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Differential Equations question

2019 · 12 Jan · Shift 2 · Q26
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  5. /2019 · 12 Jan · Shift 2 · Q26

Differential Equations question

2019 · 12 Jan · Shift 2 · Q26

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as x2−2yx{{{x^2} - 2y} \over x}xx2−2y​, then the curve also passes through the point :
  1. A
    (–1, 2)
  2. B
    (−2,1)\left( { - \sqrt 2 ,1} \right)(−2​,1)
  3. C
    (3,0)\left( { \sqrt 3 ,0} \right)(3​,0)
  4. D
    (3, 0)
View written solutionFree

Correct answer: C

  1. Form the differential equation

The slope is given by

dydx=x2−2yx.\frac{dy}{dx} = \frac{x^2 - 2y}{x}.dxdy​=xx2−2y​.

So,

dydx=x−2yx.\frac{dy}{dx} = x - \frac{2y}{x}.dxdy​=x−x2y​.

Rearrange into linear form:

dydx+2xy=x.\frac{dy}{dx} + \frac{2}{x}y = x.dxdy​+x2​y=x.
  1. Solve the linear differential equation

This is a first-order linear differential equation:

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

with

P(x)=2x,Q(x)=x.P(x)=\frac{2}{x}, \qquad Q(x)=x.P(x)=x2​,Q(x)=x.

The integrating factor is

I.F.=e∫2x dx=e2ln⁡∣x∣=x2.\text{I.F.} = e^{\int \frac{2}{x}\,dx} = e^{2\ln|x|} = x^2.I.F.=e∫x2​dx=e2ln∣x∣=x2.

Multiplying the equation by x2x^2x2:

x2dydx+2xy=x3.x^2\frac{dy}{dx} + 2xy = x^3.x2dxdy​+2xy=x3.

Left-hand side is a derivative:

ddx(x2y)=x3.\frac{d}{dx}(x^2 y) = x^3.dxd​(x2y)=x3.

Integrate:

x2y=∫x3 dx=x44+C.x^2 y = \int x^3\,dx = \frac{x^4}{4} + C.x2y=∫x3dx=4x4​+C.

Hence,

y=x24+Cx2.y = \frac{x^2}{4} + \frac{C}{x^2}.y=4x2​+x2C​.
  1. Use the given point (1,−2)(1,-2)(1,−2)

Substitute x=1x=1x=1, y=−2y=-2y=−2:

−2=14+C.-2 = \frac{1}{4} + C.−2=41​+C.

So,

C=−2−14=−94.C = -2 - \frac14 = -\frac94.C=−2−41​=−49​.

Thus the curve is

y=x24−94x2.y = \frac{x^2}{4} - \frac{9}{4x^2}.y=4x2​−4x29​.
  1. Check the options

We test which point satisfies

y=x24−94x2.y = \frac{x^2}{4} - \frac{9}{4x^2}.y=4x2​−4x29​.
  • Option A: (−1,2)(-1,2)(−1,2)

y = \frac{(-1)^2}{4} - \frac{9}{4(-1)^2} = \frac14 - \frac94 = -2 \neq 2.

Not correct. - **Option B:** $\left(-\sqrt2,1\right)$

y = \frac{2}{4} - \frac{9}{8} = \frac12 - \frac98 = -\frac58 \neq 1.

Not correct. - **Option C:** $\left(\sqrt3,0\right)$

y = \frac{3}{4} - \frac{9}{12} = \frac34 - \frac34 = 0.

Correct. - **Option D:** $(3,0)$

y = \frac{9}{4} - \frac{9}{36} = \frac94 - \frac14 = 2 \neq 0.

Notcorrect.5.∗∗Conclusion∗∗ThecurvealsopassesthroughNot correct. 5. **Conclusion** The curve also passes throughNotcorrect.5.∗∗Conclusion∗∗Thecurvealsopassesthrough

\boxed{\left(\sqrt3,0\right)}.

Sothecorrectoptionis∗∗C∗∗.So the correct option is **C**.Sothecorrectoptionis∗∗C∗∗.
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