Question

The solution of the differential equation \[\frac{dy}{dx} + \frac{2y}{x} = 0\] with y(1) = 1 is given by

पर्याय

• \[y = \frac{1}{x^2}\]

• \[x = \frac{1}{y^2}\]

• \[x = \frac{1}{y}\]

• \[y = \frac{1}{x}\]

Answer 1

\[y = \frac{1}{x^2}\]

 

We have,
\[\frac{dy}{dx} + \frac{2y}{x} = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 2y}{x}\]
\[ \Rightarrow \frac{1}{2} \times \frac{1}{y}dy = \frac{- 1}{x}dx\]
Integrating both sides, we get
\[\frac{1}{2}\int\frac{1}{y}dy = - \int\frac{1}{x}dx\]
\[ \Rightarrow \frac{1}{2}\log y = - \log x + \log C\]
\[ \Rightarrow \log y^\frac{1}{2} + \log x = \log C\]
\[ \Rightarrow \log\left( \sqrt{y}x \right) = \log C\]
\[ \Rightarrow \sqrt{y}x = C . . . . . \left( 1 \right)\]
\[\text{ As }\left( 1 \right)\text{ satisfies }y\left( 1 \right) = 1,\text{ we get }\]
\[1 = C\]
\[\text{ Putting the value of C in }\left( 1 \right),\text{ we get }\]
\[\sqrt{y}x = 1\]
\[ \Rightarrow y = \frac{1}{x^2}\]