Question

The particular solution of the differential equation $xdy+2ydx=0$, when $x=2,y=1$ is:

• A. $xy=4$
• B. $x^{2}y=4$
• C. $x{y}^2=4$
• D. $x^2{y}^2=4$

Answer 1

Answer: (B)

$x^{2}y=4$

$xdy+2ydx=0$

$\Rightarrow \frac{dy}{y}+\frac{2dx}{x}=0$

On Integrating both sides, we get

$\int \frac{dy}{y}+2\int \frac{dx}{x}=C_1$

$\log y+2\log x=\log C$

$\log y{x}^2=\log C$

$\therefore x^{2}y=C$

When $x=2,y=1$, then $C=4$

$\therefore$ Particular solution is $x^{2}y=4$