Class 12th Maths - Differential Equations Case Study Questions and Answers 2022 - 2023
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Differential Equations Case Study Questions With Answer Key
12th Standard CBSE
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Maths
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In a college hostel accommodating 1000 students, one of the hostellers came in carrying Corona virus, and the hostel was isolated. The rate at which the virus spreads is assumed to be proportional to the product of the number of infected students and remaining students. There are 50 infected students after 4 days
Based on the above information, answer the following questions
(i) If n(t) denote the number of students infected by Corona virus at any time t, then maximum value of n(t) is(a) 50 (b) 100 (c) 500 (d) 1000 (ii) \(\frac{d n}{d t}\) is proportional to
(a) n(100-n) (b) n(100+ n) (c) n(100 - n) (d) n(100 + n) (iii) The value of n( 4) is
(a) 1 (b) 50 (c) 100 (d) 1000 (iv) The most general solution of differential equation formed in given situation is
(a) \(\frac{1}{1000} \log \left(\frac{1000-n}{n}\right)=\lambda t+c\) (b) \(\log \left(\frac{n}{100-n}\right)=\lambda t+c\) (c) \(\frac{1}{1000} \log \left(\frac{n}{1000-n}\right)=\lambda t+c\) (d) None of these (v) The value of n at any time is given by
(a) \(n(t)=\frac{1000}{1+999 e^{-0.9906 t}}\) (b) \(n(t)=\frac{1000}{1-999 e^{-0.9906 t}}\) (c) \(n(t)=\frac{100}{1-999 e^{-0.996 t}}\) (d) \(n(t)=\frac{100}{999+e^{1000 t}}\) -
A thermometer reading 800P is taken outside. Five minutes later the thermometer reads 60°F. After another 5 minutes the thermometer reads 50of At any time t the thermometer reading be TOP and the outside temperature be SoF.
Based on the above information, answer the following questions.
(i) If \(\lambda\) is positive constant of proportionality, then \(\frac{d T}{d t}\) is(a) \(\lambda(T-S)\) (b) \(\lambda(T+S)\) (c) \(\lambda T S\) (d) \(-\lambda(T-S)\) (ii) The value of T(S) is
(a) 300F (b) 40oF (c) 50oF (d) 60oF (iii) The value of T(10) is
(a) 50oF (b) 40oF (c) 50oF (d) 60oF (iv) Find the general solution of differential equation formed in given situation.
(a) logT=St+c (b) \(\log (T-S)=-\lambda t+c\) (c) log S = tT + c (d) \(\log (T+S)=\lambda t+c\) (v) Find the valiie of constant of integration c in the solution of differential equation formed in given situation.
(a) log (60 -S) (b) log (80 + S) (c) log (80 - S) (d) log (60 + S) -
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 -
In a murder investigation, a corpse was found by a detective at exactly 8 p.m. Being alert, the detective measured the body temperature and found it to be 70°F. Two hours later, the detective measured the body temperature again and found it to be 60°F, where the room temperature is 50°F. Also, it is given the body temperature at the time of death was normal, i.e., 98.6°F.
Let T be the temperature of the body at any time t and initial time is taken to be 8 p.m.
Based on the above information, answer the following questions.
(i) By Newton's law of cooling,\(\frac{d T}{d t}\) is proportional to(a) T - 60 (b) T - 50 (c) T - 70 (d) T - 98.6 (ii) When t = 0, then body temperature is equal to
(a) 50°F (b) 60°F (c) 70oF (d) 98.6°F (iii) When t = 2, then body temperature is equal to
(a) 50°F (b) 60°F (c) 70oF (d) 98.6°F (iv) The value of T at any time tis
(a) \(50+20\left(\frac{1}{2}\right)^{t}\) (b) \(50+20\left(\frac{1}{2}\right)^{t-1}\) (c) \(50+20\left(\frac{1}{2}\right)^{t / 2}\) (d) None of these (v) If it is given that loge(2.43) = 0.88789 and loge(0.5) = -0.69315, then the time at which the murder occur is
(a) 7:30 p.m. (b) 5:30 p.m. (c) 6:00 p.m. (d) 5:00 p.m. -
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) \( e^{\frac{-y}{x}}+\sin x=c\) -
A differential equation is said to be in the variable separable form if it is expressible in the form j(x) dx = g(y) dy. The solution of this equation is given by \(\int f(x) d x=\int g(y) d y+c\) where c is the constant of integration.
Based on the above information, answer the following questions.
(i) If the solunon of the differential equation \(\frac{d y}{d x}=\frac{a x+3}{2 y+f}\) represents a circle, then the value of' a 'is(a) 2 (b) - 2 (c) 3 (d) - 4 (ii) The diiftfterential equation \(\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{y}\) deterrnines a family of circle with
(a) variable radii and fixed centre (0,1) (b) variable radii and fixed centre (0,-1) (c) fixed radius 1 and variable centre on x-axis (d) fixed radius 1 and variable centre on y-axis (iii) If y' = y + 1, y (0) = 1, theny (In 2) =
(a) 1 (b) 2 (c) 3 (d) 4 (iv) The solution of the differential equation \(\frac{d y}{d x}=e^{x-y}+x^{2} e^{-y}\) is
(a) \(e^{x}=\frac{y^{3}}{3}+e^{y}+c\) (b) \(e^{y}=\frac{x^{2}}{3}+e^{x}+c\) (c) \(e^{y}=\frac{x^{3}}{3}+e^{x}+c\) (d) none of these (v) If \(\frac{d y}{d x}=y \sin 2 x, y(0)=1\) then its solution is
(a) y = esin2x (b) y = sin2 (c) y = cos2x (d) y = cos2x -
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 -
If an equation is of the form \(\frac{d y}{d x}+P y=Q\) ,where P, Q are functions of x, then such equation is known as linear differential equation. Its solu~:n is given by \(y \cdot(\mathrm{I} . \mathrm{F} .)=\int \mathrm{Q} \cdot(\mathrm{I} . \mathrm{F} .) d x+c\) where \(\text { I.F. }=e^{\int P d x}\) .
Now, suppose the given equation is \((1+\sin x) \frac{d y}{d x}+y \cos x+x=0\)
Based on the above information, answer the following questions
(i) The value of P and Q respectively are(a) \(\frac{\sin x}{1+\cos x}, \frac{x}{1+\sin x}\) (b) \(\frac{\cos x}{1+\sin x}, \frac{-x}{1+\sin x}\) (c) \(\frac{-\cos x}{1+\sin x}, \frac{x}{1+\sin x}\) (d) \(\frac{\cos x}{1+\sin x}, \frac{x}{1+\sin x}\) (ii) The value of I.F. is
(a) 1 - sin x (b) cos x (c) 1 + sin x (d) 1- cosx (iii) Solution of given equation is
(a) y{1-sinx)=x+c (b) y(l + sin x) = x2+c (c) \(y(1-\sin x)=\frac{-x^{2}}{2}+c\) (d) \(y(1+\sin x)=\frac{-x^{2}}{2}+c\) (iv) If y(0) = 1, then y,equals
(a) \(\frac{2-x^{2}}{2(1+\sin x)}\) (b) \(\frac{2+x^{2}}{2(1+\sin x)}\) (c) \(\frac{2-x^{2}}{2(1-\sin x)}\) (d) \(\frac{2+x^{2}}{2(1-\sin x)}\) (v) Value of \(y\left(\frac{\pi}{2}\right)\) is
(a) \(\frac{4-\pi^{2}}{2}\) (b) \(\frac{8-\pi^{2}}{16}\) (c) \(\frac{8-\pi^{2}}{4}\) (d) \(\frac{4+\pi^{2}}{2}\)
Case Study
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Answers
Differential Equations Case Study Questions With Answer Key Answer Keys
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(i) (d) : Since, maximum number of students in hostel is 1000.
\(\therefore\) Maximum value of n(t) is 1000.
(ii) (a) : Clearly, according to given information
\(\frac{d n}{d t}=\lambda n(1000-n)\) ,where \(\lambda\) is constant of proportionality.
(iii) (b) : Since, 50 students are infected after 4 days
\(\therefore\) n(4) = 50.
(iv) (c) : We have,\(\frac{d n}{d t}=\lambda n(100-n)\)
\(\Rightarrow \int \frac{d n}{n(1000-n)}=\lambda \int d t\)
\(\Rightarrow \frac{1}{1000} \int\left(\frac{1}{1000-n}+\frac{1}{n}\right) d n=\lambda \int d t\)
\(\Rightarrow \quad \frac{1}{1000}\left[\frac{\log (1000-n)}{-1}+\log n\right]=\lambda t+C\)
\(\Rightarrow \frac{1}{1000} \log \left(\frac{n}{1000-n}\right)=\lambda t+C\)
(v) (a) : When, t = 0, n = 1
This condition is satisfied by option (a) only. -
(i) (d) :Given, at any time t the thermometer reading be ToF and the outside temperature be sop.
Then, by Newton's law of cooling, we have
\(\frac{d T}{d t} \propto(T-S) \Rightarrow \frac{d T}{d t}=-\lambda(T-S)\)
(ii) (d) : Since, after 5 minutes, thermometer reads 600F.
\(\therefore\) Value of T(5) = 600F
(iii) (a) : Clearly from given information, value of T( 10) is 50°F.
(iv) (b) : We have,\(\frac{d T}{d t}=-\lambda(T-S)\)
\(\Rightarrow \frac{d T}{T-S}=-\lambda d t \Rightarrow \int \frac{1}{T-S} d T=-\lambda \int d t\)
\(\Rightarrow \log (T-S)=-\lambda t+c\)
(v) (c) : Since, at t = 0, T = 80° -
(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) (b) : Given, T is the temperature of the body at any time t. Then, by Newton's law of cooling, we get \(\frac{d T}{d t}=k(T-50)\), where k is the constant of proportionality
(ii) (c) : From given information, we have
At 8 p.m. temperature is 70°F
\(\therefore\) At t = 0, T = 70°F
(iii) (b) : From given information, we have
At 10 p.m., temperature is 60°F.
\(\therefore\) At t = 2, T = 60° F
(iv) (c) : \(\frac{d T}{d t}=k(T-50) \Rightarrow \frac{d T}{T-50}=k d t\)
On integrating both sides, we get
\(\log |T-50|=k t+\log C \Rightarrow T-50=C e^{\wedge t}\)
Clearly, for t = 0, \(T=70^{\circ} \Rightarrow C=20\)
Thus, T - 50 = 20ekt
For \(t=2, T=60^{\circ} \Rightarrow 10=20 e^{2 k}\)
\(\Rightarrow 2 k=\log \left(\frac{1}{2}\right) \Rightarrow k=\frac{1}{2} \log \left(\frac{1}{2}\right)\)
Hence, \(T=50+20\left(\frac{1}{2}\right)^{\frac{t}{2}}\)
(v) (b) : We have, \(T=50+20\left(\frac{1}{2}\right)^{\frac{2}{2}}\) -
(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) (b) : We have,\(\frac{d y}{d x}=\frac{a x+3}{2 y+f}\)
\(\Rightarrow \quad(a x+3) d x=(2 y+f) d y\)
\(\Rightarrow \quad a \frac{x^{2}}{2}+3 x=y^{2}+f y+c\) (Integrating)
\(\Rightarrow-\frac{a}{2} x^{2}+y^{2}-3 x+f y+C=0\)
This will represent a circle, if \(\frac{-a}{2}=1 \Rightarrow a=-2\)
[\(\therefore\) In circle, coefficient of x2 = coefficient of y2]
(ii) (c) : We have,\(\frac{y d y}{\sqrt{1-y^{2}}}=d x\)
On integration, we get \(-\sqrt{1-y^{2}}=x+c\)
\(\Rightarrow \quad 1-y^{2}=(x+c)^{2} \Rightarrow(x+c)^{2}+y^{2}=1\) ,which represents a circle with radius 1 and centre on the x-axis.
(iii) (c) : \(y^{\prime}=y+1 \Rightarrow \frac{d y}{y+1}=d x\)
\(\Rightarrow \ln (y+1)=x+c\)
Now, \(y(0)=1 \Rightarrow c=\ln 2\)
\(\therefore \quad \ln \left(\frac{y+1}{2}\right)=x \Rightarrow y+1=2 e^{x}\)
So, y (In 2) = -1 + 2e1n 2 = -1 + 4 = 3
(iv) (c) : From the given differential equation, we have
\(\frac{d y}{d x}=\frac{e^{x}+x^{2}}{e^{y}} \Rightarrow e^{y} d y=\left(e^{x}+x^{2}\right) d x\)
Integrating, we get \(y=e^{x}+\frac{x^{3}}{3}+c\)
(v) (a) : We have, \(\frac{d y}{d x}=y \sin 2 x\)
\(\Rightarrow \quad \frac{d y}{y}=\sin 2 x d x \Rightarrow \log y=-\frac{\cos 2 x}{2}+c\)
Since x = 0, y = 1 therefore c = 1/2
Now, \(\log y=\frac{1}{2}(1-\cos 2 x)\)
\(\Rightarrow \log y=\sin ^{2} x \Rightarrow y=e^{\sin ^{2} x}\) -
(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 \(\text { I.F. }=e^{\int-3 d x}=e^{-3 x}\) -
(i) (b) : The given differential equation can be written as \(\frac{d y}{d x}+\frac{\cos x}{1+\sin x} y=\frac{-x}{1+\sin x}\)
Compare it with \(\frac{d y}{d x}+P y=Q\) ,we get
\(P=\frac{\cos x}{1+\sin x}\) and \(Q=\frac{-x}{1+\sin x}\)
(ii) (c) : \(\text { I.F. }=e^{\int P d x}=e^{\int \frac{\cos x}{1+\sin x} d x}\)
Put \(1+\sin x=t \Rightarrow \cos x d x=d t\)
\(\therefore \quad \text { I.F. }=e^{\int \frac{1}{t} d t}=e^{\log t}=t=1+\sin x\)
(iii) (d) : Solution of given differential equation is given by \(y \cdot(\mathrm{I} . \mathrm{F} .)=\int Q(\mathrm{I.F.}) d x+c\)
\(\Rightarrow y(1+\sin x)=\int \frac{-x}{1+\sin x} \cdot(1+\sin x) d x+c\)
\(\Rightarrow \quad y(1+\sin x)=\frac{-x^{2}}{2}+c\)
(iv) (a) : We have, y(0) = 1 i.e., x = 0, y = 1
\(\therefore \quad 1(1+\sin 0)=c \Rightarrow c=1\)
\(\therefore \quad y(1+\sin x)=\frac{-x^{2}}{2}+1=\frac{2-x^{2}}{2}\)
\(\therefore \quad y=\frac{2-x^{2}}{2(1+\sin x)}\)
(v) (b) : We have, \(y=\frac{2-x^{2}}{2(1+\sin x)}\)
\(\therefore \quad y\left(\frac{\pi}{2}\right)=\frac{2-\left(\frac{\pi}{2}\right)^{2}}{2\left(1+\sin \frac{\pi}{2}\right)}=\frac{2-\frac{\pi^{2}}{4}}{4}=\frac{8-\pi^{2}}{16}\)
Case Study
