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NCERT Exemplar · Class 12 Mathematics Differential Equations

77 questions · 77 still being checked

EXERCISE 9.3 71–77 (part 8 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 71

    General solution of dydx+ytan⁡x=sec⁡x\displaystyle \frac{d y}{d x}+y \tan x=\sec x is :
    (A)
    ysec⁡x=tan⁡x+c\displaystyle y \sec x=\tan x+c
    (B)
    ytan⁡x=sec⁡x+c\displaystyle y \tan x=\sec x+c
    (C)
    tan⁡x=ytan⁡x+c\displaystyle \tan x=y \tan x+c
    (D)
    xsec⁡x=tan⁡y+c\displaystyle x \sec x=\tan y+c

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    NCERT’s answer
    A
    (A) \(\displaystyle y\sec x=\tan x+c\)Linear equation; integrating factor:\[\text{I.F.} = e^{\int \tan x\,dx} = e^{\log|\sec x|} = \sec x \] \[\frac{d}{dx}\left(y\sec x\right) = \sec x\cdot\sec x = \sec^2 x \] \[y\sec x = \tan x + c \]

  2. Exercise 72

    Solution of the differential equation dydx+yx=sin⁡x\displaystyle \frac{d y}{d x}+\frac{y}{x}=\sin x is :
    (A)
    x(y+cos⁡x)=sin⁡x+c\displaystyle x(y+\cos x)=\sin x+c
    (B)
    x(y−cos⁡x)=sin⁡x+c\displaystyle x(y-\cos x)=\sin x+c
    (C)
    xycos⁡x=sin⁡x+c\displaystyle x y \cos x=\sin x+c
    (D)
    x(y+cos⁡x)=cos⁡x+c\displaystyle x(y+\cos x)=\cos x+c

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    NCERT’s answer
    A
    (A) \(\displaystyle x(y+\cos x)=\sin x+c\)Linear equation; integrating factor:\[\text{I.F.} = e^{\int \frac{1}{x}dx} = e^{\log x} = x \] \[\frac{d}{dx}(xy) = x\sin x \] \[xy = \int x\sin x\,dx = -x\cos x + \int \cos x\,dx = -x\cos x + \sin x + c \] \[x\,(y + \cos x) = \sin x + c \]

  3. Exercise 73

    The general solution of the differential equation (ex+1)ydy=(y+1)exdx\displaystyle \left(e^x+1\right) y d y=(y+1) e^x d x is:
    (A)
    (y+1)=k(ex+1)\displaystyle (y+1)=k\left(e^x+1\right)
    (B)
    y+1=ex+1+k\displaystyle y+1=e^x+1+k
    (C)
    y=log⁡{k(y+1)(ex+1)}\displaystyle y=\log \left\{k(y+1)\left(e^x+1\right)\right\}
    (D)
    y=log⁡{ex+1y+1}+k\displaystyle y=\log \left\{\frac{e^x+1}{y+1}\right\}+k

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    NCERT’s answer
    C
    (C) \(\displaystyle y=\log\{k(y+1)(e^x+1)\}\) — the variables separate.\[\frac{y}{y+1}\,dy=\frac{e^x}{e^x+1}\,dx \]\[\int\left(1-\frac{1}{y+1}\right)dy=\int\frac{e^x}{e^x+1}\,dx \]\[y-\log|y+1|=\log\left(e^x+1\right)+\log k \]\[y=\log\left\{k(y+1)\left(e^x+1\right)\right\} \]

  4. Exercise 74

    The solution of the differential equation dydx=ex−y+x2e−y\displaystyle \frac{d y}{d x}=e^{x-y}+x^2 e^{-y} is :
    (A)
    y=ex−y−x2e−y+c\displaystyle y=e^{x-y}-x^2 e^{-y}+c
    (B)
    ey−ex=x33+c\displaystyle e^y-e^x=\frac{x^3}{3}+c
    (C)
    ex+ey=x33+c\displaystyle e^x+e^y=\frac{x^3}{3}+c
    (D)
    ex−ey=x33+c\displaystyle e^x-e^y=\frac{x^3}{3}+c

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    NCERT’s answer
    B
    (B) \(\displaystyle e^y-e^x=\dfrac{x^3}{3}+c\) — factor out \(\displaystyle e^{-y}\), then separate.\[\frac{dy}{dx}=e^{-y}\left(e^x+x^2\right) \]\[e^y\,dy=\left(e^x+x^2\right)dx \]\[e^y=e^x+\frac{x^3}{3}+c \]\[e^y-e^x=\frac{x^3}{3}+c \]

  5. Exercise 75

    The solution of the differential equation dydx+2xy1+x2=1(1+x2)2\displaystyle \frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{1}{\left(1+x^2\right)^2} is :
    (A)
    y(1+x2)=c+tan⁡−1x\displaystyle y\left(1+x^2\right)=c+\tan ^{-1} x
    (B)
    y1+x2=c+tan⁡−1x\displaystyle \frac{y}{1+x^2}=c+\tan ^{-1} x
    (C)
    ylog⁡(1+x2)=c+tan⁡−1x\displaystyle y \log \left(1+x^2\right)=c+\tan ^{-1} x
    (D)
    y(1+x2)=c+sin⁡−1x\displaystyle y\left(1+x^2\right)=c+\sin ^{-1} x

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    NCERT’s answer
    A
    (A) \(\displaystyle y\left(1+x^2\right)=c+\tan^{-1}x\) — linear in \(\displaystyle y\), so use the integrating factor.\[\text{I.F.}=e^{\int\frac{2x}{1+x^2}dx}=e^{\log(1+x^2)}=1+x^2 \]\[y\left(1+x^2\right)=\int\frac{1+x^2}{\left(1+x^2\right)^2}\,dx \]\[y\left(1+x^2\right)=\int\frac{dx}{1+x^2}=\tan^{-1}x+c \]

  6. Exercise 76

    Fill in the blanks of the following (i to xi)
    (i)
    The degree of the differential equation d2ydx2+edydx=0\displaystyle \frac{d^2 y}{d x^2}+e^{\frac{d y}{d x}}=0 is ____.
    (ii)
    The degree of the differential equation 1+(dydx)2=x\displaystyle \sqrt{1+\left(\frac{d y}{d x}\right)^2}=x is ____.
    (iii)
    The number of arbitrary constants in the general solution of a differential equation of order three is ____.
    (iv)
    dydx+yxlog⁡x=1x\displaystyle \frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x} is an equation of the type ____.
    (v)
    General solution of the differential equation of the type dxdy+P1x=Q1\displaystyle \frac{d x}{d y}+\mathrm{P}_1 x=\mathrm{Q}_1 is given by ____.
    (vi)
    The solution of the differential equation xdydx+2y=x2\displaystyle \frac{x d y}{d x}+2 y=x^2 is ____.
    (vii)
    The solution of (1+x2)dydx+2xy−4x2=0\displaystyle \left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0 is ____.
    (viii)
    The solution of the differential equation ydx+(x+xy)dy=0\displaystyle y d x+(x+x y) d y=0 is ____.
    (ix)
    General solution of dydx+y=sin⁡x\displaystyle \frac{d y}{d x}+y=\sin x is ____.
    (x)
    The solution of differential equation cot⁡ydx=xdy\displaystyle \cot y d x=x d y is ____.
    (xi)
    The integrating factor of dydx+y=1+yx\displaystyle \frac{d y}{d x}+y=\frac{1+y}{x} is ____.

    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.

    (i)
    not defined
    \[e^{\frac{dy}{dx}}=1+\frac{dy}{dx}+\frac{1}{2!}\left(\frac{dy}{dx}\right)^2+\cdots \quad \text{(not a polynomial in } \tfrac{dy}{dx}\text{)} \]
    (ii)
    $\displaystyle 2$
    \[1+\left(\frac{dy}{dx}\right)^2=x^2 \quad \text{(squaring)} \]
    (iii)
    $\displaystyle 3$ — a general solution has as many arbitrary constants as the order.
    (iv)
    linear (first order)
    \[\frac{dy}{dx}+Py=Q,\qquad P=\frac{1}{x\log x},\quad Q=\frac1x \]
    (v)
    \(\displaystyle x\,e^{\int P_1dy}=\int Q_1\,e^{\int P_1dy}\,dy+c\)
    \[\text{I.F.}=e^{\int P_1\,dy} \]
    (vi)
    \(\displaystyle y=\dfrac{x^2}{4}+\dfrac{c}{x^2}\)
    \[\frac{dy}{dx}+\frac2x\,y=x,\qquad \text{I.F.}=e^{\int\frac2x dx}=x^2 \]
    \[x^2y=\int x^3\,dx=\frac{x^4}{4}+c \]
    (vii)
    \(\displaystyle y=\dfrac{4x^3}{3\left(1+x^2\right)}+\dfrac{c}{1+x^2}\)
    \[\frac{d}{dx}\left[\left(1+x^2\right)y\right]=\left(1+x^2\right)\frac{dy}{dx}+2xy=4x^2 \]
    \[\left(1+x^2\right)y=\int4x^2\,dx=\frac{4x^3}{3}+c \]
    (viii)
    \(\displaystyle \log|xy|+y=c\)
    \[y\,dx+x(1+y)\,dy=0 \]
    \[\frac{dx}{x}+\frac{1+y}{y}\,dy=0 \]
    \[\log|x|+\log|y|+y=c \]
    (ix)
    \(\displaystyle y=\dfrac{\sin x-\cos x}{2}+c\,e^{-x}\)
    \[\text{I.F.}=e^{\int dx}=e^x \]
    \[y\,e^x=\int e^x\sin x\,dx=\frac{e^x(\sin x-\cos x)}{2}+c \]
    (x)
    \(\displaystyle x\cos y=c\)
    \[\frac{dx}{x}=\tan y\,dy \]
    \[\log|x|=-\log|\cos y|+\log c \]
    (xi)
    \(\displaystyle \dfrac{e^x}{x}\)
    \[\frac{dy}{dx}+\left(1-\frac1x\right)y=\frac1x \]
    \[\text{I.F.}=e^{\int\left(1-\frac1x\right)dx}=e^{x-\log x}=\frac{e^x}{x} \]

  7. Exercise 77

    State True or False for the following:
    (i)
    Integrating factor of the differential of the form dxdy+p1x=Q1\displaystyle \frac{d x}{d y}+p_1 x=\mathrm{Q}_1 is given by e∫p1dy\displaystyle e^{\int p_1 d y}.
    (ii)
    Solution of the differential equation of the type dxdy+p1x=Q1\displaystyle \frac{d x}{d y}+p_1 x=\mathrm{Q}_1 is given by x\displaystyle x.I.F. =(I.F)×Q1dy\displaystyle =(\mathrm{I.F}) \times \mathrm{Q}_1 d y.
    (iii)
    Correct substitution for the solution of the differential equation of the type dydx=f(x,y)\displaystyle \frac{d y}{d x}=f(x, y), where f(x,y)\displaystyle f(x, y) is a homogeneous function of zero degree is y=vx\displaystyle y=v x.
    (iv)
    Correct substitution for the solution of the differential equation of the type dxdy=g(x,y)\displaystyle \frac{d x}{d y}=g(x, y) where g(x,y)\displaystyle g(x, y) is a homogeneous function of the degree zero is x=vy\displaystyle x=v y.
    (v)
    Number of arbitrary constants in the particular solution of a differential equation of order two is two.
    (vi)
    The differential equation representing the family of circles x2+(y−a)2=a2\displaystyle x^2+(y-a)^2=a^2 will be of order two.
    (vii)
    The solution of dydx=(yx)13\displaystyle \frac{d y}{d x}=\left(\frac{y}{x}\right)^{\frac{1}{3}} is y23−x23=c\displaystyle y^{\frac{2}{3}}-x^{\frac{2}{3}}=c.
    (viii)
    Differential equation representing the family of curves y=ex(Acos⁡x+Bsin⁡x)\displaystyle y=e^x(\mathrm{A} \cos x+\mathrm{B} \sin x) is d2ydx2−2dydx+2y=0\displaystyle \frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0
    (ix)
    The solution of the differential equation dydx=x+2yx\displaystyle \frac{d y}{d x}=\frac{x+2 y}{x} is x+y=kx2\displaystyle x+y=k x^2.
    (x)
    Solution of xdydx=y+xtan⁡yx\displaystyle \frac{x d y}{d x}=y+x \tan \frac{y}{x} is sin⁡(yx)=cx\displaystyle \sin \left(\frac{y}{x}\right)=c x
    (xi)
    The differential equation of all non horizontal lines in a plane is d2xdy2=0\displaystyle \frac{d^2 x}{d y^2}=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
    (i)
    True
    (ii)
    True
    (iii)
    True
    (iv)
    True
    (v)
    False
    (vi)
    False
    (vii)
    True
    (viii)
    True
    (ix)
    True
    (x)
    True
    (xi)
    True
    (i)
    True — multiplying by \(\displaystyle e^{\int p_1dy}\) makes the left side a derivative.
    \[\frac{d}{dy}\left(x\,e^{\int p_1dy}\right)=e^{\int p_1dy}\left(\frac{dx}{dy}+p_1x\right) \]
    (ii)
    True
    \[\frac{d}{dy}\left(x\cdot\text{I.F.}\right)=\text{I.F.}\times Q_1 \]
    \[x\cdot\text{I.F.}=\int\left(\text{I.F.}\right)Q_1\,dy \quad (+\,c) \]
    (iii)
    True
    \[y=vx\;\Rightarrow\;\frac{dy}{dx}=v+x\frac{dv}{dx},\qquad f(x,vx)=f(1,v) \]
    (iv)
    True
    \[x=vy\;\Rightarrow\;\frac{dx}{dy}=v+y\frac{dv}{dy},\qquad g(vy,y)=g(v,1) \]
    (v)
    False — a particular solution has every constant fixed, so it has none.
    (vi)
    False — only \(\displaystyle a\) is arbitrary, so the order is one.
    \[x^2+y^2-2ay=0\;\Rightarrow\;x+y\frac{dy}{dx}-a\frac{dy}{dx}=0 \]
    \[\left(x^2-y^2\right)\frac{dy}{dx}=2xy \]
    (vii)
    True
    \[y^{-\frac13}\,dy=x^{-\frac13}\,dx \]
    \[\frac32\,y^{\frac23}=\frac32\,x^{\frac23}+c' \]
    \[y^{\frac23}-x^{\frac23}=c \]
    (viii)
    True — put \(\displaystyle u=A\cos x+B\sin x\), so \(\displaystyle u''=-u\).
    \[y=e^xu,\qquad y'=e^x(u+u'),\qquad y''=e^x\left(u+2u'+u''\right)=2e^xu' \]
    \[y''=2\left(y'-y\right)\;\Rightarrow\;\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0 \]
    (ix)
    True
    \[\frac{dy}{dx}-\frac2x\,y=1,\qquad \text{I.F.}=\frac1{x^2} \]
    \[\frac{y}{x^2}=\int\frac{dx}{x^2}=-\frac1x+k \]
    \[x+y=kx^2 \]
    (x)
    True
    \[y=vx:\qquad x\frac{dv}{dx}=\tan v \]
    \[\int\cot v\,dv=\int\frac{dx}{x}\;\Rightarrow\;\log\sin v=\log x+\log c \]
    \[\sin\frac yx=cx \]
    (xi)
    True — every non-horizontal line is \(\displaystyle x=my+c\).
    \[\frac{dx}{dy}=m\;\Rightarrow\;\frac{d^2x}{dy^2}=0 \]