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Differential Equations question

2024 · 8 Apr · Shift 1 · Q36
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  5. /2024 · 8 Apr · Shift 1 · Q36

Differential Equations question

2024 · 8 Apr · Shift 1 · Q36

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let 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\gt 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=c1​f(x)+c2​, where c1c_1c1​ and c2c_2c2​ are arbitrary constants, is
  1. A
    (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​=0
  2. B
    (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​=0
  3. C
    (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​=0
  4. D
    (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​=0
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