Solve : (x2 + y2)dy/dx = 2xy

1.	Solve : (x2 + y2)dy/dx = 2xy
Answer» (x² + y²)dy/dx = 2xy dy/dx = 2xy/(x² + y²) dy/dx = (x² + y²)/2xy ...(i) Let x = vy Here, differentiating w.r.t. y, dx/dy = v.(dy/dx) + y(dv/dy) dx/dy = (v + y).(dy/dx) Here, from eq. (i), v + y(dv/dx) = (v²y² + y²)/2vy² v + y(dv/dy) = y²(v² + 1)/y²2v v + y(dv/dy) = (v² + 1)/2v y(dv/dy) = ((v² + 1)/2v) - (v/1) y(dv/dy) = (v² + 1 - 2v²)/2v y(dv/dy) = (-v² + 1)/2v y(dv/dy) = (1 - v²)/2v 2v/(1 - v²) dv = dy/y Integrating both sides ∫2v(v² - 1) dv = - ∫dy/y log\|v² - 1\| = - log y + log c log\|v² - 1\|y = log c v²y - y = c (x²/y²) x (y - y) = c (x²/y) - y = c (x²- y²)/y = c x² - y² = cy

Answer»

(x² + y²)dy/dx = 2xy

dy/dx = 2xy/(x² + y²)

dy/dx = (x² + y²)/2xy ...(i)

Let x = vy

Here, differentiating w.r.t. y,

dx/dy = v.(dy/dx) + y(dv/dy)

dx/dy = (v + y).(dy/dx)

Here, from eq. (i),

v + y(dv/dx) = (v²y² + y²)/2vy²

v + y(dv/dy) = y²(v² + 1)/y²2v

v + y(dv/dy) = (v² + 1)/2v

y(dv/dy) = ((v² + 1)/2v) - (v/1)

y(dv/dy) = (v² + 1 - 2v²)/2v

y(dv/dy) = (-v² + 1)/2v

y(dv/dy) = (1 - v²)/2v

2v/(1 - v²) dv = dy/y

Integrating both sides

∫2v(v² - 1) dv = - ∫dy/y

log|v² - 1| = - log y + log c

log|v² - 1|y = log c

v²y - y = c

(x²/y²) x (y - y) = c

(x²/y) - y = c

(x²- y²)/y = c

x² - y² = cy

Discussion