Q114 / 180 · KCET 2015 · Mathematics
The solution of differential equation
\[
x \frac{d y}{d x}+2 y=x^{2} \text { is }
\]
\[ x \frac{d y}{d x}+2 y=x^{2} \text { is } \]
- A\( y=\frac{x^{2}+C}{4 x^{2}} \)
- B\( y=\frac{x^{2}}{4}+C \)
- C\( y=\frac{x^{4}+C}{x^{2}} \)
- Correct answerD\( y=\frac{x^{4}+C}{4 x^{2}} \)
Solution
Given differential equations,
\[ \begin{array}{l} x \frac{d y}{d x}+2 y=x^{2} \\ \Rightarrow \frac{d y}{d x}+\frac{2 y}{x}=x \rightarrow(1) \end{array} \]
General differential equation of this type is given by
\[ \frac{d y}{d x}+P(x) y=Q(x) \]
So, \( P=\frac{2}{x} \) then I.F. factor is given by
\[ e^{\int P d x}=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2} \]
Then, \( x^{2} \frac{d y}{d x}+2 x y=x^{3} \)
\[ \Rightarrow \frac{d}{d x}\left(x^{2} y\right)=x^{3} \]
Integrating both the sides, we get
\[ \begin{array}{l} x^{2} y=\frac{x^{4}}{4}+C_{1} \\ \Rightarrow x^{2} y=\frac{x^{4}+C}{4} \\ \Rightarrow y=\frac{x^{4}+C}{4 x^{2}} \end{array} \]
\[ \begin{array}{l} x \frac{d y}{d x}+2 y=x^{2} \\ \Rightarrow \frac{d y}{d x}+\frac{2 y}{x}=x \rightarrow(1) \end{array} \]
General differential equation of this type is given by
\[ \frac{d y}{d x}+P(x) y=Q(x) \]
So, \( P=\frac{2}{x} \) then I.F. factor is given by
\[ e^{\int P d x}=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2} \]
Then, \( x^{2} \frac{d y}{d x}+2 x y=x^{3} \)
\[ \Rightarrow \frac{d}{d x}\left(x^{2} y\right)=x^{3} \]
Integrating both the sides, we get
\[ \begin{array}{l} x^{2} y=\frac{x^{4}}{4}+C_{1} \\ \Rightarrow x^{2} y=\frac{x^{4}+C}{4} \\ \Rightarrow y=\frac{x^{4}+C}{4 x^{2}} \end{array} \]
Hence, the correct answer is option: 'D'.