Step-by-step explanation: We have, [tex] {x}^{2} dy + xydx = \sqrt{1 – {x}^{2} {y}^{2} } dx \\ [/tex] [tex] \implies \: x(x dy + ydx) = \sqrt{1 – {x}^{2} {y}^{2} } dx \\ [/tex] [tex] \implies \: x.d(xy) = \sqrt{1 – {x}^{2} {y}^{2} } dx \\ [/tex] [tex] \implies \: \frac{d(xy) }{ \sqrt{1 – {x}^{2} {y}^{2} } }= \frac{ dx }{x}\\ [/tex] Integrating both sides, [tex] \implies \: \int\frac{d(xy) }{ \sqrt{1 – {x}^{2} {y}^{2} } }= \int \frac{ dx }{x}\\ [/tex] [tex] \implies \: \sin ^{ – 1} (xy) = ln(x)+c \\ [/tex] Reply
Step-by-step explanation:
We have,
[tex] {x}^{2} dy + xydx = \sqrt{1 – {x}^{2} {y}^{2} } dx \\ [/tex]
[tex] \implies \: x(x dy + ydx) = \sqrt{1 – {x}^{2} {y}^{2} } dx \\ [/tex]
[tex] \implies \: x.d(xy) = \sqrt{1 – {x}^{2} {y}^{2} } dx \\ [/tex]
[tex] \implies \: \frac{d(xy) }{ \sqrt{1 – {x}^{2} {y}^{2} } }= \frac{ dx }{x}\\ [/tex]
Integrating both sides,
[tex] \implies \: \int\frac{d(xy) }{ \sqrt{1 – {x}^{2} {y}^{2} } }= \int \frac{ dx }{x}\\ [/tex]
[tex] \implies \: \sin ^{ – 1} (xy) = ln(x)+c \\ [/tex]