∫0af(x)dx=e−a+4a2+a−1f(a)=−e−a+8a+1f(x)=−e−x+8x+1 Now y=C1f(x)+C2 dxdy=C1f′(x)=C1(e−x+8) ...(1) dx2d2y=−C1e−x⇒−exdx2d2y Put in equation (1) dxdy=−exdx2d2y(e−x+8)(8ex+1)dx2d2y+dxdy=0
Let f(x) be a positive function such that the area bounded by y=f(x),y=0 from x=0 to x=a>0 is e−a+4a2+a−1. Then the differential equation, whose general solution is y=c1f(x)+c2, where c1 and c2 are arbitrary constants, is
Held on 8 Apr 2024 · Verified 6 Jul 2026.
(8ex−1)dx2d2y+dxdy=0
(8ex−1)dx2d2y−dxdy=0
(8ex+1)dx2d2y−dxdy=0
(8ex+1)dx2d2y+dxdy=0
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