Limits Indeterminate Form

Evaluate: $\lim\limits_{x \rightarrow 0} \left[\dfrac{x \cos x - \sin x}{x^2 \sin x}\right]$

$\begin{aligned} \lim\limits_{x \rightarrow 0} \left[\dfrac{x \cos x - \sin x}{x^2 \sin x}\right] \;\; \left[\dfrac{0}{0} \text{ form}\right] & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\dfrac{d}{dx} \left(x \cos x - \sin x\right)}{\dfrac{d}{dx} \left(x^2 \sin x\right)}\right] \left[\text{ by L'Hopital's rule}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\cos x - x \sin x - \cos x}{2x \sin x + x^2 \cos x}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{- x \sin x}{x \left(2 \sin x + x \cos x\right)}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{- \sin x}{2 \sin x + x \cos x}\right] \;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\dfrac{d}{dx} \left(- \sin x\right)}{\dfrac{d}{dx} \left(2 \sin x + \cos x\right)}\right] \left[\text{ by L'Hopital's rule}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\cos x}{2 \cos x + \sin x}\right] \\\\ & = \dfrac{1}{2 + 0} \\\\ & = \dfrac{1}{2} \end{aligned}$

Limits Indeterminate Form

Evaluate $\;$ $\lim\limits_{y \rightarrow 0} \dfrac{y - \tan^{-1}y}{y - \sin y}$

$\begin{aligned} \lim\limits_{y \rightarrow 0} \dfrac{y - \tan^{-1}y}{y - \sin y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[y - \tan^{-1}y\right]}{\dfrac{d}{dy} \left[y - \sin y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{1 - \dfrac{1}{1 + y^2}}{1 - \cos y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[1 - \dfrac{1}{1 + y^2}\right]}{\dfrac{d}{dy} \left[1 - \cos y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{1}{\left(1 + y^2\right)^2} \times 2y}{\sin y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[\dfrac{y}{\left(1 + y^2\right)^2}\right]}{\dfrac{d}{dy} \left[\sin y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{\left(1 + y^2\right)^2 \times 1 - y \times 2 \left(1 + y^2\right) \times 2y}{\left(1 + y^2\right)^4}}{\cos y} \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\left(1 + y^2\right)^2 - 4 y^2 \left(1 + y^2\right)}{\left(1 + y^2\right)^4 \cos y} \\\\ & = 2 \times \left[\dfrac{1 - 0}{1 \times 1}\right] \\\\ & = 2 \end{aligned}$

Differential Equations

Solve the differential equation: $\;$ $\sin x \dfrac{dy}{dx} - y = \cos^2 x \times \sin x \times \tan \left(\dfrac{x}{2}\right)$

Given differential equation is: $\;$ $\sin x \dfrac{dy}{dx} - y = \cos^2 x \times \sin x \times \tan \left(\dfrac{x}{2}\right)$

i.e. $\;$ $\dfrac{dy}{dx} - \dfrac{1}{\sin x} \times y = \cos^2 x \times \tan \left(\dfrac{x}{2}\right)$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is a linear differential equation of the form $\;$ $\dfrac{dy}{dx} + P y = Q$ $\;\;\; \cdots \; (2)$

Comparing equation $(1)$ with the standard differential equation gives

$P = \dfrac{-1}{\sin x} = - \text{cosec }x$, $\;$ $Q = \cos^2 x \times \tan \left(\dfrac{x}{2}\right)$

Integrating factor I.F $= e^{\int P dx}$

$\begin{aligned} \therefore \; I.F & = e^{\int - \text{cosec }x \; dx} \\\\ & = e^{- \log \left|\tan \left(x/2\right)\right|} \\\\ & = e^{\log \left|\tan \left(x / 2\right)\right|^{-1}} \\\\ & = \left[\tan \left(\dfrac{x}{2}\right)\right]^{-1} \\\\ & = \cot \left(\dfrac{x}{2}\right) \end{aligned}$

The general solution of equation $(2)$ is

$y \; e^{\int P \; dx} = \displaystyle \int Q \; e^{\int P \; dx} \; dx$

$\therefore \;$ The solution of equation $(1)$ is

$y \; \cot \left(\dfrac{x}{2}\right) = \displaystyle \int \cos^2 x \times \tan \left(\dfrac{x}{2}\right) \times \cot \left(\dfrac{x}{2}\right) \; dx$

i.e. $\;$ $y \cot \left(\dfrac{x}{2}\right) = \displaystyle \int cos^2 x \; dx$

i.e. $\;$ $y \cot \left(\dfrac{x}{2}\right) = \displaystyle \int \left(\dfrac{1 + \cos 2x}{2}\right) \; dx$

i.e. $\;$ $y \cot \left(\dfrac{x}{2}\right) = \dfrac{1}{2} \left[\displaystyle \int dx + \displaystyle \int \cos 2x \; dx\right]$

i.e. $\;$ $y \cot \left(\dfrac{x}{2}\right) = \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$

where $\;$ $C$ is the constant of integration.

Application of Derivatives: Maxima and Minima

A figure consists of a semicircle with a rectangle on its diameter. Given that the perimeter of the figure is $20 \; m$, find its dimensions in order that its area may be maximum.

Let $r$ be the radius of the semicircle.

Let the dimensions of the rectangle be $b$ and $2r$.

Perimeter of the figure $= \pi r + 2b + 2r$

Given: $\;$ Perimeter $= 20 \; m$

i.e. $\;$ $\pi r + 2b + 2r = 20$

i.e. $\;$ $r \left(2 + \pi\right) + 2b = 20$

i.e. $\;$ $b = 10 - \dfrac{r \left(2 + \pi\right)}{2}$ $\;\;\; \cdots \; (1)$

Area of the figure $= A = \dfrac{\pi r^2}{2} + 2 r b$

i.e. $\;$ $A = \dfrac{\pi r^2}{2} + 2 r \left[10 - \dfrac{r \left(2 + \pi\right)}{2}\right]$ $\;\;\;$ [in view of equation $(1)$]

i.e. $\;$ $A = \dfrac{\pi r^2}{2} + 20 r - r^2 \left(2 + \pi\right)$ $\;\;\; \cdots \; (2)$

For maximum area, $\;$ $\dfrac{dA}{dr} = 0$ $\;\;\; \cdots \; (3)$

$\therefore \;$ Differentiating equation $(2)$ w.r.t $x$ we have,

$\dfrac{dA}{dr} = \dfrac{2 \pi r}{2} + 20 - 2r \left(2 + \pi\right)$

i.e. $\;$ $\dfrac{dA}{dr} = \pi r + 20 - r \left(4 + 2 \pi\right)$

i.e. $\;$ $\dfrac{dA}{dr} = 20 - r \left(\pi + 4\right)$ $\;\;\; \cdots \; (4)$

$\therefore \;$ $\dfrac{dA}{dr} = 0$ $\implies$ $20 - r \left(\pi + 4\right) = 0$

i.e. $\;$ $r = \dfrac{20}{\pi + 4}$

From equation $(4)$, $\;$ $\dfrac{d^2 A}{dr^2} \Big |_{r = \frac{20}{\pi + 4}} = - \left(\pi + 4\right) < 0$

$\implies$ $r = \dfrac{20}{\pi + 4}$ $\;$ gives maximum area.

Substituting the value of $r$ in equation $(1)$ gives

$b = 10 - \dfrac{20 \left(\pi + 2\right)}{2 \left(\pi + 4\right)}$

i.e. $\;$ $b = 10 \left(1 - \dfrac{\pi + 2}{\pi + 4}\right) = \dfrac{20}{\pi + 4}$

$\therefore \;$ For maximum area of the figure, $\;$ radius of semicircle $= \dfrac{20}{\pi + 4} \; m$ $\;$ and the sides of the rectangle are $\;$ $\dfrac{20}{\pi + 4} \; m$, $\;$ $\dfrac{40}{\pi + 4} \; m$

Differentiation

If $y = \left(\sin^{-1} x\right)^2$, prove that $\left(1 - x^2\right) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} - 2 = 0$

Given: $\;$ $y = \left(\sin^{-1} x\right)^2$ $\;\;\; \cdots \; (1)$

Differentiating equation $(1)$ w.r.t $x$ we have,

$\dfrac{dy}{dx} = 2 \sin^{-1} x \times \dfrac{1}{\sqrt{1 - x^2}}$

i.e. $\;$ $\left(\dfrac{dy}{dx}\right)^2 = \dfrac{4 \left(\sin^{-1}x\right)^2}{1 - x^2}$

i.e. $\;$ $\left(1 - x^2\right) \left(\dfrac{dy}{dx}\right)^2 = 4 \left(\sin^{-1}x\right)^2$

i.e. $\;$ $\left(1 - x^2\right) \left(\dfrac{dy}{dx}\right)^2 = 4 y$ $\;\;\; \cdots \; (2)$ $\;\;\;$ [in view of equation $(1)$]

Differentiating equation $(2)$ w.r.t $x$ we have,

$\left(1 - x^2\right) \times 2 \dfrac{dy}{dx} \times \dfrac{d^2 y}{dx^2} + \left(\dfrac{dy}{dx}\right)^2 \left(-2 x\right) = 4 \times \dfrac{dy}{dx}$ $\;\;\; \cdots \; (3)$

Since $\;$ $\dfrac{dy}{dx} \neq 0$, $\;$ dividing equation $(3)$ by $\dfrac{dy}{dx}$, we get,

$\left(1 - x^2\right) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} = 2$

i.e. $\;$ $\left(1 - x^2\right) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} - 2 = 0$

Hence proved.

Indefinite Integration

Evaluate: $\;$ $\displaystyle \int \cos 2x \; \log \left(\sin x\right) \; dx$

Let $\; I = \displaystyle \int \cos 2x \; \log \left(\sin x\right) \; dx$

$\left[\text{Note: Integration by parts: } \displaystyle \int u \cdot v \; dx = u \displaystyle \int v \; dx - \displaystyle \int \left\{\dfrac{du}{dx} \times \displaystyle \int v \; dx \right\} dx \right]$

$\left[ \text{Here: } u = \log \left(\sin x\right), \; v = \cos 2x \right]$

$\begin{aligned} \therefore \; I & = \log \left(\sin x\right) \int \cos 2x \; dx - \int \left\{\dfrac{d}{dx} \left[\log \left(\sin x\right)\right] \times \int \cos 2x \; dx \right\} dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{1}{\sin x} \times \cos x \times \dfrac{\sin 2x}{2} \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{\cos x \times 2 \times \sin x \times \cos x}{2 \sin x} \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \cos^2 x \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{1 + \cos 2x}{2} \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \dfrac{1}{2} \int dx - \dfrac{1}{2} \int \cos 2x \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \dfrac{x}{2} - \dfrac{\sin 2x}{4} + c \end{aligned}$

Applications of Definite Integration

Calculate the area of the figure bounded by the curve $y = \log x$, the straight line $x = 2$ and the X-axis.

The curve $y = \log x$ cuts the X-axis at the point $\left(1,0\right)$.

$\therefore \;$ The area between the curve $y = \log x$, the line $x = 2$ and the X-axis is:

$= \displaystyle \int_{1}^{2} y \; dx$

$= \displaystyle \int_{1}^{2} \log x \; dx$

$\begin{aligned} \int \log x \; dx & = \log x \int 1 \; dx - \int \left\{\dfrac{d}{dx} \left(\log x\right) \times \int 1 \; dx \right\} dx \\\\ & \left[\text{Note: Integration by parts: } \right. \\ & \left. \int u \cdot v \; dx = u \int v \; dx - \int \left\{\dfrac{du}{dx} \times \int v \; dx \right\} dx\right] \\\\ & = x \log x - \int \dfrac{1}{x} \times x \; dx \\\\ & = x \log x - \int dx \\\\ & = x \log x - x + c \end{aligned}$

$\begin{aligned} \therefore \; \displaystyle \int_{1}^{2} \log x \; dx & = \left[x \log x\right]_{1}^{2} - \left[x\right]_{1}^{2} \\\\ & = \left[2 \times \log 2 - 1 \times \log 1\right] - \left[2 - 1\right] \\\\ & = 2 \log \left(2\right) - 1 \\\\ & = \log \left(4\right) - 1 \end{aligned}$

$\therefore \;$ Required area $= \log \left(4\right) - 1$ sq units

Definite Integration

Evaluate: $\;$ $\displaystyle \int_{2}^{3} \dfrac{x^3 + 1}{x \left(x - 1\right)} \; dx$

Let $\;$ $I = \displaystyle \int_{2}^{3} \dfrac{x^3 + 1}{x \left(x - 1\right)} \; dx$

$\begin{aligned} i.e. \; I & = \int_{2}^{3} \dfrac{\left(x - 1\right) \left(x^2 + x + 1\right)}{x \left(x - 1\right)} \; dx \;\;\; \left[\text{Note: } a^3 - b^3 = \left(a - b\right) \left(a^2 + ab + b^2\right)\right] \\\\ & = \int_{2}^{3} \dfrac{x^2 + x + 1}{x} \; dx \\\\ & = \int_{2}^{3} x \; dx + \int_{2}^{3} dx + \int_{2}^{3} \dfrac{1}{x} \; dx \\\\ & = \left[\dfrac{x^2}{2}\right]_{2}^{3} + \left[x\right]_{2}^{3} + \left[\log x\right]_{2}^{3} \\\\ & = \dfrac{1}{2} \left(3^2 - 2^2\right) + \left(3 - 2\right) + \log 3 - \log 2 \\\\ & = \dfrac{7}{2} + \log \left(\dfrac{3}{2}\right) \end{aligned}$

Indefinite Integration

Evaluate: $\;$ $\displaystyle \int \sin^{-1} \left(\dfrac{2x}{1 + x^2}\right) \; dx$

Let $\;$ $I = \displaystyle \int \sin^{-1} \left(\dfrac{2x}{1 + x^2}\right) \; dx$

Let $\;$ $x = \tan \theta$

Then, $\;$ $dx = \sec^2 \theta \; d\theta$

$\begin{aligned} \therefore \; I & = \int \sin^{-1} \left(\dfrac{2 \tan \theta}{1 + \tan^2 \theta}\right) \sec^2 \theta \; d\theta \\\\ & = \int \sin^{-1} \left[\sin \left(2 \theta\right)\right] \sec^2 \theta \; d\theta \;\;\; \left[\text{Note: } \sin \left(2A\right) = \dfrac{2 \tan A}{1 + \tan^2 A}\right] \\\\ & = \int 2 \theta \; \sec^2 \theta \; d\theta \;\;\; \left[\text{Note: } \sin^{-1} \left[\sin \left(A\right)\right] = A\right] \\\\ & = 2 \left\{\theta \int \sec^2 \theta \; d\theta - \int \left[\dfrac{d}{d \theta} \left(\theta\right) \times \int \sec^2 d \theta\right] d \theta \right\} \\\\ & \left[\text{Note: Integration by parts: } \int u \cdot v \; dx = u \int v \; dx - \int \left\{\dfrac{du}{dx} \times \int v \; dx \right\} dx\right] \\\\ & = 2 \left\{\theta \; \tan \theta - \int \tan \theta \; d\theta \right\} + c \;\;\; \left[\text{Note: } \int \sec^2 x \; dx = \tan x + c\right] \\\\ & = 2 \; \theta \; \tan \theta - 2\log \left|\sec \theta\right| + c \;\;\; \left[\text{Note: } \int \tan x \; dx = \log \left|\sec x\right| + c\right] \\\\ & = 2 \; \theta \; \tan \theta - \log \left|\sec \theta\right|^2 + c \\\\ & = 2 \; \theta \; \tan \theta - \log \left|1 + \tan^2 \theta\right| + c \;\;\; \left[\text{Note: } 1 + \tan^2 x = \sec^2 x\right] \end{aligned}$

$\because \;$ $x = \tan \theta$ $\implies$ $\theta = \tan^{-1}x$

$\therefore \;$ $I = 2 \; x \; \tan^{-1}x - \log \left|1 + x^2\right| + c$

Differential Equations

If the length of subnormal of a curve is constant and the curve passes through the origin, then find the equation of the curve.

Length of subnormal of a curve at any point $P \left(x,y\right)$ is $= y \; \dfrac{dy}{dx}$

Given: Length of subnormal $=$ constant

i.e. $y \; \dfrac{dy}{dx} = k$ $\;\;\;$ [k is a constant]

i.e. $y \; dy = k \; dx$

i.e. $\displaystyle \int y \; dy = k \int dx$

i.e. $\dfrac{y^2}{2} = k \; x + c$ $\;\;\; \cdots \; (1)$ $\;\;\;$ [c is the constant of integration]

Given: The required curve passes through the origin

$\therefore$ $\;$ Putting $x = 0$, $y = 0$ in equation $(1)$ gives

$c = 0$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ becomes

$\dfrac{y^2}{2} = k \; x$ $\implies$ $y^2 = 2 \; k \; x$ $\;\;\; \cdots \; (3)$

Equation $(3)$ is the required equation of the curve.

Differential Equations

The increase in the principal amount kept at compound interest in a bank is proportional to the product of the principal amount and annual rate of interest.

1. Annual rate of interest in a bank is 5%. How many years will it take to double the principal amount?


2. At what annual rate of interest, the principal amount will double in 10 years?

Let the principal amount at any instant of time t be $= P$

Let the annual rate of interest $= R\% = \dfrac{R}{100} = r$

Increase in principal amount with time $= \dfrac{dP}{dt}$

$\therefore$ $\;$ As per question, $\;\;$ $\dfrac{dP}{dt} \propto Pr$

i.e. $\dfrac{dP}{dt} = k P r$ $\;\;\;$ [k is the constant of proportionality]

i.e. $\dfrac{dP}{P} = k \; r \; dt$

i.e. $\displaystyle \int \dfrac{dP}{p} = k \; r \int dt$

i.e. $\ln P = k \; r \; t + c$ $\;\;\;$ [c is the constant of integration] $\;\;\; \cdots \; (1)$

At time $t = 0$, initial amount $= P = P_0$

Substituting these values of $t$ and $P$ in equation $(1)$ give

$c = \ln P_0$ $\;\;\; \cdots \; (2)$

Substituting the value of $c$ from equation $(2)$ in equation $(1)$ gives

$\ln P = k \; r \; t + \ln P_0$

i.e. $\ln P - \ln P_0 = k \; r \; t$

i.e. $\ln \left(\dfrac{P}{P_0}\right) = k \; r \; t$

i.e. $\dfrac{P}{P_0} = e^{k\;r \; t}$ $\implies$ $P = P_0 \; e^{k \; r \; t}$ $\;\;\; \cdots \; (3)$

After 1 year, $t = 1$, $P = P_0 + P_0 \; r = P_0 \left(1 + r\right)$

Substituting these values of $t$ and $P$ in equation $(3)$ give

$P_0 \left(1 + r\right) = P_0 \; e^{k \; r}$

i.e. $e^{k \; r} = 1 + r$ $\implies$ $k \; r = \ln \left(1 + r\right)$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ In view of equation $(4)$, equation $(3)$ becomes

$P = P_0 \; e^{t \; \ln \left(1 + r\right)}$

i.e. $P = P_0 \; e^{\ln \left(1 + r\right)^t}$

i.e. $P = P_0 \left(1 + r\right)^t$ $\implies$ $P = P_0 \left(1 + \dfrac{R}{100}\right)^t$ $\;\;\; \cdots \; (5)$ $\;\;$ $\left[\because \; r = \dfrac{R}{100}\right]$

1. Given: Rate of interest $= R = 5\% = \dfrac{5}{100}$, $P = 2 P_0$
   
   Substituting these values in equation $(5)$ give
   
   $2 P_0 = P_0 \left(1 + \dfrac{5}{100}\right)^t$
   
   i.e. $2 = \left(1.05\right)^t$
   
   i.e. $\log \left(2\right) = t \; \log \left(1.05\right)$ $\implies$ $t = \dfrac{\log \left(2\right)}{\log \left(1.05\right)} = 14.2$
   
   $\therefore$ $\;$ It takes 14.2 years to double the principal when the rate of interest is 5%



2. Given: Time $= t = 10$ years, Principal amount $= P = 2 P_0$
   
   Substituting these values in equation $(5)$ give
   
   $2 P_0 = P_0 \left(1 + \dfrac{R}{100}\right)^{10}$
   
   i.e. $2 = \left(1 + \dfrac{R}{100}\right)^{10}$
   
   i.e. $\log \left(2\right) = 10 \; \log \left(1 + \dfrac{R}{100}\right)$
   
   i.e. $\dfrac{1}{10} \times \log \left(2\right) = \log \left(1 + \dfrac{R}{100}\right)$
   
   i.e. $\log \left(2^{0.1}\right) = \log \left(1 + \dfrac{R}{100}\right)$
   
   i.e. $2^{0.1} = 1 + \dfrac{R}{100}$ $\implies$ $R = 100 \times \left(2^{0.1} - 1\right) = 7.2$
   
   $\therefore$ $\;$ Annual rate of interest $= 7.2\%$

Differential Equations

In an experiment of culture of bacteria in a laboratory, the rate of increase of bacteria is proportional to the number of bacteria present at that time.

(i) If in one hour the number of bacteria gets doubled, then what is the number of bacteria at the end of 4 hours?

(ii) If the number of bacteria is 24,000 at the end of 3 hours, find the number of bacteria in the beginning.

Let the original number of bacteria at time $t = 0$ be $= N_0$

Let the number of bacteria at any instant of time $t$ be $= N$

Given: $\text{Rate of increase of bacteria} \propto \text{ Number present at that time}$

i.e. $\dfrac{dN}{dt} \propto N$

i.e. $\dfrac{dN}{dt} = k N$ $\hspace{2em}$ where $k$ is the constant of proportionality

i.e. $\dfrac{dN}{N} = k \; dt$

i.e. $\displaystyle \int \dfrac{dN}{N} = k \int dt$

i.e. $\log \left|N\right| = k \; t + c$ $\;\;\; \cdots \; (1)$ $\hspace{2em}$ where $c$ is the constant of integration

Applying the initial condition, when $t = 0$, $N = N_0$, equation $(1)$ becomes

$c = \log \left|N_0\right|$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ can be written as

$\log \left|N\right| = k \; t + \log \left|N_0\right|$

i.e. $\log \left|N\right| - \log \left|N_0\right| = k \; t$

i.e. $\log \left|\dfrac{N}{N_0}\right| = k \; t$

i.e. $\dfrac{N}{N_0} = e^{k t}$

i.e. $N = N_0 \; e^{kt}$ $\;\;\; \cdots \; (3)$

Now, in time $t = 1$ $\;$ hour, the number of bacteria gets doubled.

i.e. when $t = 1$, $N = 2 \; N_0$

$\therefore$ $\;$ In view of this condition equation $(3)$ becomes

$2 \; N_0 = N_0 \; e^k$

i.e. $e^{k} = 2$ $\implies$ $k = \log_e 2$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ In view of equation $(4)$, equation $(3)$ becomes

$N = N_0 \; e^{t \; \log_e 2}$ $\;\;\; \cdots \; (5)$

Given: $t = 4$ hours

$\therefore$ $\;$ We have from equation $(5)$

$\begin{aligned} N & = N_0 \; e^{4 \; \log_e 2} \\\\ & = N_0 \; e^{\log_e \left(2^4\right)} \\\\ & = N_0 \; e^{\log_e 16} \\\\ & = 16 N_0 \end{aligned}$

i.e. Number of bacteria at the end of 4 hours $=$ 16 times the original number

Given: In time $t = 3$ hours, number of bacteria $= N = 24,000$

$\therefore$ $\;$ We have from equation $(5)$

$\begin{aligned} 24000 & = N_0 \; e^{\log_e 2} \\\\ & = N_0 \; e^{\log_e \left(2^3\right)} \\\\ & = N_0 \; e^{\log_e 8} \\\\ & = 8 N_0 \end{aligned}$

$\therefore$ $N_0 = \dfrac{24000}{8} = 3000$

$\therefore$ $\;$ Original number of bacteria $= 3000$ \end{enumerate}

Differential Equations

If the X intercept of the tangent to a curve at any point is four times its Y coordinate, then find the equation of the curve.

Let the slope of the tangent at any point $P \left(x,y\right)$ be $= \dfrac{dy}{dx}$

Let the equation of the tangent be $y = x \; \dfrac{dy}{dx} + k$ $\;\;\; \cdots \; (1)$ where k is the intercept.

Given: X intercept of tangent $= 4 y$

$y = 0$ gives the X intercept made by the tangent.

$\therefore$ We have

$0 = 4y \; \dfrac{dy}{dx} + k$

i.e. $k = - 4 \; y \; \dfrac{dy}{dx}$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ can be written as

$y = x \; \dfrac{dy}{dx} - 4 \; y \; \dfrac{dy}{dx}$

i.e. $y = \left(x - 4y\right) \dfrac{dy}{dx}$

i.e. $\dfrac{dx}{dy} = \dfrac{x - 4y}{y}$

i.e. $\dfrac{dx}{dy} = \dfrac{x}{y} - 4$ $\;\;\; \cdots \; (3)$

Let $\dfrac{x}{y} = u$ $\;\;\; \cdots \; (4a)$

$\implies$ $x = u \; y$ $\;\;\; \cdots \; (4b)$

Differentiating equation $(4b)$ w.r.t $y$ gives

$\dfrac{dx}{dy} = u + y \; \dfrac{du}{dy}$ $\;\;\; \cdots \; (4c)$

$\therefore$ $\;$ In view of equations $(4a)$ and $(4c)$, equation $(3)$ becomes

$u + y \; \dfrac{du}{dy} = u - 4$

i.e. $y \; \dfrac{du}{dy} = -4$

i.e. $\dfrac{du}{4} = - \dfrac{dy}{y}$ $\;\;\; \cdots \; (5)$

Integrating equation $(5)$ gives

$\dfrac{1}{4} \displaystyle \int du = - \int \dfrac{dy}{y}$

i.e. $\dfrac{u}{4} = - \log \left|y\right| + \log \left|c\right|$ $\;$ $c \;$ is the constant of integration

i.e. $\dfrac{u}{4} = \log \left|\dfrac{c}{y}\right|$

i.e. $\dfrac{x}{4y} = \log \left|\dfrac{c}{y}\right|$ $\;\;\;$ [from equation $(4a)$]

i.e. $e^{\frac{x}{4y}} = \dfrac{c}{y}$

i.e. $y = c \; e^{\frac{-x}{4y}}$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the required equation of the curve.

Differential Equations

Solve the differential equation $\;$ $y \; dx - \left(x + 2 y^2\right) \; dy = 0$

The given differential equation is $\;$ $y \; dx - \left(x + 2 y^2\right) \; dy = 0$

i.e. $y \; \dfrac{dx}{dy} - \left(x + 2y^2\right) = 0$

i.e. $\dfrac{dx}{dy} - \dfrac{x}{y} - 2y = 0$

i.e. $\dfrac{dx}{dy} - \dfrac{x}{y} = 2 y$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is a linear differential equation in the form $\;$ $\dfrac{dx}{dy} + P \left(y\right)x = Q \left(y\right)$

Here $P \left(y\right) = -\dfrac{1}{y}$ $\;\;\; \cdots \; (2a)$; $\;$ $Q \left(y\right) = 2y$ $\;\;\; \cdots \; (2b)$

Integrating Factor (I.F) $= e^{\int P \left(y\right) \; dy}$

i.e. $I.F = e^{\int -\frac{1}{y} \; dy} = e^{- \log \left|y\right|} = e^{\log \left|\frac{1}{y}\right|} = \dfrac{1}{y}$ $\;\;\; \cdots \; (3)$

According to the general solution,

$x \; e^{\int P \left(y\right) \; dy} = \displaystyle \int Q \left(y\right) \; e^{\int P \left(y\right) \; dy} \; dy$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(3)$, the general solution of equation $(1)$ is

$\dfrac{x}{y} = \displaystyle \int 2 y \times \dfrac{1}{y} \; dy$

i.e. $\dfrac{x}{y} = 2 \displaystyle \int dy$

i.e. $\dfrac{x}{y} = 2y + c$ $\;\;\; \cdots \; (4)$

Equation $(4)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\;$ $\left(1 + y^2\right) \; dx = \left(\tan^{-1}\left(y\right) - x\right) \; dy$

The given differential equation is $\;$ $\left(1 + y^2\right) \; dx = \left(\tan^{-1}\left(y\right) - x\right) \; dy$

i.e. $\dfrac{dx}{dy} = \dfrac{\tan^{-1}\left(y\right) - x}{1 + y^2}$

i.e. $\dfrac{dx}{dy} + \dfrac{x}{1 + y^2} = \dfrac{\tan^{-1} \left(y\right)}{1 + y^2}$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is a linear differential equation in the form $\;$ $\dfrac{dx}{dy} + P \left(y\right)x = Q \left(y\right)$

Here $P \left(y\right) = \dfrac{1}{1 + y^2}$ $\;\;\; \cdots \; (2a)$; $\;$ $Q \left(y\right) = \dfrac{\tan^{-1} \left(y\right)}{1 + y^2}$ $\;\;\; \cdots \; (2b)$

Integrating Factor (I.F) $= e^{\int P \left(y\right) \; dy}$

i.e. $I.F = e^{\int \frac{dy}{1 + y^2}} = e^{\tan^{-1}\left(y\right)}$ $\;\;\; \cdots \; (3)$

According to the general solution,

$x \; e^{\int P \left(y\right) \; dy} = \displaystyle \int Q \left(y\right) \; e^{\int P \left(y\right) \; dy} \; dy$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(3)$, the general solution of equation $(1)$ is

$x \; e^{\tan^{-1} \left(y\right)} = \displaystyle \int \dfrac{\tan^{-1} \left(y\right)}{1 + y^2} \times e^{\tan^{-1}\left(y\right)} \; dy$ $\;\;\; \cdots \; (4)$

Consider $\displaystyle \int \dfrac{\tan^{-1} \left(y\right)}{1 + y^2} \times e^{\tan^{-1}\left(y\right)} \; dy$ $\;\;\; \cdots \; (5)$

Let $\tan^{-1} \left(y\right) = p$ $\;\;\; \cdots \; (6a)$

Differentiating equation $(6a)$ gives

$\dfrac{dy}{1 + y^2} = dp$ $\;\;\; \cdots \; (6b)$

$\therefore$ $\;$ In view of equations $(6a)$ and $(6b)$, equation $(5)$ becomes

$\begin{aligned} \int \dfrac{\tan^{-1} \left(y\right)}{1 + y^2} \times e^{\tan^{-1}\left(y\right)} \; dy & = \int p \; e^{p} \; dp \\\\ & = p \int e^{p} \; dp - \int \left\{\int e^{p} \; dp \times \dfrac{d}{dp} \left(p\right) \right\} \; dp \\\\ & = p \; e^{p} - \int e^{p}\; dp \\\\ & = p \; e^{p} - e^{p} + c \\\\ & = e^{p} \left(p - 1\right) + c \\\\ & = e^{\tan{-1} \left(y\right)} \left(\tan^{-1} \left(y\right) - 1\right) + c \;\;\; \cdots \; (7) \\ & \hspace{8em} \left[\text{In view of equation } (6a)\right] \end{aligned}$

$\therefore$ $\;$ We have from equations $(4)$ and $(7)$,

$x \; e^{\tan^{-1} \left(y\right)} = e^{\tan^{-1} \left(y\right)} \left(\tan^{-1} \left(y\right) - 1\right) + c$ $\;\;\; \cdots \; (8)$

Equation $(8)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\;$ $\dfrac{dy}{dx} + \dfrac{2y}{x} = e^x$

The given differential equation is $\;$ $\dfrac{dy}{dx} + \dfrac{2y}{x} = e^x$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is a linear differential equation in the form $\;$ $\dfrac{dy}{dx} + P \left(x\right)y = Q \left(x\right)$

Here $P \left(x\right) = \dfrac{2}{x}$ $\;\;\; \cdots \; (2a)$; $\;\;$ $Q \left(x\right) = e^{x}$ $\;\;\; \cdots \; (2b)$

Integrating Factor (I.F) $= e^{\int P \left(x\right) \; dx}$

$\begin{aligned} \text{i.e. I.F} = e^{\int \frac{2}{x} \; dx} & = e^{2 \log \left|x\right|} \\\\ & = e^{\log \left|x^2\right|} \\\\ & = x^2 \;\;\; \cdots \; (3) \end{aligned}$

According to the general solution,

$y \; e^{\int P \left(x\right) \; dx} = \displaystyle \int Q \left(x\right) \; e^{\int P \left(x\right) \; dx} \; dx$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(3)$, the general solution of equation $(1)$ is

$y \; x^2 = \displaystyle \int e^x \; x^2 \; dx$ $\;\;\; \cdots \; (4)$

$\begin{aligned} \text{Now, } \int x^2 \; e^x \; dx & = x^2 \int e^x \; dx - \int \left\{\int e^x \; dx \times \dfrac{d}{dx} \left(x^2\right) \right\} \; dx \\\\ & = x^2 \; e^x - \int 2 \; x \; e^x \; dx \\\\ & = x^2 \; e^x - 2 \left[x \int e^x \; dx - \int \left\{\int e^x \; dx \times \dfrac{d}{dx} \left(x\right) \right\} \; dx\right] \\\\ & = x^2 \; e^x - 2 \left[x \; e^x - \int e^x \; dx\right] \\\\ & = x^2 \; e^x - 2 \; x \; e^x + 2 \; e^x + c \;\;\; \cdots \; (5) \end{aligned}$

$\therefore$ $\;$ In view of equation $(5)$, equation $(4)$ becomes

$y \; x^2 = x^2 \; e^x - 2 \; x \; e^x + 2 \; e^x + c$

i.e. $y \; x^2 = e^x \left(x^2 - 2 \; x + 2\right) + c$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\;$ $x \; \dfrac{dy}{dx} - y = \left(1 + x\right) e^{-x}$

The given differential equation is $\;$ $x \; \dfrac{dy}{dx} - y = \left(1 + x\right) e^{-x}$

i.e. $\dfrac{dy}{dx} - \dfrac{y}{x} = \dfrac{\left(1 + x\right) e^{-x}}{x}$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is a linear differential equation in the form $\;$ $\dfrac{dy}{dx} + P \left(x\right)y = Q \left(x\right)$

Here $P \left(x\right) = - \dfrac{1}{x}$ $\;\;\; \cdots \; (2a)$; $\;\;$ $Q \left(x\right) = \dfrac{\left(1 + x\right) e^{-x}}{x}$ $\;\;\; \cdots \; (2b)$

Integrating Factor (I.F) $= e^{\int P \left(x\right) \; dx}$

$\begin{aligned} \text{i.e. I.F} = e^{\int \frac{-1}{x} \; dx} & = e^{- \log \left|x\right|} \\\\ & = e^{\log \left|\frac{1}{x}\right|} \\\\ & = \dfrac{1}{x} \;\;\; \cdots \; (3) \end{aligned}$

According to the general solution,

$y \; e^{\int P \left(x\right) \; dx} = \displaystyle \int Q \left(x\right) \; e^{\int P \left(x\right) \; dx} \; dx$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(3)$, the general solution of equation $(1)$ is

$\dfrac{y}{x} = \displaystyle \int \dfrac{\left(1 + x\right) e^{-x}}{x} \times \dfrac{1}{x} \; dx$

i.e. $\dfrac{y}{x} = \displaystyle \int \dfrac{\left(1 + x\right) e^{-x}}{x^2} \; dx$ $\;\;\; \cdots \; (4)$

$\begin{aligned} \text{Now, } \int \dfrac{\left(1 + x\right) e^{-x}}{x^2} \; dx & = \left(1 + x\right) e^{-x} \int \dfrac{1}{x^2} \; dx \\ & \hspace{4em} - \int \left\{\int \dfrac{1}{x^2} \; dx \times \dfrac{d}{dx} \left[\left(1 + x\right) e^{-x}\right] \right\} \; dx \\\\ & = \dfrac{- \left(1 + x\right) e^{-x}}{x} + \int \dfrac{1}{x} \left(- e^{-x} + e ^{-x} - x e^{-x}\right) \; dx \\\\ & = \dfrac{- \left(1 + x\right) e^{-x}}{x} - \int e^{-x} \; dx \\\\ & = \dfrac{- \left(1 + x\right) e^{-x}}{x} + e^{-x} + c \;\;\; \cdots \; (5) \end{aligned}$

$\therefore$ $\;$ In view of equation $(5)$, equation $(4)$ becomes

$\dfrac{y}{x} = \dfrac{- \left(1 + x\right) e^{-x}}{x} + e^{-x} + c$

i.e. $y = - \left(1 + x\right) e^{-x} + x e^{-x} + c x$

i.e. $y = - e^{-x} + cx$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\;\;$ $\dfrac{dy}{dx} + 2\; y = \sin x$

The given differential equation is $\;\;$ $\dfrac{dy}{dx} + 2\; y = \sin x$ $\;\;\; \cdots \; (1)$

This is a linear differential equation in the form $\;$ $\dfrac{dy}{dx} + P\left(x\right)y = Q \left(x\right)$

Here $P \left(x\right) = 2$; $Q \left(x\right) = \sin x$ $\;\;\; \cdots \; (2)$

Integrating Factor (I.F) $= e^{\int P \left(x\right) \; dx}$

i.e. $\text{I.F} = e^{\int 2 \; dx} = e^{2x}$ $\;\;\; \cdots \; (3)$

According to the general solution,

$y \; e^{\int P \left(x\right) \; dx} = \displaystyle \int Q \left(x\right) \; e^{\int P \left(x\right) \; dx} \; dx$

$\therefore$ $\;$ In view of equations $(2)$ and $(3)$, the general solution of equation $(1)$ is

$y \; e^{2x} = \displaystyle \int \sin x \; e^{2x} \; dx$ $\;\;\; \cdots \; (4)$

$\begin{aligned} \text{Now } \int \sin x \; e^{2x} \; dx & = \sin x \int e^{2x} \; dx - \int \left\{\int e^{2x} \times \dfrac{d}{dx} \left(\sin x\right) \right\} \; dx \\\\ & = \dfrac{e^{2x} \; \sin x}{2} - \dfrac{1}{2} \int e^{2x} \; \cos x \; dx \\\\ & = \dfrac{e^{2x} \; \sin x}{2} - \dfrac{1}{2} \left[\cos x \int e^{2x} \; dx \right. \\ & \left. \hspace{4em} - \int \left\{\int e^{2x} \times \dfrac{d}{dx} \left(\cos x\right) \right\} \; dx\right] \\\\ & = \dfrac{e^{2x} \; \sin x}{2} - \dfrac{1}{2} \left[\dfrac{e^{2x} \; \cos x}{2} + \dfrac{1}{2} \int e^{2x} \; \sin x \; dx\right] \\\\ & = \dfrac{e^{2x} \; \sin x}{2} - \dfrac{e^{2x} \; \cos x}{4} - \dfrac{1}{4} \int e^{2x} \; \sin x \; dx + c \end{aligned}$

i.e. $\left(1 + \dfrac{1}{4}\right) \displaystyle \int \sin x \; e^{2x} \; dx = \dfrac{e^{2x}}{2} \left(\sin x - \dfrac{\cos x}{2}\right) + c$

i.e. $\displaystyle \int \sin x \; e^{2x} \; dx = \dfrac{2}{5} \; e^{2x} \left(\sin x - \dfrac{\cos x}{2}\right) + c$

i.e. $\displaystyle \int \sin x \; e^{2x} \; dx = \dfrac{e^{2x} \left(2 \sin x - \cos x\right)}{5} + c$ $\;\;\; \cdots \; (5)$

$\therefore$ $\;$ In view of equation $(5)$, equation $(4)$ becomes

$y \; e^{2x} = \dfrac{e^{2x} \left(2 \sin x - \cos x\right)}{5} + c$

i.e. $y = \dfrac{1}{5} \left(2 \sin x - \cos x\right) + c e^{-2x}$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the general solution of the given differential equation.

Differential Equations

Find the particular solution of the differential equation $\;\;$ $\left(x^2 - 2y^2\right) \; dx + 2 \; x \; y \; dy = 0$ under the initial condition $y \left(1\right) = 1$

The given differential equation is $\;\;$ $\left(x^2 - 2y^2\right) \; dx + 2 \; x \; y \; dy = 0$

i.e. $2 \; x \; y \; dy = \left(2 y^2 - x^2\right) \; dx$

i.e. $\dfrac{dy}{dx} = \dfrac{2 y^2 - x^2}{2 \; x \; y}$

i.e. $\dfrac{dy}{dx} = \dfrac{y}{x} - \dfrac{x}{2y}$ $\;\;\; \cdots \; (1)$

Let $\dfrac{y}{x} = v$ $\;\;\; \cdots \; (2a)$

i.e. $y = v \; x$ $\;\;\; \cdots \; (2b)$

Differentiating equation $(2b)$ w.r.t $x$ gives

$\dfrac{dy}{dx} = v + x \; \dfrac{dv}{dx}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ can be written as

$v + x \; \dfrac{dv}{dx} = v - \dfrac{1}{2 \; v}$

i.e. $x \; \dfrac{dv}{dx} = - \dfrac{1}{2 \; v}$

i.e. $2 \; v \; dv = - \dfrac{dx}{x}$ $\;\;\; \cdots \; (3)$

Integrating equation $(3)$ gives

$2 \displaystyle \int v \; dv = - \int \dfrac{dx}{x}$

i.e. $2 \times \dfrac{v^2}{2} = - \log \left|x\right| + c$

i.e. $v^2 = - \log \left|x\right| + c$ $\;\;\; \cdots \; (4)$

Substituting the value of $v$ from equation $(2a)$ in equation $(4)$ gives

$\left(\dfrac{y}{x}\right)^2 = - \log \left|x\right| + c$ $\;\;\; \cdots \; (5)$

Equation $(5)$ is the general solution of the given differential equation.

The initial condition is: when $x = 1$, $y = 1$

Substituting the initial condition in equation $(5)$ gives

$1 = - \log \left|1\right| + c$ $\implies$ $c = 1$

Substituting the value of $c$ in equation $(5)$ gives

$\left(\dfrac{y}{x}\right)^2 = - \log \left|x\right| + 1$

i.e. $\log \left|x\right| = 1 - \dfrac{y^2}{x^2}$

i.e. $x = e^{\left(1 - \frac{y^2}{x^2}\right)}$

i.e. $x = \dfrac{e}{e^{\left(\frac{y^2}{x^2}\right)}}$

i.e. $x \; e^{\left(\frac{y^2}{x^2}\right)} = e$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the particular solution of the given differential equation.

Differential Equations

Find the particular solution of the differential equation $\;\;$ $\left(x^2 + y^2\right) \; dx + x \; y \; dy = 0$ under the initial condition $y \left(1\right) = 1$

The given differential equation is $\;\;$ $\left(x^2 + y^2\right) \; dx + x \; y \; dy = 0$

i.e. $\dfrac{dy}{dx} = - \dfrac{\left(x^2 + y^2\right)}{x \; y}$

i.e. $\dfrac{dy}{dx} = - \left(\dfrac{x}{y} + \dfrac{y}{x}\right)$ $\;\;\; \cdots \; (1)$

Let $\dfrac{y}{x} = v$ $\;\;\; \cdots \; (2a)$

i.e. $y = v \; x$ $\;\;\; \cdots \; (2b)$

Differentiating equation $(2b)$ w.r.t $x$ gives

$\dfrac{dy}{dx} = v + x \; \dfrac{dv}{dx}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ can be written as

$v + x \; \dfrac{dv}{dx} = - \dfrac{1}{v} - v$

i.e. $x \; \dfrac{dv}{dx} = - \left(\dfrac{1}{v} + 2v\right)$

i.e. $x \; \dfrac{dv}{dx} = \dfrac{- \left(1 + 2 v^2\right)}{v}$

i.e. $\dfrac{v \; dv}{1 + 2 v^2} = - \dfrac{dx}{x}$ $\;\;\; \cdots \; (3)$

Integrating equation $(3)$ gives

$\displaystyle \int \dfrac{v}{1 + 2 v^2} \; dv = - \int \dfrac{dx}{x}$ $\;\;\; \cdots \; (4)$

Consider $\displaystyle \int \dfrac{v \; dv}{1 + 2 v^2}$ $\;\;\; \cdots \; (5)$

Let $1 + 2 v^2 = p$ $\;\;\; \cdots \; (5a)$

Differentiating equation $(5a)$ gives

$4 \; v \; dv = dp$ $\implies$ $v \; dv = \dfrac{dp}{4}$ $\;\;\; \cdots \; (5b)$

$\therefore$ $\;$ In view of equations $(5a)$ and $(5b)$, equation $(5)$ becomes

$\displaystyle \int \dfrac{v \; dv}{1 + 2 v^2} = \dfrac{1}{} \int \dfrac{dp}{p}$ $\;\;\; \cdots \; (6)$

$\therefore$ $\;$ We have from equations $(4)$ and $(6)$

$\dfrac{1}{4} \displaystyle \int \dfrac{dp}{p} = - \int \dfrac{dx}{x}$

i.e. $\dfrac{1}{4} \; \log \left|p\right| = - \log \left|x\right| + \log \left|c_1\right|$ $\;\;\; \cdots \; (7)$

$\therefore$ $\;$ In view of equation $(5a)$, equation $(7)$ can be written as

$\dfrac{1}{4} \log \left|1 + 2v^2\right| + \log \left|x\right| = \log \left|c_1\right|$

i.e. $\log \left|x \left(1 + 2v^2\right)^{1/4}\right| = \log \left|c_1\right|$

i.e. $x \left(1 + 2 v^2\right)^{1/4} = c_1$ $\;\;\; \cdots \; (8)$

Substituting the value of $v$ from equation $(2a)$ in equation $(8)$ gives

$x \left(1 + \dfrac{2 \; y^2}{x^2}\right)^{1/4} = c_1$

i.e. $x^4 \left(1 + \dfrac{2 \; y^2}{x^2}\right) = \left(c_1\right)^4$

i.e. $x^4 \left(\dfrac{x^2 + 2\; y^2}{x^2}\right) = c$ $\;$ where $\left(c_1\right)^4 = c$

i.e. $x^2 \left(x^2 + 2 \; y^2\right) = c$ $\;\;\; \cdots \; (9)$

Equation $(9)$ is the general solution of the given differential equation.

The initial condition is: when $x = 1$, $y = 1$

Substituting the initial condition in equation $(9)$ gives

$1^2 \left(1^2 + 2 \times 1^2\right) = c$ $\implies$ $c = 3$

Substituting the value of $c$ in equation $(9)$ gives

$x^2 \left(x^2 + 2 \; y^2\right) = 3$ $\;\;\; \cdots \; (10)$

Equation $(10)$ is the particular solution of the given differential equation.