Question

What is the particular solution of the differential equation $xdy + 2ydx = 0$ when $x = 2,y = 1$.
$
  {\text{A}}{\text{. }}xy = 4 \\
  {\text{B}}{\text{. }}{x^2}y = 4 \\
  {\text{C}}{\text{. }}x{y^2} = 4 \\
  {\text{D}}{\text{. }}{x^2}{y^2} = 4 \\
 $

Answer 1

Hint: Here, the differential equation is rearranged with the help of variable separable method so that the given differential equation can be represented in the form of $f\left( y \right)dy = f\left( x \right)dx$. After obtaining this, we will be integrating this equation.

Complete step-by-step answer:

Given, differential equation is $xdy + 2ydx = 0$
By variable separable method, the differential equation can be rearranged as under
$
   \Rightarrow xdy = - 2ydx \\
   \Rightarrow \dfrac{{dy}}{y} = - \dfrac{2}{x}dx{\text{ }} \to {\text{(1)}} \\
 $
 The above differential equation can be solved by integrating the whole equation (1).
By integrating equation (1), we get
\[
   \Rightarrow \int {\dfrac{{dy}}{y}} = \int {\left( { - \dfrac{2}{x}dx} \right)} \\
   \Rightarrow \int {\dfrac{{dy}}{y}} = - 2\int {\dfrac{{dx}}{x}} \\
   \Rightarrow \log y = - 2\log x + \log c \\
   \Rightarrow \log y + 2\log x = \log c{\text{ }} \to {\text{(2)}} \\
 \]
where $\log c$ is the constant of integration
As we know that $n\log m = \log \left( {{m^n}} \right){\text{ }} \to {\text{(3)}}$
Using the formula given by equation (3) in equation (2), we get
\[ \Rightarrow \log y + \log \left( {{x^2}} \right) = \log c{\text{ }} \to {\text{(4)}}\]
Also, we know that $\log m + \log n = \log \left( {mn} \right){\text{ }} \to {\text{(5)}}$
Using the formula given by equation (5) in equation (4), we get
\[ \Rightarrow \log \left( {{x^2}y} \right) = \log c{\text{ }} \to {\text{(6)}}\]
Since, when $\log m = \log n$ then m=n
Using the above concept in equation (6), we get
\[ \Rightarrow {x^2}y = c{\text{ }} \to {\text{(7)}}\]
This equation (7) gives the general solution of the given differential solution.
The particular solution of the given differential equation is determined by putting the given values of variables x and y in the general solution and from there evaluating the value of the constant of integration.
By putting x=2 and y=1 in equation (7), we get
\[
   \Rightarrow \left( {{2^2}} \right)\left( 1 \right) = c \\
   \Rightarrow c = 4 \\
 \]
Put c=4 in equation (7), we get
\[ \Rightarrow {x^2}y = 4\]
The above equation gives the particular solution of the given differential equation when x=2, y=1.
Hence, option B is correct.

Note: The general solution of any differential equation contains the constant of integration whereas the particular solution of any differential equation is independent of any constant of integration. This particular solution is obtained by simply putting the given values of the variables in the general solution and obtaining the value of constant of integration and then substitute that value in the general solution.