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Questions:

Case Study Questions

5 x 4 = 20

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) \(\frac{4+\pi^{2}}{2}\)

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

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

(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) \(1, \frac{7}{3}\)

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) 70^oF

(d) 98.6°F

(iii) When t = 2, then body temperature is equal to

(a) 50°F

(b) 60°F

(c) 70^oF

(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 thermometer reading 800P is taken outside. Five minutes later the thermometer reads 60°F. After another 5 minutes the thermometer reads 50^of At any time t the thermometer reading be TOP and the outside temperature be S^oF.
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) 40^oF

(c) 50^oF

(d) 60^oF

(iii) The value of T(10) is

(a) 50^oF

(b) 40^oF

(c) 50^oF

(d) 60^oF

(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)

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Answers Key:

Case Study Questions

5 x 4 = 20

(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, \(y=\frac{2-x^{2}}{2(1+\sin x)}\)
\(\therefore \quad y\left(\frac{\pi}{2}\right)=\frac{2-\left(\frac{\pi}{2}\right)^{2}}{2\left(1+\sin \frac{\pi}{2}\right)}=\frac{2-\frac{\pi^{2}}{4}}{4}=\frac{8-\pi^{2}}{16}\)

(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 + 2e^{1n 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) : 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 + e^y = 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) : 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 = 20e^kt
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) (d) :Given, at any time t the thermometer reading be T^oF 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) = 60^0F
(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°