Let , be the solution of the differential equation xdy – ydx + xy (xdy + ydx) = 0, y(1) = 2. Then is equal to __________. [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let $α | x | = | y | e^{x y - β}$ , $α, β \in N$ be the solution of the differential equation xdy – ydx + xy (xdy + ydx) = 0, y(1) = 2. Then $α + β$ is equal to __________. [2024]

Ans.
(4)

We have, $α | x | = | y | e^{x y - β}$

and $\frac{x d y - y d x}{y^{2}} + \frac{x y (x d y + y d x)}{y^{2}} = 0$

$\Rightarrow - d (\frac{x}{y}) + \frac{x}{y} d (x y) = 0 \Rightarrow \int d (x y) = \int \frac{d (\frac{x}{y})}{x / y}$

$\Rightarrow x y = \ln | \frac{x}{y} | + \ln c \Rightarrow x y = \ln (| \frac{x}{y} | \cdot c)$ ... (i)

$∵$ y(1) = 2, i.e., x = 1, y = 2

$∴ From (i), 2 = \ln (| \frac{1}{2} | c) \Rightarrow c = 2 e^{2}$

$∴ x y = \ln (| \frac{x}{y} | \cdot 2 e^{2})$

$\Rightarrow e^{x y} = \frac{| x |}{| y |} \cdot 2 e^{2} \Rightarrow 2 | x | = | y | e^{x y - 2}$

Also, solution of equation is $α | x | = | y | e^{x y - β}$ (Given)

On comparing, $α = 2, β = 2$ . Hence, $α + β = 4$ .

Want Us to Email you About Special Offers & Updates?