Question

Solution of (xy⁴ + y) dx – xdy = 0 is

• A. $\frac{x^{4}}{4} + \left( \frac{x}{y} \right)^{3} = c$
• B. $\frac{x^{4}}{4} + \frac{1}{3}\left( \frac{x}{y} \right)^{2} = c$
• C. $\frac{x^{4}}{4} + 3\left( \frac{x}{y} \right)^{2} = c$
• D. None of these

Answer 1

Answer: (A)

$\frac{x^{4}}{4} + \left( \frac{x}{y} \right)^{3} = c$

Answer 2

Rearrange the diff. equation xdx + $\frac{ydx - xdy}{y^{4}}$ = 0

x³dx+ $\frac{x^{2}}{y^{2}} \cdot \frac{ydx - xdy}{y^{2}}$ = 0

x³dx+ $\frac{d}{dx}$ $\left( \frac{x}{y} \right)$ = 0

integratings $\frac{x^{4}}{4} + \frac{1}{3}\left( \frac{x}{y} \right)^{3}$ + c