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Differential Equations question

2021 · 20 Jul · Shift 2 · Q40
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  5. /2021 · 20 Jul · Shift 2 · Q40

Differential Equations question

2021 · 20 Jul · Shift 2 · Q40

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let a curve y = y(x) be given by the solution of the differential equation cos⁡(12cos⁡−1(e−x))dx=e2x−1dy\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dycos(21​cos−1(e−x))dx=e2x−1​dy. If it intersects y-axis at y = −-− 1, and the intersection point of the curve with x-axis is (α\alphaα, 0), then e α\alphaα is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. Given differential equation

    cos⁡(12cos⁡−1(e−x))dx=e2x−1 dy\cos\left(\frac12\cos^{-1}(e^{-x})\right)dx=\sqrt{e^{2x}-1}\,dycos(21​cos−1(e−x))dx=e2x−1​dy

    So,

    dydx=cos⁡(12cos⁡−1(e−x))e2x−1.\frac{dy}{dx}=\frac{\cos\left(\frac12\cos^{-1}(e^{-x})\right)}{\sqrt{e^{2x}-1}}.dxdy​=e2x−1​cos(21​cos−1(e−x))​.

  2. Simplify the denominator

    e2x−1=ex1−e−2x.\sqrt{e^{2x}-1}=e^x\sqrt{1-e^{-2x}}.e2x−1​=ex1−e−2x​.

    Hence,

    dydx=cos⁡(12cos⁡−1(e−x))ex1−e−2x.\frac{dy}{dx}=\frac{\cos\left(\frac12\cos^{-1}(e^{-x})\right)}{e^x\sqrt{1-e^{-2x}}}.dxdy​=ex1−e−2x​cos(21​cos−1(e−x))​.

  3. Use half-angle identity

    Let

    θ=cos⁡−1(e−x).\theta=\cos^{-1}(e^{-x}).θ=cos−1(e−x).

    Then cos⁡θ=e−x\cos\theta=e^{-x}cosθ=e−x. Also,

    cos⁡θ2=1+cos⁡θ2=1+e−x2.\cos\frac\theta2=\sqrt{\frac{1+\cos\theta}{2}}= \sqrt{\frac{1+e^{-x}}{2}}.cos2θ​=21+cosθ​​=21+e−x​​.

    Therefore,

    dydx=1+e−x2ex1−e−2x.\frac{dy}{dx}=\frac{\sqrt{\frac{1+e^{-x}}{2}}}{e^x\sqrt{1-e^{-2x}}}.dxdy​=ex1−e−2x​21+e−x​​​.

  4. Simplify the expression

    Since

    1−e−2x=(1−e−x)(1+e−x),1-e^{-2x}=(1-e^{-x})(1+e^{-x}),1−e−2x=(1−e−x)(1+e−x),

    we get

    dydx=1+e−x2 ex(1−e−x)(1+e−x)=12 ex1−e−x.\frac{dy}{dx}=\frac{\sqrt{1+e^{-x}}}{\sqrt{2}\,e^x\sqrt{(1-e^{-x})(1+e^{-x})}} =\frac{1}{\sqrt{2}\,e^x\sqrt{1-e^{-x}}}.dxdy​=2​ex(1−e−x)(1+e−x)​1+e−x​​=2​ex1−e−x​1​.

  5. Substitute t=e−xt=e^{-x}t=e−x

    Let

    t=e−x⇒dt=−e−xdx=−t dxt=e^{-x}\quad\Rightarrow\quad dt=-e^{-x}dx=-t\,dxt=e−x⇒dt=−e−xdx=−tdx

    so

    dx=−dtt,ex=1t.dx=-\frac{dt}{t}, \qquad e^x=\frac1t.dx=−tdt​,ex=t1​.

    Then

    dydx=12 (1/t)1−t=t21−t.\frac{dy}{dx}=\frac{1}{\sqrt{2}\,(1/t)\sqrt{1-t}}=\frac{t}{\sqrt{2}\sqrt{1-t}}.dxdy​=2​(1/t)1−t​1​=2​1−t​t​.

    Thus,

    dy=t21−tdx=t21−t(−dtt)=−dt21−t.dy=\frac{t}{\sqrt{2}\sqrt{1-t}}dx =\frac{t}{\sqrt{2}\sqrt{1-t}}\left(-\frac{dt}{t}\right) =-\frac{dt}{\sqrt{2}\sqrt{1-t}}.dy=2​1−t​t​dx=2​1−t​t​(−tdt​)=−2​1−t​dt​.

  6. Integrate

    y=∫−dt21−t.y=\int -\frac{dt}{\sqrt{2}\sqrt{1-t}}.y=∫−2​1−t​dt​.

    Put u=1−tu=1-tu=1−t, then du=−dtdu=-dtdu=−dt:

    y=12∫u−1/2du=12⋅2u+C=21−t+C.y=\frac{1}{\sqrt{2}}\int u^{-1/2}du =\frac{1}{\sqrt{2}}\cdot 2\sqrt{u}+C =\sqrt{2}\sqrt{1-t}+C.y=2​1​∫u−1/2du=2​1​⋅2u​+C=2​1−t​+C.

    Since t=e−xt=e^{-x}t=e−x,

    y=21−e−x+C.y=\sqrt{2}\sqrt{1-e^{-x}}+C.y=2​1−e−x​+C.

  7. Use the condition on the y-axis

    The curve intersects the y-axis at x=0x=0x=0, and there y=−1y=-1y=−1.

    Now,

    y(0)=21−e0+C=20+C=C.y(0)=\sqrt{2}\sqrt{1-e^0}+C=\sqrt{2}\sqrt{0}+C=C.y(0)=2​1−e0​+C=2​0​+C=C.

    So,

    C=−1.C=-1.C=−1.

    Hence the curve is

    y=21−e−x−1.y=\sqrt{2}\sqrt{1-e^{-x}}-1.y=2​1−e−x​−1.

  8. Find the x-intercept

    At the x-axis, y=0y=0y=0. Let the point be (α,0)(\alpha,0)(α,0).

    Then

    0=21−e−α−10=\sqrt{2}\sqrt{1-e^{-\alpha}}-10=2​1−e−α​−1 21−e−α=1\sqrt{2}\sqrt{1-e^{-\alpha}}=12​1−e−α​=1 1−e−α=12\sqrt{1-e^{-\alpha}}=\frac{1}{\sqrt{2}}1−e−α​=2​1​ 1−e−α=121-e^{-\alpha}=\frac121−e−α=21​ e−α=12.e^{-\alpha}=\frac12.e−α=21​.

    Therefore,

    eα=2.e^{\alpha}=2.eα=2.

  9. Final answer

    2\boxed{2}2​
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