The particular solution of the differential equation $x d y+2 y d x=0$, when $x=2$ and $y=1$ is

The particular solution of the differential equation $x d y+2 y d x=0$, when $x=2$ and $y=1$ is
  1. $x y^{2}=4$
  2. $x^{2} y=4$
  3. $x^{2} y=-4$
  4. $x y^{2}=-4$

Solution

Given D.E. is $x d y+2 y d x=0$ $\therefore x d y=-2 y d x \Rightarrow \int \frac{d y}{y}=\int-\frac{2 d x}{x}$ $\log y=-2 \log x+\log c \Rightarrow \log y+2 \log x=\log c$ $\therefore \quad \log y+\log x^{2}=\log c \Rightarrow x^{2} y=c$ when $x=2, y=1, c=4$ $\therefore$ Particular solution is $\mathrm{x}^{2} \mathrm{y}=4$

Asked in: MHT CET 2020 (13 Oct Shift 2)

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