MathematicsMedium38×since 2002Q3503Let f(x)f(x)f(x) be a positive function such that the area bounded by y=f(x),y=0y=f(x), y=0y=f(x),y=0 from x=0x=0x=0 to x=a>0x=a>0x=a>0 is e−a+4a2+a−1e^{-a}+4 a^2+a-1e−a+4a2+a−1. Then the differential equation, whose general solution is y=c1f(x)+c2y=c_1 f(x)+c_2y=c1f(x)+c2, where c1c_1c1 and c2c_2c2 are arbitrary constants, isA(8ex+1)d2ydx2−dydx=0\left(8 e^x+1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0(8ex+1)dx2d2y−dxdy=0B(8ex+1)d2ydx2+dydx=0\left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0(8ex+1)dx2d2y+dxdy=0C(8ex−1)d2ydx2−dydx=0\left(8 e^x-1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0(8ex−1)dx2d2y−dxdy=0D(8ex−1)d2ydx2+dydx=0\left(8 e^x-1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0(8ex−1)dx2d2y+dxdy=0Check answerSkip