CBSE 12th Standard Maths Subject Differential Equations Case Study Questions 2021
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CBSE 12th Standard Maths Differential Equations Case Study Questions 2021
12th Standard CBSE
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Reg.No. :
Maths
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It is known that, if the interest is compounded continuously, the principal changes at the rate equal to the produd' of the rate of bank interest per annum and the principal. Let P denotes the principal at any time t and rate of interest be r % per annum.
Based on the above information, answer the following questions.
(i) Find the value of \(\frac{d P}{d t}\) .(a) \(\frac{\operatorname{Pr}}{1000}\) (b) \(\frac{P r}{100}\) (c) \(\frac{\operatorname{Pr}}{10}\) (d) Pr (ii) fPo be the initial principal, then find the solution of differential equation formed in given situation.
(a) \(\log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100}\) (b) \(\log \left(\frac{P}{P_{0}}\right)=\frac{r t}{10}\) (c) \(\log \left(\frac{P}{P_{0}}\right)=r t\) (d) \(\log \left(\frac{P}{P_{0}}\right)=100 r t\) (iii) If the interest is compounded continuously at 5% per annum, in how many years will Rs. 100 double itself?
(a) 12.728 years (b) 14.789 years (c) 13.862 years (d) 15.872 years (iv) At what interest rate will Rs.100 double itself in 10 years? (log e2 = 0.6931).
(a) 9.66% (b) 8.239% (c) 7.341% (d) 6.931% (v) How much will Rs. 1000 be worth at 5% interest after 10 years? (e0.5 = 1.648).
(a) Rs. 1648 (b) Rs. 1500 (c) Rs. 1664 (d) Rs. 1572 -
A rumour on whatsapp spreads in a population of 5000 people at a rate proportional to the product of the number of people who have heard it and the number of people who have not. Also, it is given that 100 people initiate the rumour and a total of 500 people know the rumour after 2 days.
Based on the above information, answer the following questions
(i) If yet) denote the number of people who know the rumour at an instant t, then maximum value of yet) is(a) 500 (b) 100 (c) 5000 (d) none of these (ii) \(\frac{d y}{d t}\) is proptional to
(a) (y - 5000) (b) y(y - 500) (c) y(500 - y) (d) y(5000 - y) (iii) The value of y(0) is
(a) 100 (b) 500 (c) 600 (d) 200 (iv) The value of y(2) is
(a) 100 (b) 500 (c) 600 (d) 200 (v) The value of y at any time t is given by
(a) \(y=\frac{5000}{e^{-5000 k t}+1}\) (b) \(y=\frac{5000}{1+e^{5000 k t}}\) (c) \(y=\frac{5000}{49 e^{-5000 k t}+1}\) (d) \(y=\frac{5000}{49\left(1+e^{-5000 k t}\right)}\) -
Order: The order of a differential equation is the order of the highest order derivative appearing in the differential equation.
Degree : The degree of differential equation is the power of the highest order derivative, when differential coefficients are made free from radicals and fractions. Also, differential equation must be a polynomial equation
in derivatives for the degree to be defined.
Based on the above information, answer the following questions.
(i) Find the degree of the differential equation \(2 \frac{d^{2} y}{d x^{2}}+3 \sqrt{1-\left(\frac{d y}{d x}\right)^{2}-y}=0\)(a) 3 (b) 4 (c) 2 (d) 1 (ii) Order and degree of the differential equation \(y \frac{d y}{d x}=\frac{x}{\frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{3}}\) are respectively
(a) 1,1 (b) 1,2 (c) 1,3 (d) 1,4 (iii) Find order and degree of the equation \(y^{\prime \prime \prime}+y^{2}+e^{y^{\prime}}=0\)
(a) order = 3, degree = undefined (b) order = 1, degree = 3 (c) order = 2, degree = undefined (d) order = 1, degree = 2 (iv) Determine degree of the differential equation \((\sqrt{a+x}) \cdot\left(\frac{d y}{d x}\right)+x=0\)
(a) 3 (b) not defined (c) 1 (d) 2 (v) Order and degree of the differential equation \(\left(1+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{7}{3}}=7 \frac{d^{2} y}{d x^{2}}\) are respectively
(a) 2, 1 (b) 2,3 (c) 1,3 (d) \(1, \frac{7}{3}\) -
If the equation is of the form \(\frac{d y}{d x}=\frac{f(x, y)}{g(x, y)} \text { or } \frac{d y}{d x}=F\left(\frac{y}{x}\right)\) ,wheref (x, y), g(x, y) are homogeneous functions of the same degree in x and y, then put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\), so that the dependent variable y is changed to another variable v and then apply variable separable method. Based on the above information, answer the following questions.
(i) The general solution of \(x^{2} \frac{d y}{d x}=x^{2}+x y+y^{2}\) is(a) \(\tan ^{-1} \frac{x}{y}=\log |x|+c \) (b) \( \tan ^{-1} \frac{y}{x}=\log |x|+c\) (c) \(y=x \log |x|+c\) (d) \(x=y \log |y|+c\) (ii) Solution of the differential equation \(2 x y \frac{d y}{d x}=x^{2}+3 y^{2} \) is
(a) \( x^{3}+y^{2}=c x^{2}\) (b) \( \frac{x^{2}}{2}+\frac{y^{3}}{3}=y^{2}+c\) (c) \(x^{2}+y^{3}=c x^{2}\) (d) \( x^{2}+y^{2}=c x^{3}\) (iii) Solution of the differential equation \(\left(x^{2}+3 x y+y^{2}\right) d x-x^{2} d y=0\) is
(a) \(\frac{x+y}{x}-\log x=c\) (b) \( \frac{x+y}{x}+\log x=c\) (c) \(\frac{x}{x+y}-\log x=c\) (d) \(\frac{x}{x+y}+\log x=c\) (iv) General solution ofthe differential equation \(\frac{d y}{d x}=\frac{y}{x}\left\{\log \left(\frac{y}{x}\right)+1\right\}\) is
(a) \(\log (x y)=c\) (b) \( \log y=c x\) (c) \(\log \left(\frac{y}{x}\right)=c x\) (d) \(\log x=c y\) (v) Solution ofthe differential equation \(\left(x \frac{d y}{d x}-y\right) e^{\frac{y}{x}}=x^{2} \cos x\) is
(a) \(e^{\frac{y}{x}}-\sin x=c\) (b) \(e^{\frac{y}{x}}+\sin x=c\) (c) \(e^{\frac{-y}{x}}-\sin x=c \) (d) \( e^{\frac{-y}{x}}+\sin x=c\) -
If the equation is of the form \(\frac{d y}{d x}+P y=Q\) , where P, Q are functions of x, then the solution of the differential equation is given by \(y e^{\int P d x}=\int Q e^{\int P d x} d x+c\), where \(e^{\int P d x}\) is called the integrating factor (I.F.).
Based on the above information, answer the following questions.
(i) The integrating factor of the differential equation \(\sin x \frac{d y}{d x}+2 y \cos x=1 \text { is }(\sin x)^{\lambda}, \text { where } \lambda=\)(a) 0 (b) 1 (c) 2 (d) 3 (ii) Integrating factor of the differential equation \(\left(1-x^{2}\right) \frac{d y}{d x}-x y=1 \) is
(a) -x (b) \(\frac{x}{1+x^{2}}\) (c) \(\sqrt{1-x^{2}}\) (d) \( \frac{1}{2} \log \left(1-x^{2}\right)\) (iii) The solution of \(\frac{d y}{d x}+y=e^{-x}, y(0)=0\) is
(a) \( y=e^{x}(x-1)\) (b) \( y=x e^{-x}\) (c) \(y=x e^{-x}+1\) (d) \( y=(x+1) e^{-x}\) (iv) General solution of \(\frac{d y}{d x}+y \tan x=\sec x\) is
(a) y see x = tan x + c (b) y tan x = sec x + c (c) tan x = y tan x + c (d) x see x = tan y + c (v) The integrating factor of differential equation \(\frac{d y}{d x}-3 y=\sin 2 x\) is
(a) e3x (b) e-2x (c) e-3x (d) xe-3x
Case Study Questions
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CBSE 12th Standard Maths Differential Equations Case Study Questions 2021 Answer Keys
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(i) (b) : Here, P denotes the principal at any time t and the rate of interest be r% per annum compounded continuously, then according to the law given in the problem, we get
\(\frac{d P}{d t}=\frac{P r}{100}\)
(ii) (a) : We have, \(\frac{d P}{d t}=\frac{\operatorname{Pr}}{100}\)
\(\Rightarrow \frac{d P}{P}=\frac{r}{100} d t \Rightarrow \int \frac{1}{P} d P=\frac{r}{100} \int d t\)
\(\Rightarrow \log P=\frac{r t}{100}+C\)
At t = 0, P = Po
\(\therefore \quad C=\log P_{0}\)
So, \(\log P=\frac{r t}{100}+\log P_{0}\)
\(\Rightarrow \log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100}\)
(iii) (c) : We have, r = 5, Po = Rs. 100 and P = Rs. 200 = 2Po
Substituting these values in (2), we get
\(\log 2=\frac{5}{100} t\)
\(\Rightarrow\) t = 20 loge 2 = 20 x 0.6931 years = 13.862 years
(iv) (d) : We have
Po = Rs. 100, P = Rs. Rs. 200 = 2Po and t = 10 years
Substituting these values in (2), we get
\(\log 2=\frac{10 r}{100} \Rightarrow r=10 \log 2=10 \times 0.6931=6.931\)
(v) (a) : We have
Po = Rs. 1000, r = 5 and t = 10
Substituting these values in (2), we get
\(\log \left(\frac{P}{1000}\right)=\frac{5 \times 10}{100}=\frac{1}{2}=0.5 \Rightarrow \frac{P}{1000}=e^{0.5}\)
\(\Rightarrow\) P = 1000 x 1.648 = Rs. 1648 -
(i) (c) : Since, size of population is 5000.
\(\therefore\) Maximum value of y(t) is 5000.
(ii) (d) : Clearly, according to given information
\(\frac{d y}{d t}=k y(5000-y)\) ,where k is the constant of proportionality.
(iii) (a): Since, rumour is initiated with 100 people.
\(\therefore\) When t = 0, then y = 100
Thus y(O) = 100
(iv) (b) : Since, rumour is spread in 500 people, after
2 days.
\(\therefore\) When t = 2, then y = 500.
Thus, y(2) = 500
(v) (c) : We know that, when t = 0, then y = 100
This condition is satisfied by option (c) only. -
(i) (c) : We have, \(2 \frac{d^{2} y}{d x^{2}}+3 \sqrt{1-\left(\frac{d y}{d x}\right)^{2}-y}=0\)
\(\therefore \quad 2 \frac{d^{2} y}{d x^{2}}=-3 \sqrt{1-\left(\frac{d y}{d x}\right)^{2}-y}\)
Squaring both sides, we get
\(4\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=9\left[1-\left(\frac{d y}{d x}\right)^{2}-y\right]\)
Here, highest order derivative is \(\frac{d^{2} y}{d x^{2}}\) and its power is 2. So, its degree is 2.
(ii) (d) : We have, \(y \frac{d y}{d x}=\frac{x}{\frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{3}}\)
\(\Rightarrow y\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d y}{d x}\right)^{4}=x\)
\(\Rightarrow\) Here, highest order derivative is \(\frac{d y}{d x}\) is So , its order is 1 and degree is 4.
(iii) (a) : We have,\(y^{\prime \prime \prime}+y^{2}+e^{y^{\prime}}=0\)
\(\frac{d^{3} y}{d x^{3}}+y^{2}+e^{(d y / d x)}=0\)
Highest order derivative is \(\frac{d^{3} y}{d x^{3}}\) .So, its order is 3.
Also, the given differential cannot be expressed as a polynomial. So, its degree is not defined.
(iv) (c) : The given differential equation is,
\(\sqrt{a+x} \cdot\left(\frac{d y}{d x}\right)+x=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{\sqrt{a+x}}\)
Clearly, degree = 1
(v) (b) : We have \(y \frac{d y}{d x}=\frac{x}{\frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{3}}\)
\(\Rightarrow y\left(\frac{d y}{d x}\right)^{2}+y\left(\frac{d y}{d x}\right)^{4}=x\)
\(\Rightarrow\) Here, highest order derivative is \(\frac{d y}{d x}\) ,So , its order is 1 and degree is 4.
(iii) (a) : We have, y'" +y2 + ey = 0
\(\frac{d^{3} y}{d x^{3}}+y^{2}+e^{(d y / d x)}=0\)
Highest order derivative is \(\frac{d^{3} y}{d x^{3}}\) So, its order is 3.
Also, the given differential cannot be expressed as a polynomial. So, its degree is not defined
(iv) (c) :The given differential equation is,
\(\sqrt{a+x} \cdot\left(\frac{d y}{d x}\right)+x=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{\sqrt{a+x}}\)
Clearly, degree = 1.
(v) (b) : We have \(\left(1+\left(\frac{d y}{d x} \mid\right)^{3}\right)^{\frac{1}{3}}=7 \frac{d^{2} y}{d x^{2}}\)
\(\therefore\) Order is 2 and degree is 3. -
(i) (b): We have, \(\frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore v+x \frac{d v}{d x}=\frac{x^{2}+x \cdot v x+v^{2} x^{2}}{x^{2}}=1+v+v^{2}\)
\(\Rightarrow x \frac{d v}{d x}=1+v^{2} \Rightarrow \int \frac{d v}{1+v^{2}}=\int \frac{d x}{x}+c\)
\(\Rightarrow \tan ^{-1} v=\log |x|+c \Rightarrow \tan ^{-1} \frac{y}{x}=\log |x|+c\)
(ii) (d): We have, \( 2 x y \frac{d y}{d x}=x^{2}+3 y^{2} \Rightarrow \frac{d y}{d x}=\frac{x^{2}+3 y^{2}}{2 x y}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore v+x \frac{d v}{d x}=\frac{x^{2}+3 v^{2} x^{2}}{2 v x^{2}} \Rightarrow x \frac{d v}{d x}=\frac{1+3 v^{2}}{2 v}-v\)
\(\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{2 v_{*}} \Rightarrow \int \frac{2 v}{1+v^{2}} d v=\int \frac{d x}{x}+\log c\)
\(\Rightarrow \log \left|1+v^{2}\right|=\log |x|+\log |c| \Rightarrow \log \left|v^{2}+1\right|=\log |x c|\)
\(\Rightarrow \quad v^{2}+1=x c \Rightarrow \frac{y^{2}}{2}+1=x c \Rightarrow x^{2}+y^{2}=x^{3} c\)
(iii) (d): We have, \(\left(x^{2}+3 x y+y^{2}\right) d x-x^{2} d y=0\)
\(\Rightarrow \frac{x^{2}+3 x y+y^{2}}{x^{2}}=\frac{d y}{d x}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore \frac{x^{2}+3 x^{2} v+x^{2} v^{2}}{x^{2}}=\left(v+x \frac{d v}{d x}\right)\)
\(\Rightarrow 1+3 v+v^{2}=v+x \frac{d v}{d x} \Rightarrow 1+2 v+v^{2}=x \frac{d v}{d x}\)
\(\Rightarrow \int \frac{d x}{x}-\int(v+1)^{-2} d v=c \Rightarrow \log x+\frac{1}{v+1}=c\)
\(\Rightarrow \log x+\frac{x}{x+y}=c\)
(iv) (c): We have, \(\frac{d y}{d x}=\frac{y}{x}\left\{\log \left(\frac{y}{x}\right)+1\right\}\)
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\therefore v+x \frac{d v}{d x}=v\{\log (v)+1\} \Rightarrow x \frac{d v}{d x}=v \log v\)
\(\Rightarrow \int \frac{d v}{v \log v}=\int \frac{d x}{x} \Rightarrow \log |\log v|=\log |x|+\log |c|\)
\(\Rightarrow \log \left(\frac{y}{x}\right)=c x\)
(v) (a): We have,\(\left(x \frac{d y}{d x}-y\right) e^{\frac{y}{x}}=x^{2} \cos x\)
\(\Rightarrow\left(\frac{d y}{d x}-\frac{y}{x}\right) e^{\frac{y}{x}}=x \cos x\)
\(\text { Put } y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(\Rightarrow\left(v+x \frac{d v}{d x}-v\right) e^{v}=x \cos x \Rightarrow x e^{v} \frac{d v}{d x}=x \cos x\)
\(\Rightarrow \int e^{v} d v=\int \cos x d x \Rightarrow e^{v}=\sin x+c\)
\(\Rightarrow e^{\frac{y}{x}}-\sin x=c\) -
(i) (c) : The given differential equation can be written as \( \frac{d y}{d x}+2 y \cot x=\operatorname{cosec} x\)
\(\therefore \text { I.F. }=e^{\int 2 \cot x d x}=e^{2 \log |\sin x|}=(\sin x)^{2} \)
\(\therefore \quad \lambda=2\)
(ii) (c) : We have, \(\left(1-x^{2}\right) \frac{d y}{d x}-x y=1\)
\(\Rightarrow \frac{d y}{d x}-\frac{x}{1-x^{2}} \cdot y=\frac{1}{1-x^{2}} \)
\(\therefore \text { I.F. }=e^{-\int \frac{x}{1-x^{2}} d x}=e^{\frac{1}{2} \int \frac{-2 x}{1-x^{2}} d x}\)
\(=e^{\frac{1}{2} \log \left(1-x^{2}\right)}=e^{\log \left(1-x^{2}\right)^{\frac{1}{2}}}=\sqrt{1-x^{2}}\)
(iii) (b) : We have, \(\frac{d y}{d x}+y=e^{-x} \)
It is a linear differential equation with I.F. = \(e^{\int d x}=e^{x}\)
Now, solution is \(y \cdot e^{x}=\int e^{x} \cdot e^{-x} d x+c\)
\(\Rightarrow y e^{x}=\int d x+c \Rightarrow y e^{x}=x+c \Rightarrow y=x e^{-x}+c e^{-x}\)
\(\because y(0)=0 \Rightarrow c=0 \quad \therefore y=x e^{-x}\)
(iv) (a) : We have,\( \frac{d y}{d x}+y \tan x=\sec x\)
It is a linear differential equation with I.F. = \(e^{\int \tan x d x}=e^{\log |\sec x|}=\sec x\)
Now, solution is \(y \sec x=\int \sec ^{2} x d x+c\)
\(\Rightarrow y \sec x=\tan x+c\)
(v) (c) : We have, \(\frac{d y}{d x}-3 y=\sin 2 x\)
It is a linear differential equation with \(\text { I.F. }=e^{\int-3 d x}=e^{-3 x}\)
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