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Differential Equations question

2021 · 25 Feb · Shift 1 · Q29
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  5. /2021 · 25 Feb · Shift 1 · Q29

Differential Equations question

2021 · 25 Feb · Shift 1 · Q29

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is x2−4x+y+8x−2{{{x^2} - 4x + y + 8} \over {x - 2}}x−2x2−4x+y+8​, then this curve also passes through the point :
  1. A
    (4, 4)
  2. B
    (5, 5)
  3. C
    (5, 4)
  4. D
    (4, 5)
View written solutionFree

Correct answer: B

  1. Form the differential equation

The slope of the tangent is given by

dydx=x2−4x+y+8x−2.\frac{dy}{dx} = \frac{x^2-4x+y+8}{x-2}.dxdy​=x−2x2−4x+y+8​.

Rewrite the numerator:

x2−4x+y+8=(x2−4x+4)+y+4=(x−2)2+y+4.x^2-4x+y+8 = (x^2-4x+4) + y + 4 = (x-2)^2 + y + 4.x2−4x+y+8=(x2−4x+4)+y+4=(x−2)2+y+4.

So,

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

Thus,

dydx−1x−2y=x−2+4x−2.\frac{dy}{dx} - \frac{1}{x-2}y = x-2 + \frac{4}{x-2}.dxdy​−x−21​y=x−2+x−24​.
  1. Solve the linear differential equation

This is a linear differential equation of the form

dydx+P(x)y=Q(x),\frac{dy}{dx} + P(x)y = Q(x),dxdy​+P(x)y=Q(x),

with

P(x)=−1x−2.P(x) = -\frac{1}{x-2}.P(x)=−x−21​.

The integrating factor is

I.F.=e∫−1x−2 dx=e−ln⁡∣x−2∣=1∣x−2∣.\text{I.F.} = e^{\int -\frac{1}{x-2}\,dx} = e^{-\ln|x-2|} = \frac{1}{|x-2|}.I.F.=e∫−x−21​dx=e−ln∣x−2∣=∣x−2∣1​.

We may take

I.F.=1x−2.\text{I.F.} = \frac{1}{x-2}.I.F.=x−21​.

Multiplying the equation by 1x−2\frac{1}{x-2}x−21​:

1x−2dydx−y(x−2)2=1+4(x−2)2.\frac{1}{x-2}\frac{dy}{dx} - \frac{y}{(x-2)^2} = 1 + \frac{4}{(x-2)^2}.x−21​dxdy​−(x−2)2y​=1+(x−2)24​.

The left-hand side is

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

Hence,

ddx(yx−2)=1+4(x−2)2.\frac{d}{dx}\left(\frac{y}{x-2}\right) = 1 + \frac{4}{(x-2)^2}.dxd​(x−2y​)=1+(x−2)24​.

Integrating,

yx−2=∫(1+4(x−2)2)dx.\frac{y}{x-2} = \int \left(1 + \frac{4}{(x-2)^2}\right)dx.x−2y​=∫(1+(x−2)24​)dx.

Now,

∫1 dx=x,\int 1\,dx = x,∫1dx=x,

and

∫4(x−2)2 dx=4∫(x−2)−2dx=−4x−2.\int \frac{4}{(x-2)^2}\,dx = 4\int (x-2)^{-2}dx = -\frac{4}{x-2}.∫(x−2)24​dx=4∫(x−2)−2dx=−x−24​.

So,

yx−2=x−4x−2+C.\frac{y}{x-2} = x - \frac{4}{x-2} + C.x−2y​=x−x−24​+C.

Multiplying by (x−2)(x-2)(x−2),

y=x(x−2)−4+C(x−2).y = x(x-2) - 4 + C(x-2).y=x(x−2)−4+C(x−2).

That is,

y=x2−2x−4+Cx−2C.y = x^2 - 2x - 4 + Cx - 2C.y=x2−2x−4+Cx−2C.
  1. Use the condition that the curve passes through the origin

Since the curve passes through (0,0)(0,0)(0,0),

0=0−0−4+0−2C.0 = 0 - 0 - 4 + 0 - 2C.0=0−0−4+0−2C.

So,

−4−2C=0⇒C=−2.-4 - 2C = 0 \quad \Rightarrow \quad C = -2.−4−2C=0⇒C=−2.

Substitute back:

y=x2−2x−4−2x+4=x2−4x.y = x^2 - 2x - 4 - 2x + 4 = x^2 - 4x.y=x2−2x−4−2x+4=x2−4x.

Thus the curve is

y=x2−4x.y = x^2 - 4x.y=x2−4x.
  1. Check the given options

  • For (4,4)(4,4)(4,4): y=42−4⋅4=16−16=0≠4y = 4^2 - 4\cdot 4 = 16 - 16 = 0 \ne 4y=42−4⋅4=16−16=0=4 Not on the curve.

  • For (5,5)(5,5)(5,5): y=52−4⋅5=25−20=5y = 5^2 - 4\cdot 5 = 25 - 20 = 5y=52−4⋅5=25−20=5 So (5,5)(5,5)(5,5) lies on the curve.

  • For (5,4)(5,4)(5,4): y=25−20=5≠4y = 25 - 20 = 5 \ne 4y=25−20=5=4 Not on the curve.

  • For (4,5)(4,5)(4,5): y=16−16=0≠5y = 16 - 16 = 0 \ne 5y=16−16=0=5 Not on the curve.

  1. Conclusion

The curve also passes through

(5,5).(5,5).(5,5).

So the correct option is B.

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