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NCERT Exemplar · Class 12 Mathematics Differential Equations

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EXERCISE 9.3 51–60 (part 6 of 8)

  1. Choose the correct answer from the given four options in each of the Exercises from $\displaystyle 34$ to $\displaystyle 75$ (M.C.Q)

    Exercise 51

    The degree of the differential equation d2ydx2+(dydx)3+6y5=0\displaystyle \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3+6 y^5=0 is :
    (A)
    1\displaystyle 1 (B) 2\displaystyle 2 (C) 3\displaystyle 3 (D) 5\displaystyle 5

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    NCERT’s answer
    A
    (A) \(\displaystyle 1\): the highest-order derivative \(\displaystyle \frac{d^2y}{dx^2}\) occurs to the first power; the other powers do not matter. \[\frac{d^2y}{dx^2} = -\left(\frac{dy}{dx}\right)^3 - 6y^5 \]

  2. Exercise 52

    The solution of dydx+y=e−x,y(0)=0\displaystyle \frac{d y}{d x}+y=e^{-x}, y(0)=0 is :
    (A)
    y=ex(x−1)\displaystyle y=e^x(x-1)
    (B)
    y=xe−x\displaystyle y=x e^{-x}
    (C)
    y=xe−x+1\displaystyle y=x e^{-x}+1
    (D)
    y=(x+1)e−x\displaystyle y=(x+1) e^{-x}

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    NCERT’s answer
    B
    (B) \(\displaystyle y = xe^{-x}\): linear equation, solved with the integrating factor. \[\text{I.F.} = e^{\int 1\,dx} = e^x \] \[\frac{d}{dx}\left(ye^x\right) = e^{-x}\cdot e^x = 1 \] \[ye^x = x + C \] \[y(0) = 0 \;\Rightarrow\; 0 = 0 + C \;\Rightarrow\; C = 0 \] \[y = xe^{-x} \]

  3. Exercise 53

    Integrating factor of the differential equation dydx+ytan⁡x−sec⁡x=0\displaystyle \frac{d y}{d x}+y \tan x-\sec x=0 is:
    (A)
    cos⁡x\displaystyle \cos x
    (B)
    sec⁡x\displaystyle \sec x
    (C)
    ecos⁡x\displaystyle e^{\cos x}
    (D)
    esec⁡x\displaystyle e^{\sec x}

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    NCERT’s answer
    B
    (B) \(\displaystyle \sec x\) \[\frac{dy}{dx} + (\tan x)\,y = \sec x \] \[\text{I.F.} = e^{\int \tan x\,dx} = e^{\ln|\sec x|} = \sec x \]

  4. Exercise 54

    The solution of the differential equation dydx=1+y21+x2\displaystyle \frac{d y}{d x}=\frac{1+y^2}{1+x^2} is:
    (A)
    y=tan⁡−1x\displaystyle y=\tan ^{-1} x
    (B)
    y−x=k(1+xy)\displaystyle y-x=k(1+x y)
    (C)
    x=tan⁡−1y\displaystyle x=\tan ^{-1} y
    (D)
    tan⁡(xy)=k\displaystyle \tan (x y)=k

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    NCERT’s answer
    B
    (B) \(\displaystyle y - x = k(1+xy)\): separate the variables and combine the two inverse tangents. \[\frac{dy}{1+y^2} = \frac{dx}{1+x^2} \] \[\tan^{-1}y - \tan^{-1}x = C \] \[\tan^{-1}\frac{y-x}{1+xy} = C \] \[\frac{y-x}{1+xy} = \tan C = k \] \[y - x = k(1+xy) \]

  5. Exercise 55

    The integrating factor of the differential equation dydx+y=1+yx\displaystyle \frac{d y}{d x}+y=\frac{1+y}{x} is:
    (A)
    xex\displaystyle \frac{x}{e^x}
    (B)
    exx\displaystyle \frac{e^x}{x}
    (C)
    xex\displaystyle x e^x
    (D)
    ex\displaystyle e^x

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    NCERT’s answer
    B
    (B) \(\displaystyle \dfrac{e^x}{x}\): write the equation in the form \(\displaystyle \frac{dy}{dx}+Py=Q\) first. \[\frac{dy}{dx} + y = \frac{1}{x} + \frac{y}{x} \] \[\frac{dy}{dx} + \left(1 - \frac{1}{x}\right)y = \frac{1}{x} \] \[\text{I.F.} = e^{\int\left(1-\frac1x\right)dx} = e^{x-\ln x} = \frac{e^x}{x} \]

  6. Exercise 56

    y=aemx+be−mx\displaystyle y=a e^{m x}+b e^{-m x} satisfies which of the following differential equation?
    (A)
    dydx+my=0\displaystyle \frac{d y}{d x}+m y=0
    (B)
    dydx−my=0\displaystyle \frac{d y}{d x}-m y=0
    (C)
    d2ydx2−m2y=0\displaystyle \frac{d^2 y}{d x^2}-m^2 y=0
    (D)
    d2ydx2+m2y=0\displaystyle \frac{d^2 y}{d x^2}+m^2 y=0

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    NCERT’s answer
    C
    (C) \(\displaystyle \frac{d^2y}{dx^2} - m^2y = 0\): differentiate twice and compare with \(\displaystyle y\). \[\frac{dy}{dx} = mae^{mx} - mbe^{-mx} \] \[\frac{d^2y}{dx^2} = m^2ae^{mx} + m^2be^{-mx} = m^2y \] \[\frac{d^2y}{dx^2} - m^2y = 0 \]

  7. Exercise 57

    The solution of the differential equation cos⁡xsin⁡ydx+sin⁡xcos⁡ydy=0\displaystyle \cos x \sin y d x+\sin x \cos y d y=0 is :
    (A)
    sin⁡xsin⁡y=c\displaystyle \frac{\sin x}{\sin y}=c
    (B)
    sin⁡xsin⁡y=c\displaystyle \sin x \sin y=c
    (C)
    sin⁡x+sin⁡y=c\displaystyle \sin x+\sin y=c
    (D)
    cos⁡xcos⁡y=c\displaystyle \cos x \cos y=c

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    NCERT’s answer
    B
    (B) \(\displaystyle \sin x \sin y = c\). Divide by \(\displaystyle \sin x \sin y\) and the variables separate.\[\frac{\cos x}{\sin x}\,dx + \frac{\cos y}{\sin y}\,dy = 0 \] \[\log|\sin x| + \log|\sin y| = \log|c| \] \[\sin x \sin y = c \]

  8. Exercise 58

    The solution of xdydx+y=ex\displaystyle x \frac{d y}{d x}+y=e^x is:
    (A)
    y=exx+kx\displaystyle y=\frac{e^x}{x}+\frac{k}{x}
    (B)
    y=xex+cx\displaystyle y=x e^x+c x
    (C)
    y=xex+k\displaystyle y=x e^x+k
    (D)
    x=eyy+ky\displaystyle x=\frac{e^y}{y}+\frac{k}{y}

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    NCERT’s answer
    A
    (A) \(\displaystyle y=\dfrac{e^x}{x}+\dfrac{k}{x}\). The left side is an exact derivative.\[x\frac{dy}{dx} + y = \frac{d}{dx}(xy) = e^x \] \[xy = \int e^x\,dx = e^x + k \] \[y = \frac{e^x}{x} + \frac{k}{x} \]

  9. Exercise 59

    The differential equation of the family of curves x2+y2−2ay=0\displaystyle x^2+y^2-2 a y=0, where a\displaystyle a is arbitrary constant, is:
    (A)
    (x2−y2)dydx=2xy\displaystyle \left(x^2-y^2\right) \frac{d y}{d x}=2 x y
    (B)
    2(x2+y2)dydx=xy\displaystyle 2\left(x^2+y^2\right) \frac{d y}{d x}=x y
    (C)
    2(x2−y2)dydx=xy\displaystyle 2\left(x^2-y^2\right) \frac{d y}{d x}=x y
    (D)
    (x2+y2)dydx=2xy\displaystyle \left(x^2+y^2\right) \frac{d y}{d x}=2 x y

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    NCERT’s answer
    A
    (A) \(\displaystyle (x^2-y^2)\dfrac{dy}{dx}=2xy\). Differentiate once and eliminate \(\displaystyle a\).\[x^2 + y^2 = 2ay \] \[2x + 2y\frac{dy}{dx} = 2a\frac{dy}{dx} \] \[2a = \frac{x^2+y^2}{y} \] \[2x + 2y\frac{dy}{dx} = \frac{x^2+y^2}{y}\cdot\frac{dy}{dx} \] \[2xy = \left(x^2 + y^2 - 2y^2\right)\frac{dy}{dx} \] \[\left(x^2 - y^2\right)\frac{dy}{dx} = 2xy \]

  10. Exercise 60

    Family y=Ax+A3\displaystyle y=\mathrm{A} x+\mathrm{A}^3 of curves will correspond to a differential equation of order
    (A)
    3\displaystyle 3 (B) 2\displaystyle 2 (C) 1\displaystyle 1 (D) not defined

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    NCERT’s answer
    C
    (C) \(\displaystyle 1\). The family has one arbitrary constant \(\displaystyle A\), so eliminating it needs one differentiation.\[y = Ax + A^3 \] \[\frac{dy}{dx} = A \] \[y = x\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^3 \]