Find the equation of the curve passing through the point (1, -1) whos

1.	Find the equation of the curve passing through the point (1, -1) whose differential equation is xy(dy/dx) = (x + 2)(y + 2).
Answer» Given differential equation is xy(dy/dx) = (x + 2)(y + 2) ∴ (y/(y + 2)) dy = ((x + 2)/x) dx, y ≠ -2, x ≠ 0 or, (1 - (2/(y + 2))) dy = (1 + (2/x)) dy Integrating, ∫(1 - (2/(y + 2))) dy = ∫(1 + (2/x)) dx or, y - 2 log\|y + 2\| = x + 2 log\|x\| + k ...(i) Since, curve passes through (1, -1) So, by(1) -1 - 2log 1 = 1 + 2log 1 + k ∴ k = -2 Hence, required equation of the curve is y - 2 log\|y + 2\| = x + 2log\|x\| - 2

Find the equation of the curve passing through the point (1, -1) whose differential equation is xy(dy/dx) = (x + 2)(y + 2).

Answer»

Given differential equation is

xy(dy/dx) = (x + 2)(y + 2)

∴ (y/(y + 2)) dy = ((x + 2)/x) dx, y ≠ -2, x ≠ 0

or, (1 - (2/(y + 2))) dy = (1 + (2/x)) dy

Integrating,

∫(1 - (2/(y + 2))) dy = ∫(1 + (2/x)) dx

or, y - 2 log|y + 2| = x + 2 log|x| + k ...(i)

Since, curve passes through (1, -1)

So, by(1)

-1 - 2log 1 = 1 + 2log 1 + k

∴ k = -2

Hence, required equation of the curve is

y - 2 log|y + 2| = x + 2log|x| - 2