Q.

If a curve y = y(x) passes through the point (1,π2) and satisfies the differential equation (7x4cot y–excosec y)dxdy=x5, x≥1, then at x = 2, the value of cos y is:          [2025]

1 2e2+e128  
2 2e2–e128  
3 2e2–e64  
4 2e2+e64  

Ans.

(2)

We have, (7x4cot y–excosec y)dxdy=x5

⇒ dydx=7x4cot yx5–excosec yx5

⇒ dydx-7xcot y=–exx5cosec y

⇒ sin y·dydx–cos y(7x)=–exx5

If –cos y=t ⇒ sin y·dydx=dtdx

⇒ dtdx+t(7x)=–exx5

Here, I.F.=x7

⇒ t·x7=–∫x2·exdx ⇒ cos y·x7=x2ex–2∫xexdx

⇒ cos y·x7=x2ex–2xex+2ex+c

at point (1,π2), we get c = –e

Thus, the value of cos y at x = 2 is

cos y=2e2–e128