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Differential Equations question

2023 · 11 Apr · Shift 1 · Q29
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  5. /2023 · 11 Apr · Shift 1 · Q29

Differential Equations question

2023 · 11 Apr · Shift 1 · Q29

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be a solution curve of the differential equation. (1−x2y2)dx=ydx+xdy\left(1-x^{2} y^{2}\right) d x=y d x+x d y(1−x2y2)dx=ydx+xdy. If the line x=1x=1x=1 intersects the curve y=y(x)y=y(x)y=y(x) at y=2y=2y=2 and the line x=2x=2x=2 intersects the curve y=y(x)y=y(x)y=y(x) at y=αy=\alphay=α, then a value of α\alphaα is :
  1. A
    1+3e22(3e2−1)\frac{1+3 e^{2}}{2\left(3 e^{2}-1\right)}2(3e2−1)1+3e2​
  2. B
    3e22(3e2−1)\frac{3 e^{2}}{2\left(3 e^{2}-1\right)}2(3e2−1)3e2​
  3. C
    1−3e22(3e2+1)\frac{1-3 e^{2}}{2\left(3 e^{2}+1\right)}2(3e2+1)1−3e2​
  4. D
    3e22(3e2+1)\frac{3 e^{2}}{2\left(3 e^{2}+1\right)}2(3e2+1)3e2​
View written solutionFree

Correct answer: A

  1. Rewrite the differential equation

Given

(1−x2y2) dx=y dx+x dy.(1-x^2y^2)\,dx = y\,dx + x\,dy.(1−x2y2)dx=ydx+xdy.

Move terms together:

(1−x2y2−y) dx=x dy.(1-x^2y^2-y)\,dx = x\,dy.(1−x2y2−y)dx=xdy.

So

xdydx=1−y−x2y2.x\frac{dy}{dx}=1-y-x^2y^2.xdxdy​=1−y−x2y2.
  1. Use the substitution u=xyu=xyu=xy

Let

u=xy.u=xy.u=xy.

Then

dudx=xdydx+y.\frac{du}{dx}=x\frac{dy}{dx}+y.dxdu​=xdxdy​+y.

From the differential equation,

xdydx=1−y−x2y2.x\frac{dy}{dx}=1-y-x^2y^2.xdxdy​=1−y−x2y2.

Hence

dudx=(1−y−x2y2)+y=1−u2.\frac{du}{dx}=(1-y-x^2y^2)+y=1-u^2.dxdu​=(1−y−x2y2)+y=1−u2.

Therefore,

dudx=1−u2.\frac{du}{dx}=1-u^2.dxdu​=1−u2.
  1. Separate variables
du1−u2=dx.\frac{du}{1-u^2}=dx.1−u2du​=dx.

Integrate:

∫du1−u2=∫dx.\int \frac{du}{1-u^2}=\int dx.∫1−u2du​=∫dx.

Using

∫du1−u2=12ln⁡∣1+u1−u∣,\int \frac{du}{1-u^2}=\frac12\ln\left|\frac{1+u}{1-u}\right|,∫1−u2du​=21​ln​1−u1+u​​,

we get

12ln⁡∣1+u1−u∣=x+C.\frac12\ln\left|\frac{1+u}{1-u}\right|=x+C.21​ln​1−u1+u​​=x+C.

So

ln⁡∣1+u1−u∣=2x+C1.\ln\left|\frac{1+u}{1-u}\right|=2x+C_1.ln​1−u1+u​​=2x+C1​.

Thus,

1+u1−u=Ce2x.\frac{1+u}{1-u}=Ce^{2x}.1−u1+u​=Ce2x.
  1. Use the condition x=1,y=2x=1, y=2x=1,y=2

At x=1x=1x=1, y=2y=2y=2, so

u=xy=2.u=xy=2.u=xy=2.

Substitute into the solution:

1+21−2=Ce2⇒3−1=Ce2⇒C=−3e−2.\frac{1+2}{1-2}=Ce^2 \quad\Rightarrow\quad \frac{3}{-1}=Ce^2 \quad\Rightarrow\quad C=-3e^{-2}.1−21+2​=Ce2⇒−13​=Ce2⇒C=−3e−2.

Hence

1+u1−u=−3e2x−2.\frac{1+u}{1-u}=-3e^{2x-2}.1−u1+u​=−3e2x−2.
  1. Use the condition at x=2x=2x=2

At x=2x=2x=2, let y=αy=\alphay=α. Then

u=xy=2α.u=xy=2\alpha.u=xy=2α.

So

1+2α1−2α=−3e2.\frac{1+2\alpha}{1-2\alpha}=-3e^2.1−2α1+2α​=−3e2.

Now solve for α\alphaα:

1+2α=−3e2(1−2α)1+2\alpha=-3e^2(1-2\alpha)1+2α=−3e2(1−2α) 1+2α=−3e2+6e2α1+2\alpha=-3e^2+6e^2\alpha1+2α=−3e2+6e2α 1+3e2=2α(3e2−1)1+3e^2=2\alpha(3e^2-1)1+3e2=2α(3e2−1) α=1+3e22(3e2−1).\alpha=\frac{1+3e^2}{2(3e^2-1)}.α=2(3e2−1)1+3e2​.
  1. Match with the options
α=1+3e22(3e2−1)\alpha=\frac{1+3e^2}{2(3e^2-1)}α=2(3e2−1)1+3e2​

which is Option A.

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