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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
View written solutionFree

Correct answer: B

  1. Use the given area function to find f(x)f(x)f(x)

The area under y=f(x)y=f(x)y=f(x) from x=0x=0x=0 to x=ax=ax=a is

∫0af(x) dx=e−a+4a2+a−1.\int_0^a f(x)\,dx = e^{-a}+4a^2+a-1.∫0a​f(x)dx=e−a+4a2+a−1.

Differentiate both sides with respect to aaa:

f(a)=dda(e−a+4a2+a−1).f(a)=\frac{d}{da}\left(e^{-a}+4a^2+a-1\right).f(a)=dad​(e−a+4a2+a−1).

So,

f(a)=−e−a+8a+1.f(a)=-e^{-a}+8a+1.f(a)=−e−a+8a+1.

Hence,

f(x)=8x+1−e−x.f(x)=8x+1-e^{-x}.f(x)=8x+1−e−x.
  1. Form the differential equation whose general solution is y=c1f(x)+c2y=c_1f(x)+c_2y=c1​f(x)+c2​

Given

y=c1f(x)+c2.y=c_1 f(x)+c_2.y=c1​f(x)+c2​.

Differentiate once:

y′=c1f′(x).y'=c_1 f'(x).y′=c1​f′(x).

Differentiate again:

y′′=c1f′′(x).y''=c_1 f''(x).y′′=c1​f′′(x).

Now compute derivatives of f(x)f(x)f(x):

f(x)=8x+1−e−x,f(x)=8x+1-e^{-x},f(x)=8x+1−e−x,

so

f′(x)=8+e−x,f'(x)=8+e^{-x},f′(x)=8+e−x,

and

f′′(x)=−e−x.f''(x)=-e^{-x}.f′′(x)=−e−x.

Thus,

y′=c1(8+e−x),y'=c_1(8+e^{-x}),y′=c1​(8+e−x), y′′=c1(−e−x).y''=c_1(-e^{-x}).y′′=c1​(−e−x).

Eliminate c1c_1c1​. From y′′=−c1e−xy''=-c_1 e^{-x}y′′=−c1​e−x,

c1=−exy′′.c_1=-e^x y''.c1​=−exy′′.

Substitute into y′y'y′:

y′=(−exy′′)(8+e−x).y'=(-e^x y'')(8+e^{-x}).y′=(−exy′′)(8+e−x).

Simplify:

y′=−(8ex+1)y′′.y'=-(8e^x+1)y''.y′=−(8ex+1)y′′.

Therefore,

(8ex+1)y′′+y′=0.(8e^x+1)y''+y'=0.(8ex+1)y′′+y′=0.

So the required differential equation is

(8ex+1)d2ydx2+dydx=0.\boxed{(8e^x+1)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0}.(8ex+1)dx2d2y​+dxdy​=0​.
  1. Match with the options

This is exactly Option B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, the derived answer agrees with the stored answer.

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