Let y = y(x) be the solution of the differential equation, xy' − y = x²(xcos x + sin x), x > 0. If y(π) = π, then y''(π/2) + y(π/2) is equal to:
Let y = y(x) be the solution of the differential equation, xy' − y = x²(xcos x + sin x), x > 0. If y(π) = π, then y''(π/2) + y(π/2) is equal to:
Option 1 -
1 + π²/2
Option 2 -
2 + π²/4
Option 3 -
2 + π/2
Option 4 -
1 + π²/4
Asked by Shiksha User
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1 Answer
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Correct Option - 3
Detailed Solution:x dy/dx - y = x² (xcosx + sinx), x > 0
dy/dx - y/x = x (xcosx+sinx) ⇒ dy/dx + Py = QSo, I.F. = e^ (∫-1/x dx) = 1/|x| = 1/x (x > 0)
Thus, y/x = ∫ 1/x (x (xcosx+sinx)dx
⇒ y/x = xsinx + C
? y (π) = π ⇒ C = 1
So, y = x²sinx + x ⇒ (y)? /? = π²/4 + π/2
Also, dy/dx = x²cosx + 2xsinx + 1
⇒ d²y/dx² = -x²sinx + 4xcosx + 2sinx
⇒ [d²y/dx²] at π/2 = -π²/4 + 2
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