9}{4}$
  • D
    $\frac{13}{4}$

Solution

$\frac{d y}{d x}+\frac{y}{x}=x^2$ On comparing to $\frac{d y}{d x}+P(x) y=Q(x)$ $$ \begin{aligned} & P(x)=\frac{1}{x}, Q(x)=x^2 \\ & \mathrm{IF}=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\ln x}=x \\ & y \cdot(\mathrm{IF})=\int Q(x)(\mathrm{IF}) d x+C \\ & y \cdot x=\int x^2 x d x+C=\frac{x^4}{4}+C \\ & y=\frac{x^3}{4}+\frac{C}{x} \\ & y(2)=\frac{2^3}{4}+\frac{C}{2}=2+\frac{C}{2} \\ & y(1)=\frac{1}{4}+\frac{C}{1}=\frac{1}{4}+C \\ & 2 y(2)-y(1)=2\left(2+\frac{C}{2}\right)-\left(\frac{1}{4}+C\right) \\ &=4+C-\frac{1}{4}-C=\frac{15}{4} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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