Question

The general solution of the differential equation (1 + tan y) (dx – dy) + 2xdy = 0 is –

• A. x(sin y + cos y) = sin y + ce^y
• B. x(sin y + cos y) = sin y + ce^–y
• C. y(sin x + cos x) = sin x + ce^x
• D. None of these

Answer 1

Answer: (B)

x(sin y + cos y) = sin y + ce^–y

Answer 2

We have, (1 + tan y) (dx – dy ) + 2x dy = 0

Ž (1 + tan y) dx = (1 + tan y – 2x) dy

Ž $\frac{dx}{dy}$ + $\frac{2}{1 + \tan y}$x = 1, which is linear in x.

I.F. = $e^{2\int_{}^{}\frac{dy}{1 + \tan y}}$ = $e^{\int_{}^{}\frac{2\cos y}{\sin y + \cos y}dy}$

= $e^{\int_{}^{}\left( 1 + \frac{\cos y - \sin y}{\sin y + \cos y} \right)dy}$= e^{y + log(cos y + sin y)}

= (cos y + sin y) e^y.

So, the solution is

xe^y (sin y + cos y) = $\int_{}^{}e^{y}$ (sin y + cos y) dy + c

i.e. xe^y (sin y + cos y) = e^y sin y + c.

i.e. x (sin y + cos y) = sin y + ce^–y