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1.	(x sin y\(Y/x\)ath widget
Answer» (x sin y/x - y cos y/x) dx + x cos y/x dy = 0 ∴ dy/dx \(=\frac{y\,cos\,y/x\,-\,x\,sin\,y/x}{x\,cos\,y/x}\) ⇒ dy/dx \(=\frac{y/x\,cos\,y/x\,-\,sin\,y/x}{cos\,y/x}\) ...(1) Let y = vx ∴ dy/dx = v + x dv/dx ∴ From (1), v + x dv/dx \(=\frac{v\,cos\,v-sin\,v}{cos\,v}\) ⇒ x dv/dx \(=\frac{v\,cos\,v-sin\,v}{cos\,v}-v\) \(=\frac{v\,cos\,v-sin\,v-v\,cos\,v}{cos\,v}\) = -sin v/cos v = -tan v ⇒ cot v dv = -1/x dx ⇒ ∫ cos v/sin v dv = -∫1/x dx ⇒ log sin v = -log x + log c ⇒ log sin v = log c/x ⇒ sin v = c/x ⇒ x sin v = c ⇒ x sin y/x = c ...(2) (By putting v = y/x) Equation (2) represent the solution of given differential equation.

(x sin y\(Y/x\)ath widget

Answer»

(x sin y/x - y cos y/x) dx + x cos y/x dy = 0

∴ dy/dx \(=\frac{y\,cos\,y/x\,-\,x\,sin\,y/x}{x\,cos\,y/x}\)

⇒ dy/dx \(=\frac{y/x\,cos\,y/x\,-\,sin\,y/x}{cos\,y/x}\) ...(1)

Let y = vx

∴ dy/dx = v + x dv/dx

∴ From (1), v + x dv/dx \(=\frac{v\,cos\,v-sin\,v}{cos\,v}\)

⇒ x dv/dx \(=\frac{v\,cos\,v-sin\,v}{cos\,v}-v\)

\(=\frac{v\,cos\,v-sin\,v-v\,cos\,v}{cos\,v}\)

= -sin v/cos v = -tan v

⇒ cot v dv = -1/x dx

⇒ ∫ cos v/sin v dv = -∫1/x dx

⇒ log sin v = -log x + log c

⇒ log sin v = log c/x

⇒ sin v = c/x

⇒ x sin v = c

⇒ x sin y/x = c ...(2) (By putting v = y/x)

Equation (2) represent the solution of given differential equation.

Discussion