Question

Solution of the equation xdy – [y + xy³ (1 + log x)] dx = 0 is –

• A. $\frac{- x^{2}}{y^{2}}$= $\left( \frac{2}{3} + \log x \right)$ + C
• B. $\frac{x^{2}}{y^{2}}$= $\left( \frac{2}{3} + \log x \right)$ + C
• C. $\frac{- x^{2}}{y^{2}}$= $\left( \frac{2}{3} + \log x \right)$ + C
• D. None of these

Answer 1

Answer: (A)

$\frac{- x^{2}}{y^{2}}$= $\left( \frac{2}{3} + \log x \right)$ + C

Answer 2

We have,

x dy – y dx = xy³ (1 + log x) dx

Ž – $\left( \frac{ydx - xdy}{y^{2}} \right)$ = xy (1 + log x) dx

Ž – d $\left( \frac{x}{y} \right)$ = xy (1 + log x) dx

Ž – $\frac{x}{y}$ d$\left( \frac{x}{y} \right)$ = x² (1 + log x) dx

Integrating, we get

– $\frac{\left( \frac{x}{y} \right)^{2}}{2}$= (1 + log x) $\frac{x^{3}}{3}$– $\int_{}^{}\frac{x^{3}}{3}$.$\frac{1}{x}$ dx

Ž – $\frac{x^{2}}{2y^{2}}$ = $\frac{x^{3}}{3}$ (1 + log x) – $\frac{x^{3}}{9}$ + $\frac{C}{2}$

Ž – $\frac{x^{2}}{y^{2}}$ = $\left( \frac{2}{3} + \log x \right)$ + C