Question

The solution of the differential equation e^–x (y + 1) dy + (cos² x – sin 2x)ydx = 0 subjected to the condition that y = 1, when x = 0 is ______.

पर्याय

• y + log y + e^x cos² x = 2

• log (y + 1) + e^x cos² x = 1

• y + log y = e^x cos² x

• (y + 1) + e^x cos² x = 2

Answer 1

The solution of the differential equation e^–x (y + 1) dy + (cos² x – sin 2x)ydx = 0 subjected to the condition that y = 1, when x = 0 is y + log y + e^x cos² x = 2.

Explanation:

Given equation can be rewritten as

`(1 + 1/y)dy` = –e^x (cos² x – sin 2x)dx

On integrating both sides, we get

y + log y = `-e^x cos^2 x + int e^x sin2x  dx - int e^x sin2x  dx + C`

`\implies` y + log y = –e^x cos² x + C

At x = 0 and y = 1,

1 + 0 = –e⁰ cos 0 + C

`\implies` C = 2  ...[given]

∴ Required solutions is

y + log y = –e^x cos² x + 2

`\implies` y + log y + e^x cos² x = 2