Q.

Let α|x|=|y|exy–β, α, β∈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|exy–β

and xdy–ydxy2+xy(xdy+ydx)y2=0

⇒ –d(xy)+xyd(xy)=0 ⇒ ∫d(xy)=∫d(xy)x/y

⇒ xy=ln|xy|+ln c ⇒ xy=ln(|xy|·c)          ... (i)

∵  y(1) = 2, i.e., x = 1, y = 2

∴  From (i), 2=ln(|12|c) ⇒ c=2e2

∴  xy=ln(|xy|·2e2)

⇒ exy=|x||y|·2e2 ⇒ 2|x|=|y|exy–2

Also, solution of equation is α|x|=|y|exy–β          (Given)

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