Question

The curve, which satisfies the differential equation

$\frac{xdy - ydx}{xdy + ydx}$ = y² sin (xy) and passes through (0, 1), is given by

• A. y (1 −cosxy) + x = 0
• B. sin xy− x = 0
• C. sin y + y = 0
• D. cosxy− 2y = 0

Answer 1

Answer: (A)

y (1 −cosxy) + x = 0

Answer 2

Differential equation can be written as

$\left( \frac{xdy - ydx}{x^{2}} \right)\left( \frac{x^{2}}{y^{2}} \right)$ = (x dy + y dx) sin xy

or d$\left( \frac{y}{x} \right)\left( \frac{x^{2}}{y^{2}} \right)$ = d (xy) sin xy

Integrating both the sides we get,

−$\frac{1}{y/x}$ = − cos xy + c ⇒ $\frac{x}{y}$ = cos (xy) − c.

For x = 0, y = 1, we get c = 1