Solution of the differential equation
$c o s x d y = y (s i n x - y) d x, 0 < x < \frac{π}{2}$ is

y s e c x = t a n x + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y t a n x = s e c x + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

t a n x = (s e c x + c) y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

s e c x = (t a n x + c) y

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $s e c x = (t a n x + c) y$
$c o s x d y = y (s i n x - y) d x \frac{d y}{d x} = y t a n x - y^{2} s e c x \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} t a n x = - s e c x$ ....(i)
Let $\frac{1}{y} = t \Rightarrow - \frac{1}{y^{2}} \frac{d y}{d x} = \frac{d t}{d x}$
From equations (i)
$- \frac{d t}{d x} - t (t a n x) = - s e c x \Rightarrow \frac{d t}{d x} + (t a n x) t = s e c x$
I.F. $= e^{\int t a n x d x} = (e)^{l o g | s e c x |} s e c x$
Solution: $t (I . F) = \int (I . F) s e c x d x \Rightarrow \frac{1}{y} s e c x = t a n x + c$

Suggest Corrections

Similar questions

The solution of the differential equation is

[AISSE 1990]

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i)

(ii)

(iii)

(iv)

Q. Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y =

\frac{π}{2}

, when x =

\frac{π}{2}

Q. Show that y = e^x (A cos x + B sin x) is the solution of the differential equation

\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0

Q. Find the particular solution of differential equation :

\frac{d y}{d x} = - \frac{x + y cos x}{1 + sin x}

, given that

y = 1

when

x = 0

Solution of the differential equation cosxdy=y(sin x−y)dx, 0<x<π2 is

Solution of the differential equation
$c o s x d y = y (s i n x - y) d x, 0 < x < \frac{π}{2}$ is