Differential Equations - Test Papers

 CBSE Test Paper 01

Chapter 9 Differential Equations

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1. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge2 = 0.6931).
   1. 9.93%
   2. 7.93%
   3. 6.93%
   4. 8.93%
2. General solution ofcos2xdydx+y=tanx(0⩽x<π2)$co{s}^{2}x\frac{dy}{dx}+y=\tan x\left(0⩽x<\frac{\pi }{2}\right)$ is 
   1. y=(tanx−1)+Ce−tanx$y=\left(\tan x-1\right)+C{e}^{-\tan x}$
   2. y=(tanx+1)+Ce−tanx$y=\left(\tan x+1\right)+C{e}^{-\tan x}$
   3. y=(tanx+1)−Ce−tanx$y=\left(\tan x+1\right)-C{e}^{-\tan x}$
   4. y=(tanx−1)−Ce−tanx$y=\left(\tan x-1\right)-C{e}^{-\tan x}$
3. The number of arbitrary constants in the general solution of a differential equation of fourth order are: 
   1. 3
   2. 2
   3. 1
   4. 4
4. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs1000 is deposited with this bank, how much will it worth after 10 years (e0.5= 1.648).$\left(e^{0.5}=1.648\right).$ 
   1. Rs 1848
   2. Rs 1648
   3. Rs 1748
   4. Rs 1948
5. What is the order of differential equation : d3ydx3+d2ydx2+(dydx)2=ex$\frac{d^{3}y}{d{x}^3}+\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=e^x$. 
   1. 2
   2. 3
   3. 1
   4. 0
6. F(x, y) = x2+y2√+yx$\frac{\sqrt{x^2+y^2}+y}{x}$ is a homogeneous function of degree ________.
7. The degree of the differential equation 1+(dydx)2−−−−−−−−√=x$\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=x$ is ________.
8. The order of the differential equation of all circles of given radius a is ________.
9. Verify that the function is a solution of the corresponding differential equation y=xsinx;xy,=y+xx2−y2−−−−−−√$y=x\sin x;x{y}^{,}=y+x\sqrt{x^2-y^2}$. 
10. Find order and degree. d4ydx2+sin(y′′′)=0$\frac{d^{4}y}{d{x}^2}+\sin \left(y^{‴}\right)=0$. 
11. Write the solution of the differential equation dydx=2−y$\frac{dy}{dx}=2^{-y}$. 
12. Verify that the given function (explicit) is a solution of the corresponding differential equation: y = x2 + 2x + C : y' - 2x - 2 = 0. 
13. Find the differential equation of all non-horizontal lines in a plane. 
14. Verify that the function is a solution of the corresponding differential equation
    y=1+x2−−−−−√;y′=xy1+x2$y=\sqrt{1+x^2};y^{\prime }=\frac{xy}{1+x^2}$. 
15. Solve the following differential equation.
    (y+3x2)dxdy=x$\left(y+3x^2\right)\frac{dx}{dy}=x$

     
16. Solve the differential equation (1 + y2) tan-1x dx + 2y (1 + x2) dy = 0. 
17. Find the particular solution of the differential equation (1 + e2x)dy + (1 + y2)ex dx = 0, given that y = 1, when x = 0. 
18. Solve (1+exy)dx+exy(1−xy)dy=0$\left(1+e^{\frac{x}{y}}\right)dx+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)dy=0$. 

CBSE Test Paper 01
Chapter 9 Differential Equations

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Solution

1. 3. 6.93%
      Explanation: Let P be the principal at any time t. then,
      dPdt=rP100⇒dPdt=P100$\frac{dP}{dt}=\frac{rP}{100}\Rightarrow \frac{dP}{dt}=\frac{P}{100}$
      ⇒∫1PdP=∫r100dt$\Rightarrow \int \frac{1}{P}dP=\int \frac{r}{100}dt$
      ⇒logP=r100t+logc$\Rightarrow \log P=\frac{r}{100}t+\log c$
      ⇒logPc=r100t$\Rightarrow \log \frac{P}{c}=\frac{r}{100}t$
      ⇒P=cer100$\Rightarrow P=c{e}^{\frac{r}{100}}$
      When P = 100 and t = 0., then, c = 100, therefore, we have:
      ⇒P=100 er╱100$\Rightarrow P=100e^{{}^r╱{}_{100}}$
      Now, let t = T, when P = 100., then;
      ⇒200=100eT100$\Rightarrow 200=100e^{\frac{T}{100}}$
      ⇒eT100=2$\Rightarrow e^{\frac{T}{100}}=2$
      ⇒T=100log2$\Rightarrow T=100\log 2$ = 100(0.6931) = 6.93%
2. 1. y=(tanx−1)+Ce−tanx$y=\left(\tan x-1\right)+C{e}^{-\tan x}$
      Explanation: dydx+sec2x.y=tanx.sec2x⇒P=sec2x,Q=tanx.sec2x$\frac{dy}{dx}+{\sec }^{2}x.y=\tan x.{\sec }^{2}x\Rightarrow P={\sec }^{2}x,Q=\tan x.{\sec }^{2}x$
      ⇒I.F.=e∫sec2xdx=etanx$\Rightarrow I.F.=e^{{\int }_{}^{}{\sec }^{2}xdx}=e^{\tan x}$
      ⇒y.etanx=∫tanxsec2xetanxdx⇒y.etanx=(tanx−1)etanx+C$\Rightarrow y.e^{\tan x}={\int }_{}^{}\tan x{\sec }^{2}x{e}^{\tan x}dx\Rightarrow y.e^{\tan x}=(\tan x-1)e^{\tan x}+C$
      ⇒y=(tanx−1)+Ce−tanx$\Rightarrow y=(\tan x-1)+C{e}^{-\tan x}$
3. 4. 4
      Explanation: 4, because the no. of arbitrary constants is equal to order of the differential equation.
4. 2. Rs 1648
      Explanation: Here P is the principal at time t
      dPdt=5P100⇒dPdt=P20$\frac{dP}{dt}=\frac{5P}{100}\Rightarrow \frac{dP}{dt}=\frac{P}{20}$
      ⇒∫1PdP=∫120dt$\Rightarrow {\int }_{}^{}\frac{1}{P}dP={\int }_{}^{}\frac{1}{20}dt$
      ⇒logP=120t+logc$\Rightarrow \log P=\frac{1}{20}t+\log c$
      ⇒logPc=120t$\Rightarrow \log \frac{P}{c}=\frac{1}{20}t$
      ⇒P=ce1100$\Rightarrow P=c{e}^{\frac{1}{100}}$
      When P = 1000 and t = 0 ., then ,
      c = 1000, therefore, we have :
      ⇒P=1000eT100$\Rightarrow P=1000e^{\frac{T}{100}}$
      ⇒A=1000e510$\Rightarrow A=1000e^{\frac{5}{10}}$
      ⇒e510=A$\Rightarrow e^{\frac{5}{10}}=A$
      ⇒A=1000log0.5$\Rightarrow A=1000\log 0.5$
      = 1000(1.648)
      = 1648
5. 1. 3
      Explanation: Order = 3. Since the third derivative is the highest derivative present in the equation. i.e.d3ydx3$\frac{d^{3}y}{d{x}^3}$
6. Zero
7. not defined
8. 2
9. y=x.sinx$y=x.\sin x$...(1)
   y,=x.cosx+sinx.1$y^{,}=x.\cos x+\sin x.1$
   ⇒xy,=x2cosx+x.sinx$\Rightarrow x{y}^{,}=x^2\cos x+x.\sin x$
   xy,=x21−sin2x−−−−−−−√+x.sinx$x{y}^{,}=x^2\sqrt{1-{\sin }^{2}x}+x.\sin x$
   xy,=x21−(yx)2−−−−−−−√+x.sinx$x{y}^{,}=x^2\sqrt{1-{\left(\frac{y}{x}\right)}^2}+x.\sin x$ [∵yx=sinx]$[\because \frac{y}{x}=\sin x]$
   xy,=x2x2−y2√x+x.sinx$x{y}^{,}=x^2\frac{\sqrt{x^2-y^2}}{x}+x.\sin x$
   xy,=xx2−y2−−−−−−√+y$x{y}^{,}=x\sqrt{x^2-y^2}+y$
   Hence proved.
10. order = 4 ,degree = not defined
11. Given differential equation is
    dydx=2−y$\frac{dy}{dx}=2^{-y}$
    on separating the variables, we get
    2ydy = dx
    On integrating both sides, we get
    ∫2ydy=∫dx$\int 2^{y}dy=\int dx$
    ⇒2ylog2=x+C1$\Rightarrow \frac{2^y}{\log 2}=x+C_1$
    ⇒$\Rightarrow$ 2y = x log 2 + C1 log 2
    ∴$\therefore$ 2y = x log 2 + C, where C = C1 log 2
12. Given: y = x2 + 2x + C ...(i)
    To prove: y is a solution of the differential equation y' - 2x - 2 = 0 ...(ii)
    Proof:From, eq. (i),
    y' = 2x + 2
    L.H.S. of eq. (ii),
    = y' - 2x - 2
    = (2x + 2) - 2x - 2
    = 2x + 2 - 2x - 2 = 0 = R.H.S.
    Hence, y given by eq. (i) is a solution of y' - 2x - 2 = 0.
13. The general equation of all non-horizontal lines in a plane is ax + by = c, where a≠0$a\ne 0$.
    differentiating both sides w.r.t. y on both sides,we get
    adydx+b=0$a\frac{dy}{dx}+b=0$
    Again, differentiating both sides w.r.t. y, we get
    ad2xdy2=0⇒d2xdy2=0.$a\frac{d^{2}x}{d{y}^2}=0\Rightarrow \frac{d^{2}x}{d{y}^2}=0.$
14. y=1+x2−−−−−√$y=\sqrt{1+x^2}$ ......(i)
    y′=121+x2√.2x$y^{\prime }=\frac{1}{2\sqrt{1+x^2}}.2x$ ......(ii)
    (ii)÷(i)$(ii)÷(i)$,we get,
    ⇒y′y=x1+x2√1+x2√$\Rightarrow \frac{y^{\prime }}{y}=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+x^2}}$
    ⇒y′y=x1+x2$\Rightarrow \frac{y^{\prime }}{y}=\frac{x}{1+x^2}$
    y′=xy1+x2$y^{\prime }=\frac{xy}{1+x^2}$
    Hence given value of y is the solution of given differential equation.
15. According to the question,we have to solve the differential equation ,
    (y+3x2)dxdy=x⇒dydx=yx+3x$\left(y+3x^2\right)\frac{dx}{dy}=x\Rightarrow \frac{dy}{dx}=\frac{y}{x}+3x$
    ⇒dydx−yx=3x$\Rightarrow \frac{dy}{dx}-\frac{y}{x}=3x$
    which is a linear differential equation of the form
    dydx+Py=Q$\frac{dy}{dx}+Py=Q$.
    Here, P=−1x$P=\frac{-1}{x}$and Q = 3x
    ∴IF=e∫Pdx=e∫−1xdx=e−log|x|=elogx−1=x−1$\therefore \mathrm{I}\mathrm{F}=e^{\int Pdx}=e^{\int -\frac{1}{x}dx}=e^{-\log |x|}=e^{\log x^{-1}}=x^{-1}$
    ⇒IF=x−1=1x$\Rightarrow \mathrm{I}\mathrm{F}=x^{-1}=\frac{1}{x}$
    The solution of linear differential equation is given by
    y×IF=∫(Q×IF)dx+C$y\times IF=\int (Q\times IF)dx+C$
    ⇒y×1x=∫(3x×1x)dx+C$\Rightarrow y\times \frac{1}{x}=\int \left(3x\times \frac{1}{x}\right)dx+C$
    ⇒yx=∫3dx+C⇒yx=3x+C$\Rightarrow \frac{y}{x}=\int 3dx+C\Rightarrow \frac{y}{x}=3x+C$
    ∴y=3x2+Cx$\therefore y=3x^2+Cx$
    which is the required solution.
16. Given differential equation is
    (1 + y2) tan-1x dx + 2y (1 + x2) dy = 0
    ⇒(1+y2)tan−1xdx=−2y(1+x2)dy$\Rightarrow \left(1+y^2\right){\tan }^{-1}xdx=-2y\left(1+x^2\right)dy$
    ⇒tan−1xdx1+x2=−2y1+y2dy$\Rightarrow \frac{{\tan }^{-1}xdx}{1+x^2}=-\frac{2y}{1+y^2}dy$
    On integrating both sides, we get
    ∫tan−1x1+x2dx=−∫2y1+y2dy$\int \frac{{\tan }^{-1}x}{1+x^2}dx=-\int \frac{2y}{1+y^2}dy$
    Put tan−1x=tinLHS,${\tan }^{-1}x=tinLHS,$ we get
    11+x2dx=dt$\frac{1}{1+x^2}dx=dt$
    and put 1 + y2 = u in RHS, we get
    2ydy = du
    ⇒∫tdt=−∫1u⇒t22=−logu+C$\Rightarrow \int tdt=-\int \frac{1}{u}\Rightarrow \frac{t^2}{2}=-\log u+C$
    ⇒12(tan−1x)2=−log(1+y2)+C$\Rightarrow \frac{1}{2}{\left({\tan }^{-1}x\right)}^2=-\log \left(1+y^2\right)+C$
    ⇒12(tan−1x)2+log(1+y2)=C$\Rightarrow \frac{1}{2}{\left({\tan }^{-1}x\right)}^2+\log \left(1+y^2\right)=C$
17. Given differential equation is,
    (1 + e2x)dy + (1 + y2)ex dx = 0
    Above equation may be written as
    dy1+y2=−ex1+e2xdx$\frac{dy}{1+y^2}=\frac{-e^x}{1+e^{2x}}dx$
    On integrating both sides, we get
    ∫dy1+y2=−∫ex1+e2xdx$\int \frac{dy}{1+y^2}=-\int \frac{e^x}{1+e^{2x}}dx$
    On putting ex = t ⇒$\Rightarrow$ ex dx = dt in RHS, we get
    tan−1y=−∫11+t2dt${\tan }^{-1}y=-\int \frac{1}{1+t^2}dt$
    ⇒tan−1y=−tan−1t+C$\Rightarrow {\tan }^{-1}y=-{\tan }^{-1}t+C$
    ⇒tan−1y=−tan−1(ex)+C$\Rightarrow {\tan }^{-1}y=-{\tan }^{-1}\left(e^x\right)+C$ ...(i) [put t = ex]
    Also, given that y = 1, when x = 0.
    On putting above values in Eq. (i), we get
    tan-11 = -tan-1(e0) + C
    ⇒tan−11=−tan−11+C[∵e0=1]$\Rightarrow {\tan }^{-1}1=-{\tan }^{-1}1+C\left[\because e^0=1\right]$
    ⇒2tan−11=C$\Rightarrow 2{\tan }^{-1}1=C$
    ⇒2tan−1(tanπ4)=C$\Rightarrow 2{\tan }^{-1}\left(\tan \frac{\pi }{4}\right)=C$
    ⇒C=2×π4=π2$\Rightarrow C=2\times \frac{\pi }{4}=\frac{\pi }{2}$
    On putting C=π2$C=\frac{\pi }{2}$ in Eq. (i), we get
    tan−1y=−tan−1ex+π2${\tan }^{-1}y=-{\tan }^{-1}{e}^x+\frac{\pi }{2}$
    ⇒y=tan[π2−tan−1(ex)]=cot[tan−1(ex)]$\Rightarrow y=\tan \left[\frac{\pi }{2}-{\tan }^{-1}\left(e^x\right)\right]=\cot \left[{\tan }^{-1}\left(e^x\right)\right]$
    =cot[cot−1(1ex)][∵tan−1x=cot−11x]$=\cot \left[{\cot }^{-1}\left(\frac{1}{e^x}\right)\right]\left[\because {\tan }^{-1}x={\cot }^{-1}\frac{1}{x}\right]$
    ∴y=1ex$\therefore y=\frac{1}{e^x}$
    which is the required solution.
18. (1+exy)dx+exy(1−xy)dy=0$\left(1+e^{\frac{x}{y}}\right)dx+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)dy=0$
    ⇒dxdy=−ex/y(1−xy)1+ex/y$\Rightarrow \frac{dx}{dy}=-\frac{e^{x/y}\left(1-\frac{x}{y}\right)}{1+e^{x/y}}$
    ⇒dxdy=ex/y(xy−1)1+ex/y$\Rightarrow \frac{dx}{dy}=\frac{e^{x/y}\left(\frac{x}{y}-1\right)}{1+e^{x/y}}$........(1)
    Let x = vy, then,
    dxdy=v+ydvdy$\frac{dx}{dy}=v+y\frac{dv}{dy}$
    Put dxdy$\frac{dx}{dy}$ in eq (1),we get,
    v+ydvdy=ev(v−1)ev+1$v+y\frac{dv}{dy}=\frac{e^v(v-1)}{e^v+1}$
    ⇒ydvdy=vev−evev+1−v$\Rightarrow y\frac{dv}{dy}=\frac{v{e}^v-e^v}{e^v+1}-v$
    ⇒ydvdy=vev−ev−vev−vev+1$\Rightarrow y\frac{dv}{dy}=\frac{v{e}^v-e^v-v{e}^v-v}{e^v+1}$
    ⇒−∫dyy=∫ev+1v+evdv$\Rightarrow -\int \frac{dy}{y}=\int \frac{e^v+1}{v+e^v}dv$
    ⇒log(ev+v)=−log(y)+c$\Rightarrow \log (e^v+v)=-\log (y)+c$
    ⇒log((ev+v).y)=c$\Rightarrow \log ((e^v+v).y)=c$
    ⇒(ev+v)y=ec$\Rightarrow (e^v+v)y=e^c$
    ⇒(ev+v)y=A$\Rightarrow (e^v+v)y=A$ [Putting ec = A]
    ⇒(ex/y+xy)y=A$\Rightarrow \left(e^{x/y}+\frac{x}{y}\right)y=A$
    ⇒yex/y+x=A