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Differential Equations question

2024 · 30 Jan · Shift 2 · Q52
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  5. /2024 · 30 Jan · Shift 2 · Q52

Differential Equations question

2024 · 30 Jan · Shift 2 · Q52

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let Y=Y(X)Y=Y(X)Y=Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y−y=Y′(x)(X−x)Y-y=Y^{\prime}(x)(X-x)Y−y=Y′(x)(X−x) and the co-ordinate axes, where (x,y)(x, y)(x,y) is any point on the curve, is always −y22Y′(x)+1,Y′(x)eq0\frac{-y^2}{2 Y^{\prime}(x)}+1, Y^{\prime}(x) eq 02Y′(x)−y2​+1,Y′(x)eq0. If Y(1)=1Y(1)=1Y(1)=1, then 12Y(2)12 Y(2)12Y(2) equals ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 20

  1. Equation of the tangent and its intercepts

At any point (x,y)(x,y)(x,y) on the curve, the tangent is

Y−y=Y′(x)(X−x).Y-y=Y'(x)(X-x).Y−y=Y′(x)(X−x).

Let m=Y′(x)m=Y'(x)m=Y′(x). Then the tangent at (x,y)(x,y)(x,y) is

Y−y=m(X−x).Y-y=m(X-x).Y−y=m(X−x).

Rewriting,

Y=mX+(y−mx).Y=mX+(y-mx).Y=mX+(y−mx).

Its intercepts with the axes are:

  • YYY-intercept: put X=0X=0X=0 Y0=y−mx.Y_0=y-mx.Y0​=y−mx.
  • XXX-intercept: put Y=0Y=0Y=0 0−y=m(X0−x)  ⟹  X0=x−ym.0-y=m(X_0-x) \implies X_0=x-\frac{y}{m}.0−y=m(X0​−x)⟹X0​=x−my​.

So the area enclosed by this line and the coordinate axes is

A=12(x−ym)(y−mx).A=\frac12\left(x-\frac{y}{m}\right)(y-mx).A=21​(x−my​)(y−mx).
  1. Use the given area condition

We are given that this area is always

−y22Y′(x)+1=1−y22m.\frac{-y^2}{2Y'(x)}+1=1-\frac{y^2}{2m}.2Y′(x)−y2​+1=1−2my2​.

Hence,

12(x−ym)(y−mx)=1−y22m.\frac12\left(x-\frac{y}{m}\right)(y-mx)=1-\frac{y^2}{2m}.21​(x−my​)(y−mx)=1−2my2​.

Multiply by 222:

(x−ym)(y−mx)=2−y2m.\left(x-\frac{y}{m}\right)(y-mx)=2-\frac{y^2}{m}.(x−my​)(y−mx)=2−my2​.

Now expand the left side:

(x−ym)(y−mx)=xy−mx2−y2m+xy=2xy−mx2−y2m.\left(x-\frac{y}{m}\right)(y-mx)=xy-mx^2-\frac{y^2}{m}+xy =2xy-mx^2-\frac{y^2}{m}.(x−my​)(y−mx)=xy−mx2−my2​+xy=2xy−mx2−my2​.

So,

2xy−mx2−y2m=2−y2m.2xy-mx^2-\frac{y^2}{m}=2-\frac{y^2}{m}.2xy−mx2−my2​=2−my2​.

The terms −y2m-\dfrac{y^2}{m}−my2​ cancel, giving

2xy−mx2=2.2xy-mx^2=2.2xy−mx2=2.

Thus,

m=2xy−2x2=2(xy−1)x2.m=\frac{2xy-2}{x^2}=\frac{2(xy-1)}{x^2}.m=x22xy−2​=x22(xy−1)​.

Since m=dydxm=\dfrac{dy}{dx}m=dxdy​, the differential equation is

dydx=2(xy−1)x2=2yx−2x2.\frac{dy}{dx}=\frac{2(xy-1)}{x^2}=\frac{2y}{x}-\frac{2}{x^2}.dxdy​=x22(xy−1)​=x2y​−x22​.
  1. Solve the differential equation

Rewrite as

dydx−2xy=−2x2.\frac{dy}{dx}-\frac{2}{x}y=-\frac{2}{x^2}.dxdy​−x2​y=−x22​.

This is a linear differential equation.

Its integrating factor is

I.F.=e∫−2/x dx=e−2ln⁡x=x−2.\text{I.F.}=e^{\int -2/x\,dx}=e^{-2\ln x}=x^{-2}.I.F.=e∫−2/xdx=e−2lnx=x−2.

Multiplying throughout by x−2x^{-2}x−2,

x−2dydx−2x−3y=−2x−4.x^{-2}\frac{dy}{dx}-2x^{-3}y=-2x^{-4}.x−2dxdy​−2x−3y=−2x−4.

The left side is

ddx(yx−2)=−2x−4.\frac{d}{dx}\left(yx^{-2}\right)=-2x^{-4}.dxd​(yx−2)=−2x−4.

Integrate:

yx−2=∫−2x−4 dx=23x−3+C.yx^{-2}=\int -2x^{-4}\,dx=\frac{2}{3}x^{-3}+C.yx−2=∫−2x−4dx=32​x−3+C.

Hence,

y=23x+Cx2.y=\frac{2}{3x}+Cx^2.y=3x2​+Cx2.
  1. Use the initial condition Y(1)=1Y(1)=1Y(1)=1

Given y(1)=1y(1)=1y(1)=1,

1=23+C  ⟹  C=13.1=\frac{2}{3}+C \implies C=\frac13.1=32​+C⟹C=31​.

Therefore,

Y(x)=23x+x23.Y(x)=\frac{2}{3x}+\frac{x^2}{3}.Y(x)=3x2​+3x2​.
  1. Find Y(2)Y(2)Y(2) and then 12Y(2)12Y(2)12Y(2)
Y(2)=23⋅2+223=13+43=53.Y(2)=\frac{2}{3\cdot 2}+\frac{2^2}{3}=\frac13+\frac43=\frac53.Y(2)=3⋅22​+322​=31​+34​=35​.

Thus,

12Y(2)=12⋅53=20.12Y(2)=12\cdot \frac53=20.12Y(2)=12⋅35​=20.
  1. Comparison with stored answer

The derived answer is 202020, which matches the stored correct answer.

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