Question

The equation of the curve whose slope is given by \[\frac{dy}{dx} = \frac{2y}{x}; x > 0, y > 0\] and which passes through the point (1, 1) is

Options

• x² = y

• y² = x

• x² = 2y

• y² = 2x

Answer 1

x² = y

 

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