SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 12 Mathematics Differential Equations

77 questions · 77 still being checked

EXERCISE 9.3 41–50 (part 5 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 41

    Solution of the differential equation tan⁡ysec⁡2xdx+tan⁡xsec⁡2ydy=0\displaystyle \tan y \sec ^2 x d x+\tan x \sec ^2 y d y=0 is :
    (A)
    tan⁡x+tan⁡y=k\displaystyle \tan x+\tan y=k
    (B)
    tan⁡x−tan⁡y=k\displaystyle \tan x-\tan y=k
    (C)
    tan⁡xtan⁡y=k\displaystyle \frac{\tan x}{\tan y}=k
    (D)
    tan⁡x.tan⁡y=k\displaystyle \tan x . \tan y=k

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    D
    (D) \(\displaystyle \tan x \cdot \tan y = k\)\[\frac{\sec^2 x}{\tan x}\,dx + \frac{\sec^2 y}{\tan y}\,dy = 0 \quad \text{(divide by } \tan x\tan y\text{)} \] \[\log|\tan x| + \log|\tan y| = \log k \] \[\tan x \cdot \tan y = k \]

  2. Exercise 42

    Family y=Ax+A3\displaystyle y=\mathrm{A} x+\mathrm{A}^3 of curves is represented by the differential equation of degree :
    (A)
    1\displaystyle 1 (B) 2\displaystyle 2 (C) 3\displaystyle 3 (D) 4\displaystyle 4

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (C) \(\displaystyle 3\)\[y = Ax + A^3 \;\Rightarrow\; \frac{dy}{dx} = A \] \[y = x\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^3 \] The highest derivative is \(\displaystyle \frac{dy}{dx}\), raised to the power $\displaystyle 3$: order $\displaystyle 1$, degree 3.

  3. Exercise 43

    Integrating factor of xdydx−y=x4−3x\displaystyle \frac{x d y}{d x}-y=x^4-3 x is :
    (A)
    x\displaystyle x (B) log⁡x\displaystyle \log x
    (C)
    1x\displaystyle \frac{1}{x}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac{1}{x}\)\[\frac{dy}{dx} - \frac{1}{x}\,y = x^3 - 3 \] \[\text{I.F.} = e^{\int -\frac{1}{x}\,dx} = e^{-\log x} = \frac{1}{x} \]

  4. Exercise 44

    Solution of dydx−y=1,y(0)=1\displaystyle \frac{d y}{d x}-y=1, y(0)=1 is given by
    (A)
    xy=−ex\displaystyle x y=-e^x
    (B)
    xy=−e−x\displaystyle x y=-e^{-x}
    (C)
    xy=−1\displaystyle x y=-1
    (D)
    y=2ex−1\displaystyle y=2 e^x-1

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    D
    (D) \(\displaystyle y = 2e^x - 1\)\[\text{I.F.} = e^{\int -1\,dx} = e^{-x} \] \[y\,e^{-x} = \int e^{-x}\,dx = -e^{-x} + C \;\Rightarrow\; y = Ce^{x} - 1 \] \[y(0) = 1 \;\Rightarrow\; 1 = C - 1 \;\Rightarrow\; C = 2 \] \[y = 2e^x - 1 \]

  5. Exercise 45

    The number of solutions of dydx=y+1x−1\displaystyle \frac{d y}{d x}=\frac{y+1}{x-1} when y(1)=2\displaystyle y(1)=2 is :
    (A)
    none
    (B)
    one
    (C)
    two
    (D)
    infinite

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (A) none\[\frac{dy}{y+1} = \frac{dx}{x-1} \] \[\log|y+1| = \log|x-1| + \log|C| \;\Rightarrow\; y + 1 = C(x-1) \] \[x = 1,\; y = 2: \quad 3 = C\cdot 0 = 0 \] This is impossible for every \(\displaystyle C\), so no solution passes through \(\displaystyle (1, 2)\).

  6. Exercise 46

    Which of the following is a second order differential equation?
    (A)
    (y′)2+x=y2\displaystyle \left(y^{\prime}\right)^2+x=y^2
    (B)
    y′y′+y=sin⁡x\displaystyle y^{\prime} y^{\prime}+y=\sin x
    (C)
    y′′+(y′)2+y=0\displaystyle y^{\prime \prime}+\left(y^{\prime}\right)^2+y=0
    (D)
    y′=y2\displaystyle y^{\prime}=y^2

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (C) \(\displaystyle y'' + (y')^2 + y = 0\)It is the only equation containing \(\displaystyle y''\). In (A), (B) and (D) the highest derivative is \(\displaystyle y'\), so they are first order.

  7. Exercise 47

    Integrating factor of the differential equation (1−x2)dydx−xy=1\displaystyle \left(1-x^2\right) \frac{d y}{d x}-x y=1 is
    (A)
    −x\displaystyle -x (B) x1+x2\displaystyle \frac{x}{1+x^2}
    (C)
    1−x2\displaystyle \sqrt{1-x^2}
    (D)
    12log⁡(1−x2)\displaystyle \frac{1}{2} \log \left(1-x^2\right)

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    C
    (C) \(\displaystyle \sqrt{1-x^2}\)\[\frac{dy}{dx} - \frac{x}{1-x^2}\,y = \frac{1}{1-x^2} \] \[\int -\frac{x}{1-x^2}\,dx = \frac{1}{2}\log(1-x^2) \] \[\text{I.F.} = e^{\frac{1}{2}\log(1-x^2)} = \sqrt{1-x^2} \]

  8. Exercise 48

    tan⁡−1x+tan⁡−1y=c\displaystyle \tan ^{-1} x+\tan ^{-1} y=c is the general solution of the differential equation:
    (A)
    dydx=1+y21+x2\displaystyle \frac{d y}{d x}=\frac{1+y^2}{1+x^2}
    (B)
    dydx=1+x21+y2\displaystyle \frac{d y}{d x}=\frac{1+x^2}{1+y^2}
    (C)
    (1+x2)dy+(1+y2)dx=0\displaystyle \left(1+x^2\right) d y+\left(1+y^2\right) d x=0
    (D)
    (1+x2)dx+(1+y2)dy=0\displaystyle \left(1+x^2\right) d x+\left(1+y^2\right) d y=0

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    C
    (C) \(\displaystyle (1+x^2)\,dy + (1+y^2)\,dx = 0\)\[\tan^{-1}x + \tan^{-1}y = c \] \[\frac{1}{1+x^2} + \frac{1}{1+y^2}\,\frac{dy}{dx} = 0 \] \[(1+y^2)\,dx + (1+x^2)\,dy = 0 \]

  9. Exercise 49

    The differential equation ydydx+x=c\displaystyle y \frac{d y}{d x}+x=c represents :
    (A)
    Family of hyperbolas
    (B)
    Family of parabolas
    (C)
    Family of ellipses
    (D)
    Family of circles

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    D
    (D) Family of circles: separating the variables and integrating gives a circle equation. \[y\,dy = (c-x)\,dx \] \[\frac{y^2}{2} = cx - \frac{x^2}{2} + k \] \[x^2 + y^2 - 2cx = 2k \] \[(x-c)^2 + y^2 = c^2 + 2k \]

  10. Exercise 50

    The general solution of excos⁡ydx−exsin⁡ydy=0\displaystyle e^x \cos y d x-e^x \sin y d y=0 is :
    (A)
    excos⁡y=k\displaystyle e^x \cos y=k
    (B)
    exsin⁡y=k\displaystyle e^x \sin y=k
    (C)
    ex=kcos⁡y\displaystyle e^x=k \cos y
    (D)
    ex=ksin⁡y\displaystyle e^x=k \sin y

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    A
    (A) \(\displaystyle e^x\cos y = k\): the left side is an exact differential. \[d\left(e^x\cos y\right) = e^x\cos y\,dx - e^x\sin y\,dy \] \[d\left(e^x\cos y\right) = 0 \] \[e^x\cos y = k \]