Q81 / 180 · KCET 2014 · Mathematics
The general solution of the differential equation \( \frac{d y}{d x}+\frac{y}{x}=3 x \) is
- A\( y=x+\frac{C}{x} \)
- B\( y=x^{2}+\frac{C}{x} \)
- C\( y=x-\frac{C}{x} \)
- Correct answerD\( y=x^{2}-\frac{C}{x} \)
Solution
Given differential equation is,
\[ \begin{array}{l} \frac{d y}{d x}+\frac{y}{x}=3 x \rightarrow(1) \\ \Rightarrow x \frac{d y}{d x}+y=3 x^{2} \\ \Rightarrow \frac{d}{d x}(x y)=3 x^{2} \rightarrow(2) \end{array} \]
On integrating, we get;
\[ \begin{array}{l} x y=x^{3}+c \\ \Rightarrow y=\frac{x^{3}+c}{x} \end{array} \]
After integrating equation (2), we have
\[ \begin{array}{l} x y=x^{3}-c \\ \Rightarrow y=x^{2}-\frac{C}{x} \end{array} \]
\[ \begin{array}{l} \frac{d y}{d x}+\frac{y}{x}=3 x \rightarrow(1) \\ \Rightarrow x \frac{d y}{d x}+y=3 x^{2} \\ \Rightarrow \frac{d}{d x}(x y)=3 x^{2} \rightarrow(2) \end{array} \]
On integrating, we get;
\[ \begin{array}{l} x y=x^{3}+c \\ \Rightarrow y=\frac{x^{3}+c}{x} \end{array} \]
After integrating equation (2), we have
\[ \begin{array}{l} x y=x^{3}-c \\ \Rightarrow y=x^{2}-\frac{C}{x} \end{array} \]
Hence, the correct answer is option: 'D'.