Case study (4 Marks)

Question 14 Marks

A differential equation is said to be in the variable separable form if it is expressible in the form $f(x) dx = g(y) dy.$
The solution of this equation is given by
$\int\text{f(x)dx}=\int\text{g(y)dy}+\text{c},$ where $c$ is the constant of integration.
Based on the above information, answer the following questions.

If the solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{ax+3}}{\text{2y+f}}$ represents a circle, then the value of $'a\ '$ is:

$2$
$-2$
$3$
$-4$

The differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{\text{y}}$ determines a family of circle with.

Variable radii and fixed centre $(0, 1)$
Variable radii and fixed centre $(0, -1)$
Fixed radius $1$ and variable centre on $x-$ axis
Fixed radius $1$ and variable centre on $y-$ axis

If $= y'+ 1, y(0) = 1,$ then $y (In 2) =$

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x-y}+\text{x}^2\text{e}^\text{-y}$ is:

$\text{e}^\text{x}=\frac{\text{y}^3}{3}+\text{e}^\text{y}+\text{c}$
$\text{e}^\text{y}=\frac{\text{x}^2}{3}+\text{e}^\text{x}+\text{c}$
$\text{e}^\text{y}=\frac{\text{x}^3}{3}+\text{e}^\text{x}+\text{c}$
None of these

If $\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x},\ \text{y}(0)=1,$ then its solution is:

$\text{y}=\text{e}^{\sin^2}\text{x}$
$\text{y}={\sin^2}\text{x}$
$\text{y}={\cos^2}\text{x}$
$\text{y}=\text{e}^{\cos^2}\text{x}$

Answer

$(b)\ -2$

We have, $\frac{\text{dy}}{\text{dx}}=\frac{\text{ax+3}}{\text{2y+f}}$
$\Rightarrow\ \ (\text{ax+3})\text{dx}=(2\text{y}+\text{f})\text{dy}$
$\Rightarrow\text{a}\frac{\text{x}^2}{2}+\text{3x}=\text{y}^2+\text{fy}+\text{c} \ ($Integrating$)$
$\Rightarrow-\frac{\text{a}}{2}\text{x}^2+\text{y}^2-\text{3x}+\text{fy}+\text{C}=0$
This will represent a circle, if $\frac{-\text{a}}{2}=1$
$\Rightarrow\text{a}=-2$
$[\therefore$ In circle, coefficient of $x^2 =$ coefficient of $y^2)$

$(c)$ Fixed radius $1$ and variable centre on $x-$ axis

We have, $\frac{\text{ydy}}{\sqrt{1-\text{y}^2}}=\text{dx}$
On integration, we get $-\sqrt{1-\text{y}^2}=\text{x+c}$
$\Rightarrow 1 - y^2 = (x + c)^2$
$\Rightarrow ^(x + c)^{2 }+ y^{2 }= 1$ which represents a circle with radius $I$ and centre on the $x-$ axis.

$(c)\ 3$

$\text{y}'=\text{y}+1$
$\Rightarrow\frac{\text{dy}}{\text{y}+1}=\text{dx}$
$\Rightarrow In (y + 1) = x + c \ ($integrating$)$
Now $, y(0) = 1.$
$\Rightarrow c = In 2$
$\therefore \ \text{In}\Bigg(\frac{\text{y}+1}{2}\Bigg)=\text{x}$
$\Rightarrow y + 1 = 2e^x$
So $, y (In 2) = -1 + 2e^{In 2} = -1 + 4 = 3$

$(c)\ \text{e}^\text{y}=\frac{\text{x}^3}{3}+\text{e}^\text{x}+\text{c}$

From the given differential equation, we have
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}+\text{x}^\text{2}}{\text{e}^\text{y}}$
$\Rightarrow\ \text{e}^\text{y}\text{dy}=(\text{e}^\text{x}+\text{x}^2)\text{dx}$
Integrating, we get $\text{e}^\text{y}=\text{e}^\text{x}+\frac{\text{x}^3}{3}+\text{c}$

$(a)\ \text{y}=\text{e}^{\sin^2}\text{x}$

We have, $\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{y}}=\sin2\text{x}\ \text{dx}$
$\Rightarrow\ \log\text{y}=-\frac{\cos2\text{x}}{2}+\text{c}$
Since $x = 0, y = 1$
therefore $\text{C}=\frac{1}{2}$
Now $, \log\text{y}=\frac{1}{2}(1-\cos2\text{x})$
$\Rightarrow\ \log\text{y}=\sin^2\text{x}$
$\Rightarrow\text{y}=\text{e}^{\sin^2}\text{x}$

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Question 24 Marks

It is known that, if the interest is compounded continuously, the principal changes at the rate equal to the product 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 question.

Find the value of $\frac{\text{dP}}{\text{dt}}.$

$\frac{\text{Pr}}{1000}$
$\frac{\text{Pr}}{100}$
$\frac{\text{Pr}}{10}$
$\text{Pr}$

If $P_0$ be the initial principal, then find the solution of differential equation formed in given situation.

$\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{100}$
$\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{10}$
$\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\text{rt}$
$\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=100\text{rt}$

If the interest is compounded continuously at $5\%$ per annum, in how many years will $₹\ 100$ double itself?

$12.728$ years
$14.789$ years
$13.862$ years
$15.872$ years

At what interest rate will $₹\ 100$ double itself in $10$ years? $(\log_\text{e}2 = 0.6931 ).$

$9.66\%$
$8.239\%$
$7.341\%$
$6.931\%$

How much will $₹\ 1000$ be worth at $5\%$ interest after $10$ years? $(e^{0.5} = 1.648)$.

$₹\ 1648$
$₹\ 1500$
$₹\ 1664$
$₹\ 1572$

Answer

$(b) \frac{\text{Pr}}{100}$

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
$\frac{\text{dP}}{\text{dt}}=\frac{\text{Pr}}{100}$

$(a) \log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{100}$

We have, $\frac{\text{dP}}{\text{dt}}=\frac{\text{Pr}}{100}$
$\Rightarrow\frac{\text{dP}}{\text{P}}=\frac{\text{r}}{100}\text{dt}$
$\Rightarrow\int\frac{\text{1}}{\text{P}}\text{dP}=\frac{\text{r}}{100}\int\text{dt}$
$\Rightarrow\log\text{P}=\frac{\text{rt}}{100}+\text{C}...\text{(i)}$
$\text{At t}=0,\ \ \text{P}=\text{P}_0.$
$\therefore\ \ \text{C}=\log\text{P}_0$
So, $\log\text{P}=\frac{\text{rt}}{100}+\log\text{p}_0$
$\Rightarrow\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{100}...\text{(ii)}$

$(c) 13.862$ years

We have, $r = 5, P_{0 }= ₹\ 100$ and $P = ₹\ 200 = 2P_0$ Substituting these values in $(2),$ we get
$\log2=\frac{5}{100}\text{t}$
$\Rightarrow\text{t}=20\log_\text{e}$
$2 = 20 \times 0.6931\ \text{years} = 13.862\ \text{years}$

$(d) 6.931\%$

We have,
$P_0 = ₹\ 100, P = ₹\ 200 = 2P_{0 }$ and $t = 10$ years Substituting these values in $(2),$ we get
$\log2\frac{10\text{r}}{100}\Rightarrow\text{r}=10\log2=10\times0.6931=6.931$

$(a)\ ₹ 1648$

We have $P_0 = ₹\ 1000, r = 5$ and $t = 10$ Substituting these values in $(2),$ we get
$\log\Big(\frac{\text{P}}{1000}\Big)=\frac{5\times10}{100}=\frac{1}{2}=0.5$
$\Rightarrow\frac{\text{P}}{1000}=\text{e}^{0.5}$
$\Rightarrow\text{ P} = 1000 × 1.648 = ₹ \ 1648$

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Question 34 Marks

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.

Find the degree of the differential equation $2\frac{\text{d}^2\text{y}}{\text{dx}^2}+3\sqrt{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}=0}.$

Order and degree of the differential equation $\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3}$ are respectively.

$1, 1$
$1, 2$
$1, 3$
$1, 4$

Find order and degree of the equation $y + y^2 + e^{y} = 0.$

Order $= 3,$ degree $=$ undefined.
Order $= 1,$ degree $= 3.$
Order $= 2,$ degree $=$ undefined.
Order $= 1,$ degree $= 2.$

Determine degree of the differential equation $(\sqrt{\text{a+x}})\times\Big(\frac{\text{dy}}{\text{dx}}\Big)+\text{x}=0.$

$3$
Not defined
$1$
$2$

Order and degree of the differential equation $\Bigg(1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3\Bigg)^\frac{7}{3}=7\frac{\text{d}^2\text{y}}{\text{dx}^2}$ are respectively.

$2, 1$
$2, 3$
$1, 3$
$1,\ \frac{7}{3}$

Answer

$(c)\ 3$

We have, $2\frac{\text{d}^2\text{y}}{\text{dx}^2}+3\sqrt{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}=0}.$
$\therefore\ \ 2\frac{\text{d}^2\text{y}}{\text{dx}^2}=-3\sqrt{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}}$
Squaring both sides, we get
$4\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2=9\Bigg[1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}\Bigg]$
Here, highest order derivative is $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ and its power is $2.$ So, its degree is $2.$

$(d)\ 1, 4$

We have, $\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3}$
$\Rightarrow\ \ \text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^4=\text{x}$
$\Rightarrow $ Here, highest order derivative is $\frac{\text{d}\text{y}}{\text{dx}}.$ So, its order is $I$ and degree is $4.$

$(a)$ Order $= 3,$ degree $=$ undefined.

We have, $y + y^2 + e^{y} = 0.$
$\frac{\text{d}^3\text{y}}{\text{dy}^3}+\text{y}^2+\text{e}^\frac{\text{dy}}{\text{dx}}=0$
Highest order derivative is $\frac{\text{d}^3\text{y}}{\text{dy}^3}.$
So, its order is $3$. Also, the given differential cannot be expressed as a polynomial.
So, its degree is not defined.

$(c)\ 1$

The given differential equation is,
$\sqrt{\text{a+x}}\times\Big(\frac{\text{dy}}{\text{dx}}\Big)+\text{x}=0$
$\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\sqrt{\text{a+x}}}$
Clearly, degree $= 1.$

$(b)\ 2, 3$

We have, $\Bigg(1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3\Bigg)^\frac{7}{3}=7\frac{\text{d}^2\text{y}}{\text{dx}^2}$
$\Rightarrow\ \ \Bigg(1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3\Bigg)^7=\Bigg(7\frac{\text{d}^2\text{y}}{\text{dx}^2}\Bigg)^3$
$\therefore$ Order is $2$ and degree is $3.$

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Question 44 Marks

If an equation is of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ where P, Qare functions of x, then such equation is known as linear differential equation. Its solution is given by
$\text{y}\times\text{(I.F.)}=\int\text{Q}\times\text{(I.F.)}\text{dx}+\text{c},$ where $\text{I.F.}=\text{e}^{\int\text{pdx}}.$
Now, suppose the given equation is $(1+\sin\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}+\text{x}=0.$
Based on the above information, answer the following questions.

The value of P and Q respectively are:

$\frac{\sin\text{x}}{1+\cos\text{x}},\ \frac{\text{x}}{1+\sin\text{x}}$
$\frac{\cos\text{x}}{1+\sin\text{x}},\ \frac{\text{-x}}{1+\sin\text{x}}$
$\frac{-\cos\text{x}}{1+\sin\text{x}},\ \frac{\text{x}}{1+\sin\text{x}}$
$\frac{\cos\text{x}}{1+\sin\text{x}},\ \frac{\text{x}}{1+\sin\text{x}}$

The value of I.F is:

$1-\sin\text{x}$
$\cos\text{x}$
$1+\sin\text{x}$
$1-\cos\text{x}$

Solution of given equation is:

$\text{y}(1-\sin\text{x})=\text{x+c}$
$\text{y}(1+\sin\text{x})=-\text{x}^2+\text{c}$
$\text{y}(1-\sin\text{x})=\frac{\text{-x}^2}{2}\text{+c}$
$\text{y}(1+\sin\text{x})=\frac{\text{-x}^2}{2}\text{+c}$

If y(0) = 1, then y equals

$\frac{2-\text{x}^2}{2(1+\sin\text{x})}$
$\frac{2+\text{x}^2}{2(1+\sin\text{x})}$
$\frac{2-\text{x}^2}{2(1-\sin\text{x})}$
$\frac{2+\text{x}^2}{2(1-\sin\text{x})}$

Value of is $\text{y}\Big(\frac{\pi}{2}\Big)$ is:

$\frac{4-\pi^2}{2}$
$\frac{8-\pi^2}{16}$
$\frac{8-\pi^2}{4}$
$\frac{4+\pi^2}{2}$

Answer

(b) $\frac{\cos\text{x}}{1+\sin\text{x}},\ \frac{\text{-x}}{1+\sin\text{x}}$

Solution:
he given differential equation can be written as $\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}}{1+\sin\text{x}}\text{y}=\frac{\text{-x}}{1+\sin\text{x}}$
Compare it with $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q},$ we get $\text{P}=\frac{\cos\text{x}}{1+\sin\text{x}}$ and $\text{Q}=\frac{\text{-x}}{1+\sin\text{x}}$

Solution:
$\text{I.F.}=\text{e}^{\int\text{pdx}}=\text{e}^{\int\frac{\cos\text{x}}{1+\sin\text{x}}\text{dx}}$
Put $1+\sin\text{x}=\text{t}\Rightarrow\cos\text{x}\ \text{dx}=\text{dt}$
$\therefore\ \text{I.F.}=\text{e}^{\int\frac{1}{\text{t}}\text{dt}}=\text{e}^{\log\text{t}}=\text{t}=1+\sin\text{x}$

(d) $\text{y}(1+\sin\text{x})=\frac{\text{-x}^2}{2}\text{+c}$

Solution:
Solution of given differential equation is given by $\text{y}\times\text{(I.F.)}=\int\text{Q}\text{(I.F.)}\ \text{dx}+\text{c}$
$\Rightarrow\ \text{y}(1+\sin\text{x})=\int\frac{\text{-x}}{1+\sin\text{x}}\times(1+\sin\text{x})\ \text{dx + c}$
$\Rightarrow\ \text{y}(1+\sin\text{x})=\frac{\text{-x}^2}{2}+\text{ c}$

(a) $\frac{2-\text{x}^2}{2(1+\sin\text{x})}$

Solution:
We have, y(0) = 1 i.e., x = 0, y = 1
$\therefore\ 1(1+\sin0)=\text{c}\Rightarrow\text{c}=1$
$\therefore\ \ \text{y}(1+\sin\text{x})=\frac{\text{-x}^2}{2}+1=\frac{2-\text{x}^2}{2}$
$\therefore\ \ \ \text{y}=\frac{2-\text{x}^2}{2(1+\sin\text{x})}$

(b) $\frac{8-\pi^2}{16}$

Solution:
We have, $\text{y}=\frac{2-\text{x}^2}{2(1+\sin\text{x})}$
$\therefore\ \text{y}\Big(\frac{\pi}{2}\Big)=\frac{2-\Big(\frac{\pi}{2}\Big)^2}{2\Bigg(1+\sin\frac{\pi}{2}\Bigg)}=\frac{2-\frac{\pi^2}{4}}{4}=\frac{8-\pi^2}{16}$

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Question 54 Marks

If the equation is of the form $\frac{\text{dy}}{\text{dx}}=\frac{\text{f(x, y)}}{\text{g(x, y)}}$or $\frac{\text{dy}}{\text{dx}}=\text{F}\Big(\frac{\text{y}}{\text{x}}\Big),$ where $f(x, y), g(x, y)$ are homogeneous functions of the same degree in $x$ and $y,$ then put $y = vx$ and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\Big(\frac{\text{dv}}{\text{dx}}\Big),$
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.

The general solution of $\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{xy}+\text{y}^2$ is:

$\tan^{-1}\frac{\text{x}}{\text{y}}=\log|\text{x}|+\text{c}$
$\tan^{-1}\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{c}$
$\text{y}=\text{x}\log|\text{x}|+\text{c}$
$\text{x}=\text{y}\log|\text{y}|+\text{c}$

Solution of the differential equation $2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2+3\text{y}^2$ is:

$x^3 + y^2 = cx^2$
$\frac{\text{x}^2}{2}+\frac{\text{y}^3}{3}=\text{y}^2+\text{c}$
$x^2 + y^3 = cx^2$
$x^2 + y^2 = cx^3$

General solution of the differential equation $(x^2 + 3xy + y^2) dx - x^2$ dy $= 0$ is:

$\frac{\text{x+y}}{\text{y}}-\log\text{x = c}$
$\frac{\text{x+y}}{\text{y}}+\log\text{x = c}$
$\frac{\text{x}}{\text{x+y}}-\log\text{x = c}$
$\frac{\text{x}}{\text{x+y}}+\log\text{x = c}$

General solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\bigg\{\log\Big(\frac{\text{y}}{\text{x}}\Big)+1\bigg\}$ is:

$\log(\text{xy})=\text{c}$
$\log\text{y}=\text{cx}$
$\log\frac{\text{y}}{\text{x}}=\text{cx}$
$\log\text{x}=\text{cy}$

Solution of the differential equation $\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)\text{e}^\frac{\text{y}}{\text{x}}=\text{x}^2\ \cos\text{x}$ is:

$\text{e}^\frac{\text{y}}{\text{x}}-\sin\text{x = c}$
$\text{e}^\frac{\text{y}}{\text{x}}+\sin\text{x = c}$
$\text{e}^\frac{\text{-y}}{\text{x}}-\sin\text{x = c}$
$\text{e}^\frac{\text{-y}}{\text{x}}+\sin\text{x = c}$

Answer

$(b)\ \tan^{-1}\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{c}$

We have, $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+\text{xy}+\text{y}^2}{\text{x}^2}$
Put $y = vx$ and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\frac{\text{dv}}{\text{dx}}$
$\therefore\ \text{v+x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+\text{x}\times\text{vx}+\text{v}^2\text{x}^2}{\text{x}^2}=1+\text{v}+\text{v}^2$
$\Rightarrow \text{x}\frac{\text{dv}}{\text{dx}}=1+\text{v}^2$
$\Rightarrow\int\frac{\text{dv}}{1+\text{v}^2}=\int\frac{\text{dx}}{\text{x}}+\text{c}$
$\Rightarrow\tan^{-1}\text{v}=\log|\text{x}|+\text{c}$
$\Rightarrow\tan^{-1}\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{c}$

$(d)\ x^2 + y^2 = cx^3$

We have,
$2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2+3\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+3\text{y}^2}{2\text{xy}}$
Put $y = vx$ and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\frac{\text{dv}}{\text{dx}}$
$\therefore\ \text{v+x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+3\text{v}^2\text{x}^2}{2\text{vx}^2}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+3\text{v}^2}{2\text{v}}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{2\text{v}}$
$\Rightarrow\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{\text{dx}}{\text{x}}+\log\text{c}$
$\Rightarrow\log|1+\text{v}^2|=\log|\text{x}|+\log|\text{c}|$
$\Rightarrow\log|\text{v}^2+1|=\log|\text{xc}|$
$\Rightarrow\text{v}^2+1=\text{xc}$
$\Rightarrow\frac{\text{y}^2}{\text{x}^2}+1=\text{xc}$
$\Rightarrow\text{x}^2+\text{y}^2=\text{x}^3\text{c}$

$(d)\ \frac{\text{x}}{\text{x+y}}+\log\text{x = c}$

We have,
$(x^2 + 3xy + y^2) dx - x^2 dy = 0$
$\Rightarrow\frac{\text{x}^2+3\text{xy}+\text{y}^2}{\text{x}^2}=\frac{\text{dy}}{\text{dx}}$
Put $y= vx$ and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\frac{\text{dv}}{\text{dx}}$
$\therefore\ \frac{\text{x}^2+3\text{x}^2\text{v}+\text{x}^2\text{v}^2}{\text{x}^2}=\Big(\text{v+x}\frac{\text{dv}}{\text{dx}}\Big)$
$\Rightarrow1+3\text{v}+\text{v}^2=\text{v+x}\frac{\text{dv}}{\text{dx}}$
$\Rightarrow1+2\text{v}+\text{v}^2=\text{x}\frac{\text{dv}}{\text{dx}}$
$\Rightarrow\int\frac{\text{dx}}{\text{x}}-\int(\text{v+1})^{-2}=\text{dv}=\text{c}$
$\log{\text{x}}+\frac{1}{\text{v+1}}=\text{c}$
$\Rightarrow\log\text{x}+\frac{\text{x}}{\text{x+y}}=\text{c}$

$(c)\ \log\frac{\text{y}}{\text{x}}=\text{cx}$

We have, $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\bigg\{\log\Big(\frac{\text{y}}{\text{x}}\Big)+1\bigg\}$
Put $y = vx$ and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\frac{\text{dv}}{\text{dx}}$
$\therefore\ \text{v+x}\frac{\text{dv}}{\text{dx}}=\text{v}\{\log(\text{v}+1\}$
$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}\log\text{v}$
$\Rightarrow\int\frac{\text{dv}}{\text{v}\log\text{v}}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\log|\log\text{v}|=\log|\text{x}|+\log|\text{c}|$
$\Rightarrow\log\Big(\frac{\text{y}}{\text{x}}\Big)=\text{cx}$

$(a)\ \text{e}^\frac{\text{y}}{\text{x}}-\sin\text{x = c}$

We have, $\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)\text{e}^\frac{\text{y}}{\text{x}}=\text{x}^2\ \cos\text{x}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}\Big)\text{e}^\frac{\text{y}}{\text{x}}=\text{x}\ \cos\text{x}$
Put $y = vx$ and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\frac{\text{dv}}{\text{dx}}$
$\Rightarrow\Big(\text{v+x}\frac{\text{dv}}{\text{dx}}-\text{v}\Big)\text{e}^\text{v}=\text{x}\cos\text{x}$
$\Rightarrow\text{xe}^\text{v}\frac{\text{dv}}{\text{dx}}=\text{x}\cos\text{x}$
$\Rightarrow\int\text{e}^\text{v}\text{dv}=\int\cos\text{x}\ \text{dx}$
$\Rightarrow\text{e}^\text{v}=\sin\text{x + c}$
$\Rightarrow\ \text{e}^\frac{\text{y}}{\text{x}}-\sin\text{x = c}$

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Question 64 Marks

If the equation is of the form $\frac{\text{dy}}{\text{dx}}=\text{py}=\text{Q},$ where P, Q are functions of x, then the solution of the differential equation is given by $\text{ye}^{\int\text{pdx}}=\int\text{Q e}^{\int\text{pdx}}\text{dx}+\text{c},$ where $\text{e}^{\int\text{pdx}}$ is called the integrating factor (I.F.).
Based on the above information, answer the following questions.

The integrating factor of the differential equation $\sin\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}\cos\text{x}=1$ is $(\sin\text{x})^\lambda,$ where $\lambda=$

Integrating factor of the differential equation $(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}-\text{xy}=1$ is:

$-\text{x}$
$\frac{\text{x}}{1+\text{x}^2}$
$\sqrt{1-\text{x}^2}$
$\frac{1}{2}\log(1-\text{x}^2)$

The solution of $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-\text{x}},\text{ y}(0)=0,$ is:

$\text{y}=\text{e}^\text{x}(\text{x}-1)$
$\text{y}=\text{xe}^{-\text{x}}$
$\text{y}=\text{xe}^{-\text{x}}+1$
$\text{y}=(\text{x}+1)\text{e}^{-\text{x}}$

General solution of $\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\sec\text{x}$ is:

$\text{y}\sec\text{y}=\tan\text{x}+\text{c}$
$\text{y}\tan\text{x}=\sec\text{x}+\text{c}$
$\tan\text{x}=\text{y}\tan\text{x}+\text{c}$
$\text{x}\sec\text{x}=\tan\text{y}+\text{c}$

The integrating factor of differential equation $\frac{\text{dy}}{\text{dx}}-3\text{y}=\sin2\text{x}$ is:

$\text{e}^{3\text{x}}$
$\text{e}^{-2\text{x}}$
$\text{e}^{-3\text{x}}$
$\text{xe}^{-3\text{x}}$

Answer

Solution:
The given differential equation can be written as $\frac{\text{dy}}{\text{dx}}+2\text{y}\cot\text{x}=\text{cosec x}$
$\therefore\ \text{I.F}=\text{e}^{\int2\cot\text{xdx}}=\text{e}^{2\log|\sin\text{x}|}=(\sin\text{x})^2$
$\therefore\ \lambda=2$

Solution:
We have, $(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}-\text{xy}=1$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}-\frac{\text{x}}{1-\text{x}^2}\cdot\text{y}=\frac{1}{1-\text{x}^2}$
$\therefore\ \text{I.F.}=\text{e}^{-\int\frac{\text{x}}{1-\text{x}^2}\text{dx}}=\text{e}^{\frac{1}{2}\int\frac{-2\text{x}}{1-\text{x}^2}\text{dx}}$
$=\text{e}^{\frac{1}{2}\log(1-\text{x}^2)}=\text{e}^{\log(1-\text{x}^2)^\frac{1}{2}}=\sqrt{1-\text{x}^2}$

(b) $\text{y}=\text{xe}^{-\text{x}}$

Solution:
We have, $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-\text{x}}$
It is a linear differential equation with $\text{I.F.}=\text{e}^{\int\text{dx}}=\text{e}^\text{x}$
Now, solution is $\text{y}\cdot\text{e}^\text{x}=\int\text{e}^{-\text{x}}\text{dx}+\text{c}$
$\Rightarrow\text{ye}^\text{x}=\int\text{dx}+\text{c}$
$\Rightarrow\text{ye}^\text{x}=\text{x}+\text{c}$
$\Rightarrow\text{y}=\text{xe}^{-\text{x}}+\text{ce}^{-\text{x}}$
$\because\ \text{y}(0)=0\Rightarrow\text{c}=0$
$\therefore\ \text{y}=\text{xe}^{-\text{x}}$

(a) $\text{y}\sec\text{y}=\tan\text{x}+\text{c}$

Solution:
We have, $\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\sec\text{x}$
It is a linear differential equation with,
$\text{I.F.}=\text{e}^{\int\tan\text{xdx}}=\text{e}^{\log|\sec\text{x}|}=\sec\text{x}$
Now, solution is $\text{y}\sec\text{x}=\int\sec^2\text{x dx}+\text{c}$
$\Rightarrow\text{y}\sec\text{x}=\tan\text{x}+\text{c}$

Solution:
We have, $\frac{\text{dy}}{\text{dx}}3\text{y}=\sin2\text{x}$
It is a linear differential equation with,
$\text{I.F.}=\text{e}^{\int-3\text{dx}}=\text{e}^{-3\text{x}}$

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Question 74 Marks

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.

If n(I) denote the number of students infected by Corona virus at any time I, then maximum value of n(I) is:

50
100
500
1000

$\frac{\text{dn}}{\text{dt}}$ is proporuona to:

n(1000 - n)
n(100 + n)
n(100 - n)
n(100 + n)

The value of n(4) is:

1
50
100
1000

The most general solution of differential equation formed in given situation is:

$\frac{1}{1000}\log\Big(\frac{1000-\text{n}}{\text{n}}\Big)=\lambda\text{t}+\text{c}$
$\log\Big(\frac{\text{n}}{100-\text{n}}\Big)=\lambda\text{t}+\text{c}$
$\frac{1}{1000}\log\Big(\frac{\text{n}}{1000-\text{n}}\Big)=\lambda\text{t}+\text{c}$
None of these.

The value of n at any time is given by:

$\text{n(t)}=\frac{1000}{1+999\text{e}^{-0.9906\text{t}}}$
$\text{n(t)}=\frac{1000}{1-999\text{e}^{-0.9906\text{t}}}$
$\text{n(t)}=\frac{100}{1-999\text{e}^{-0.9906\text{t}}}$
$\text{n(t)}=\frac{100}{1+999\text{e}^{-0.9906\text{t}}}$

Answer

(d) 1000

Solution:
Since, maximum number of students in hostel is 1000
$\therefore$ Maximum value of n(t) is 1000.

(a) n(1000 - n)

Solution:
Clearly, according to given information,
$\frac{\text{dn}}{\text{dt}}=\lambda\text{n}(1000-\text{n})$ where $\lambda$ is constant of proportionality.

(b) 50

Solution:
Since, 50 students are infected after 4 days.
$\therefore$ n(4) = 50.

Solution:
We have, $\frac{\text{dn}}{\text{dt}}=\lambda\text{n}(1000-\text{n})$
$\Rightarrow\ \ \int\frac{\text{dn}}{\text{n}(1000-\text{n})}=\lambda\int\text{dt}$
$\Rightarrow\frac{1}{1000}\int\Big(\frac{1}{1000-\text{n}}+\frac{1}{\text{n}}\Big)\text{dn}=\lambda\int\text{dt}$
$\Rightarrow\frac{1}{1000}\Big[\frac{\log(1000-\text{n})}{-1}+\log\text{n}\Big]=\lambda\text{t+c}$
$\Rightarrow\frac{1}{1000}\log\Big(\frac{\text{n}}{1000-\text{n}}\Big)=\lambda\text{t}+\text{c}$

(a) $\text{n(t)}=\frac{1000}{1+999\text{e}^{-0.9906\text{t}}}$

Solution:
When, t = 0, n = 1
This condition is satisfied by option (a) only.

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Question 84 Marks

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 $70^\circ F$ Two hours later, the detective measured the body temperature again and found it to be $60^\circ F,$ where the room temperature is $50^\circ F$ Also, it is given the body temperature at the time of death was normal, i.e., $98.6^\circ F.$
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.

By Newton's law of cooling, $\frac{\text{dT}}{\text{dt}}$ is proportional to:

$T - 60$
$T - 50$
$T - 70$
$T - 98.6$

When $t = 0,$ then body temperature is equal to:

$50^\circ F$
$60^\circ F$
$70^\circ F$
$98.6^\circ F$

When $t = 2,$ then body temperature is equal to:

$50^\circ F$
$60^\circ F$
$70^\circ F$
$98.6^\circ F$

The value of $T$ at any time $t$ is:

$50+20\Big(\frac{1}{2}\Big)^\text{t}$
$50+20\Big(\frac{1}{2}\Big)^\text{t-1}$
$50+20\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$
None of these

If it is given that $\log_\text{e} (2.43) = 0.88789$ and $\log_\text{e} (0.5) = -0.69315,$ then the time at which the murder occur is:

$7:30 \ p.m.$
$5:30 \ p.m.$
$6:00 \ p.m.$
$5:00 \ p.m.$

Answer

$(b) T - 50$

Given, $T$ is the temperature of the body at any time $t.$
Then, by Newton's law of cooling,
we get $\frac{\text{dT}}{\text{dt}}=\text{k(T} - \text{50)},$
where $k$ is the constant of proportionatity.

$(c) 70^\circ F$

From given information, we have
At $8 \ p.m. $ temperature is $70^\circ F$
$\therefore$ At $t = 0, T = 70^\circ F$

$(b) 60^\circ F$

From given information, we have
At $10 \ p.m.,$ temperature is $60^\circ F$
$\therefore$ At $t = 2, T = 60^\circ F$

$(c) 50+20\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$

$\frac{\text{dT}}{\text{dt}}=\text{k}(\text{T}-50)$
$\Rightarrow\frac{\text{dT}}{\text{T}-50}=\text{k}\ \text{dt}$
On integrating both sides, we get
$\log|\text{T}-50|=\text{kt}+\log\text{C}$
$\Rightarrow T - 50 = Ce^{kt}$
Clearly, for $t = 0, T = 70^\circ$
$\Rightarrow C = 20$
Thus, $T - 50 = 20e^{kt}$
For $t = 2, T = 60^\circ$
$\Rightarrow 10 = 20e^{2k}$
$\Rightarrow2\text{k}=\log\Big(\frac{1}{2}\Big)$
$\Rightarrow\text{k}=\frac{1}{2}\log\Big(\frac{1}{2}\Big)$
Hence, $\text{T}=50+20\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$

$(b) 5:30 \ p.m.$

We have, $\text{T}=50+20\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$
Now, $\text{98.6}=50+20\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$
$\Rightarrow\frac{48.6}{20}=\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$
$\Rightarrow\log\Big(\frac{48.6}{20}\Big)=\frac{\text{t}}{2}\log\Big(\frac{1}{2}\Big)$
$\Rightarrow\frac{\text{t}}{2}=\frac{\log\Big(\frac{48.6}{20}\Big)}{\log\Big(\frac{1}{2}\Big)}$
$\Rightarrow{\text{t}}={2}=\begin{pmatrix}\frac{\log\Big(\frac{48.6}{20}\Big)}{\log\Big(\frac{1}{2}\Big)}\end{pmatrix}$
$=-2.56$
So, it appears that the person was murdered $2.5$ hours before $8 \ p.m.$ i.e., about $5 : 30 \ p.m.$

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Question 94 Marks

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.

If y(t) denote the number of people who know the rumour at an instant t, then maximum value of y(t) is:

500
100
5000
None of these

$\frac{\text{dn}}{\text{dt}}$ is proportional to:

(y - 5000)
y(y - 500)
y(500 - y)
y(5000 - y)

The value of y(0) is:

The value of y(2) is:

The value of y at any time t is given by:

$\text{y}=\frac{5000}{_\text{e}-5000\text{kt}_{+1}}$
$\text{y}=\frac{5000}{_\text{1+e}-5000\text{kt}}$
$\text{y}=\frac{5000}{_\text{49e}-5000\text{kt}_{+1}}$
$\text{y}=\frac{5000}{_\text{49}{(_\text{1+e}}-5000\text{kt})}$

Answer

Solution:
Since, size of population is 5000.
$\therefore$ Maximum value of y(t) is 5000.

(d) y(5000 - y)

Solution:
Clearly, according to given information, $\frac{\text{dy}}{\text{dt}}=\text{ky}(5000-\text{y}),$ where k is the constant of proportionality.

(a) 100

Solution:
Since, rumour is initiated with 100 people.
$\therefore$ When t = 0, then y = 100
Thus y(0) = 100

(b) 500

Solution:
Since, rumour is spread in 500 people, after 2 days.
$\therefore$ When t = 2, then y = 500.
Thus, y(2) = 500

Solution:
We know that, when t = 0, then y = 100 This condition is satisfied by option (c) only.

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Question 104 Marks

A thermometer reading 80°F is taken outside. Five minutes later the thermometer reads 60°F. After another 5 minutes the thermometer reads 50°F. At any time t the thermometer reading be T°F and the outside temperature be S°F.

Based on the above information, answer the following questions.

If $\lambda$ is posinve constant of propornonality, then $\frac{\text{dn}}{\text{dt}}$ is:

$\lambda(\text{T - S})$
$\lambda(\text{T + S})$
$\lambda\text{T S}$
$-\lambda(\text{T - S})$

The value of T(S) is:

30°F
40°F
5D°F
60°F

The value of T(10) is:

50°F
60°F
80°F
90°F

Find the general solution of differential equation fanned in given situation.

$\log\text{T}=\text{St + c}$
$\log(\text{T}-\text{S})=\lambda{\text{t + c}}$
$\log\text{T}=\text{tT + c}$
$\log(\text{T + S})=\lambda{\text{t + c}}$

Find the value of constant of integration c in the solution of differential equation formed in given situation.

$\log(60\ -\ \text{S})$
$\log(80\ +\ \text{S})$
$\log(80\ -\ \text{S})$
$\log(60\ +\ \text{S})$

Answer

(d) $-\lambda(\text{T - S})$

Solution:
Given, at any time t the thermometer reading be T°F and the outside temperature be S°F. Then, by Newton's law of cooling, we have
$\frac{\text{dT}}{\text{dt}}\propto(\text{T -S })\Rightarrow\frac{\text{dT}}{\text{dt}}=-\lambda(\text{T - S})$

(d) 60°F

Solution:
Since, after 5 minutes, thermometer reads 60"F
$\therefore$ Value of T(S) = 60°F

(a) 50°F

Solution:
Clearly from given information, value of T(10) is 5O°F.

(b) $\log(\text{T - S})=\lambda{\text{t + c}}$

Solution:
We have, $\frac{\text{dT}}{\text{dt}}=-\lambda({\text{T - S}})$
$\Rightarrow\frac{\text{dT}}{\text{T - S}}=-\lambda\text{dt}\int\frac{1}{\text{T - S}}\text{dt}=-\lambda\int\text{dt}$
$\Rightarrow\log(\text{T - S})=-\lambda\text{t + c}$

Solution:
Since, at t = 0, T = 80°F
$\therefore\ \log(80\ -\ \text{S})=0\ +\ \text{c}$
$\Rightarrow\text{c}=\log(80\ -\ \text{S})$

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MATHS Case study (4 Marks)

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