Question

\[2x\frac{dy}{dx} = 3y, y\left( 1 \right) = 2\]

Answer 1

We have, 
\[2x\frac{dy}{dx} = 3y, y\left( 1 \right) = 2\]
\[ \Rightarrow \frac{2}{y}dy = \frac{3}{x}dx\]
Integrating both sides, we get 
\[2\int\frac{1}{y}dy = 3\int\frac{1}{x}dx\]
\[ \Rightarrow 2 \log \left| y \right| = 3 \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| y \right|^2 = \log \left| x \right|^3 + \log C\]
\[ \Rightarrow y^2 = C x^3 . . . . (1)\]
It is given that at x = 1, y = 2 . 
Substituting the values of x and y in (1), we get
\[C = 4\]
Now, substituting the value of C in (1), we get
\[ \Rightarrow y^2 = 4 x^3 \]
\[\text{Hence,} y^2 = 4 x^3\text{ is the required solution }. \]