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Differential Equations question

2021 · 27 Aug · Shift 1 · Q28
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  5. /2021 · 27 Aug · Shift 1 · Q28

Differential Equations question

2021 · 27 Aug · Shift 1 · Q28

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let us consider a curve, y = f(x) passing through the point (−-− 2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x2. Then :
  1. A
    x2+2xf(x)−12=0{x^2} + 2xf(x) - 12 = 0x2+2xf(x)−12=0
  2. B
    x3+xf(x)+12=0{x^3} + xf(x) + 12 = 0x3+xf(x)+12=0
  3. C
    x3−3xf(x)−4=0{x^3} - 3xf(x) - 4 = 0x3−3xf(x)−4=0
  4. D
    x2+2xf(x)+4=0{x^2} + 2xf(x) + 4 = 0x2+2xf(x)+4=0
View written solutionFree

Correct answer: C

  1. Given differential equation

We are told that the curve is y=f(x)y=f(x)y=f(x) and f(x)+xf′(x)=x2.f(x)+x f'(x)=x^2.f(x)+xf′(x)=x2. Also, it passes through (−2,2)(-2,2)(−2,2), so f(−2)=2.f(-2)=2.f(−2)=2.

  1. Recognize the left-hand side

Notice that ddx[xf(x)]=f(x)+xf′(x).\frac{d}{dx}[x f(x)] = f(x)+x f'(x).dxd​[xf(x)]=f(x)+xf′(x). So the differential equation becomes ddx[xf(x)]=x2.\frac{d}{dx}[x f(x)] = x^2.dxd​[xf(x)]=x2.

  1. Integrate both sides

Integrating with respect to xxx, xf(x)=∫x2 dx=x33+C.x f(x)=\int x^2\,dx = \frac{x^3}{3}+C.xf(x)=∫x2dx=3x3​+C. Hence, f(x)=x23+Cx.f(x)=\frac{x^2}{3}+\frac{C}{x}.f(x)=3x2​+xC​. But it is better to keep it as xf(x)=x33+C.x f(x)=\frac{x^3}{3}+C.xf(x)=3x3​+C.

  1. Use the point (−2,2)(-2,2)(−2,2)

Substitute x=−2x=-2x=−2 and f(−2)=2f(-2)=2f(−2)=2: (−2)(2)=(−2)33+C.(-2)(2)=\frac{(-2)^3}{3}+C.(−2)(2)=3(−2)3​+C. So, −4=−83+C-4=-\frac{8}{3}+C−4=−38​+C C=−4+83=−43.C=-4+\frac{8}{3}=-\frac{4}{3}.C=−4+38​=−34​.

Thus, xf(x)=x33−43.x f(x)=\frac{x^3}{3}-\frac{4}{3}.xf(x)=3x3​−34​. Multiply by 333: 3xf(x)=x3−4.3x f(x)=x^3-4.3xf(x)=x3−4. Rearrange: x3−3xf(x)−4=0.x^3-3x f(x)-4=0.x3−3xf(x)−4=0.

  1. Match with the options

This is exactly Option C: x3−3xf(x)−4=0.x^3-3x f(x)-4=0.x3−3xf(x)−4=0.

Therefore, the correct answer is C.

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