MHT CET · Maths · Differential Equations
The solution of differential equation \(x^{2} \frac{d y}{d x}=y^{2}+x y\) is
- A \(\frac{x}{y}+\log |x|=c\)
- B \(\frac{y}{x}+\log |x|=c\)
- C \(\frac{x}{y}-\log |x|=c\)
- D \(\frac{y}{x}-\log |x|=c\)
Answer & Solution
Correct Answer
(A) \(\frac{x}{y}+\log |x|=c\)
Step-by-step Solution
Detailed explanation
We have \(x^{2} \frac{d y}{d x}=y^{2}+x y\)
\(\begin{array}{l}
\therefore \frac{d y}{d x}=\frac{y^{2}+x y}{x^{2}} ...(1) \\
\therefore \text { Put } y=u x \\
\therefore \frac{d y}{d x}=u+x \frac{d u}{d x}
\end{array}\)
\(\therefore\) Equation (1) becomes
\(u+x \frac{d u}{d x}=\frac{u^{2} x^{2}+x(u x)}{x^{2}} \Rightarrow u+x \frac{d u}{d x}=u^{2}+u\)
\(\begin{array}{l}
\therefore \mathrm{x} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}^{2} \\
\therefore \int \frac{\mathrm{du}}{\mathrm{u}^{2}}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\
\therefore \frac{-1}{\mathrm{u}}=\log |\mathrm{x}|+\mathrm{c}_{1} \Rightarrow \quad \frac{-\mathrm{x}}{\mathrm{y}}=\log |\mathrm{x}|+\mathrm{c}_{1} \\
\therefore \frac{\mathrm{x}}{\mathrm{y}}+\log |\mathrm{x}|=\mathrm{c}
\end{array}\)
\(\begin{array}{l}
\therefore \frac{d y}{d x}=\frac{y^{2}+x y}{x^{2}} ...(1) \\
\therefore \text { Put } y=u x \\
\therefore \frac{d y}{d x}=u+x \frac{d u}{d x}
\end{array}\)
\(\therefore\) Equation (1) becomes
\(u+x \frac{d u}{d x}=\frac{u^{2} x^{2}+x(u x)}{x^{2}} \Rightarrow u+x \frac{d u}{d x}=u^{2}+u\)
\(\begin{array}{l}
\therefore \mathrm{x} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}^{2} \\
\therefore \int \frac{\mathrm{du}}{\mathrm{u}^{2}}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\
\therefore \frac{-1}{\mathrm{u}}=\log |\mathrm{x}|+\mathrm{c}_{1} \Rightarrow \quad \frac{-\mathrm{x}}{\mathrm{y}}=\log |\mathrm{x}|+\mathrm{c}_{1} \\
\therefore \frac{\mathrm{x}}{\mathrm{y}}+\log |\mathrm{x}|=\mathrm{c}
\end{array}\)
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