Let y = y ( x ) be the solution of the differential equation d y   d x = 2 y + 2 sin x - 5 x - 2 cos x…

Let y=y(x) be the solution of the differential equation dy dx=2y+2sinx-5x-2cosx such that y(0)=7. Then y(π) is equal to
  1. 7eπ2+5
  2. eπ2+5
  3. 2eπ2+5
  4. 3eπ2+5

Solution

dydx-2xy=22sin x-5x-2cos x

IF=e-x2

So, y e-x2=∫e-x22x2sin x-5-2cos x dx

⇒y·e-x2=e-x25-2sinx+C

y=5-2 sinx+C·ex2

Given at x=0, y=7

⇒7=5+C

⇒C=2

So, y=5-2 sin x+2ex2

Now at x=π

y=5+2eπ2

Asked in: JEE Main 2021 (27 Aug Shift 1)

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