Let y be the number of bacteria at any instant t.
It is given that the rate of growth of the bacteria is proportional to the number present.
\(\therefore \frac{d y}{d t} \propto y\)
\(\Rightarrow \frac{d y}{d t}=k y \text { (where } k \text { is a constant) }\)
\(\Rightarrow \frac{d y}{y}=k d t\)
Integrating both sides, we get:
\(\int \frac{d y}{y}=k \int d t \)
\(\Rightarrow \log y=k t+\mathrm{C} \)
Let y₀ be the number of bacteria at t = 0.
\(\Rightarrow \log y_{0}=C\)
Substituting the value of C in equation (1), we get:
\(\log y=k t+\log y_{0} \)
\(\Rightarrow \log y-\log y_{0}=k t \)
\(\Rightarrow \log \left(\frac{y}{y_{0}}\right)=k t \)
\(\Rightarrow k t=\log \left(\frac{y}{y_{0}}\right) \)
Also, it is given that the number of bacteria increases by 10% in 2 hours.
\(\Rightarrow y=\frac{110}{100} y_{0} \)
\(\Rightarrow \frac{y}{y_{0}}=\frac{11}{10} \)
Substituting this value in equation (2), we get:
\(k \cdot 2=\log \left(\frac{11}{10}\right) \)
\(\Rightarrow k=\frac{1}{2} \log \left(\frac{11}{10}\right) \)
\(\)Therefore, equation (2) becomes:
\(\frac{1}{2} \log \left(\frac{11}{10}\right) \cdot t=\log \left(\frac{y}{y_{0}}\right) \)
\(\Rightarrow t=\frac{2 \log \left(\frac{y}{y_{0}}\right)}{\log \left(\frac{11}{10}\right)} \)
Now, let the time when the number of bacteria increases from 100000 to 200000 be t₁
\(\Rightarrow y=2 y_{0} \text { at } t=t_{1}\)
From equation (4), we get:
\(t_{1}=\frac{2 \log \left(\frac{y}{y_{0}}\right)}{\log \left(\frac{11}{10}\right)}=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}\)
\(\text { Hence, in } \frac{2 \log 2}{\log \left(\frac{11}{10}\right)}\)hours the number of bacteria increases from 100000 to 200000 .

\(\frac{d y}{d x}=y \tan x \)
\(\Rightarrow \frac{d y}{y}=\tan x d x\)
Integrating both side we get
\(\int \frac{d y}{y}=-\int \tan x d x \)
\(\Rightarrow \log y=\log (\sec x)+\log C \)
\(\Rightarrow \log y=\log (\mathrm{C} \sec x) \)
\(\Rightarrow y=\mathrm{Csec} x \)
\(\text { Now, } y=1 \text { when } x=0 . \)
\(\Rightarrow 1=\mathrm{C} \times \sec 0 \)
\(\Rightarrow 1=\mathrm{C} \times 1 \)
\(\Rightarrow \mathrm{C}=1\)
Substituting C = 1 in equation (1), we get:
\(y=\sec x\)

From the figure, the area of the region ABA′B′A bounded by the ellipse
\(=4\left(\begin{array}{l} \text { area of the region AOBAin the first quadrantbounded } \\ \text { by thecurve, } x-\text { axis and the ordinates } x=0, x=a \end{array}\right)\)
(as the ellipse is symmetrical about both x-axis and y-axis)
\(=4 \int_{0}^{a} y d x \text { (taking verticalstrips) }\)
\(\text { Now } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { gives } y=\pm \frac{b}{a} \sqrt{a^{2}-x^{2}} \text { , }\)but as the region AOBA lies in the first quadrant, y is taken as positive. So, the required area is
\(=4 \int_{0}^{a} \frac{b}{a} \sqrt{a^{2}-x^{2}} d x \)
\(=\frac{4 b}{a}\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a} \)
\(=\frac{4 b}{a}\left[\left(\frac{a}{2} \times 0+\frac{a^{2}}{2} \sin ^{-1} 1\right)-0\right] \)
\(=\frac{4 b}{a} \frac{a^{2}}{2} \frac{\pi}{2}=\pi a b \)

Alternatively, considering horizontal strips as shown in the Figure, the area of the ellipse is
\(=4 \int_0^b x d y=4 \frac{a}{b} \int_0^b \sqrt{b^2-y^2} d y\)
\(=\frac{4 a}{b}\left[\frac{y}{2} \sqrt{b^2-y^2}+\frac{b^2}{2} \sin ^{-1} \frac{y}{b}\right]_0^b\)
\( =\frac{4 a}{b}\left[\left(\frac{b}{2} \times 0+\frac{b^2}{2} \sin ^{-1} 1\right)-0\right] \\ =\frac{4 a}{b} \frac{b^2}{2} \frac{\pi}{2}=\pi a b \)

From , the whole area enclosed by the given circle = 4 (area of the region AOBA bounded by the curve, x-axis and the ordinates x = 0 and x = a) [as the circle is symmetrical about both x-axis and y-axis]

Clearly, area of region in I quadrant
\(=4 \int_0^a y d x\) (taking vertical strips)
\(=4 \int_0^a \sqrt{a^2-x^2} d x\) 
Since \(x^2+y^2=a^2 \text { gives } \quad y= \pm \sqrt{a^2-x^2}\) 
As the region AOBA lies in the first quadrant, y is taken as positive. Integrating, we get the whole area enclosed by the given circle 
Now, required area = 4 x Area of region in I quadrant
\(=4\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a\) 
\(=4\left[\left(\frac{a}{2} \times 0+\frac{a^2}{2} \sin ^{-1} 1\right)-0\right]=4\left(\frac{a^2}{2}\right)\left(\frac{\pi}{2}\right)=\pi a^2\)
Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the region enclosed by circle
\( =4 \int_0^a x d y=4 \int_0^a \sqrt{a^2-y^2} d y \)
\(=4\left[\frac{y}{2} \sqrt{a^2-y^2}+\frac{a^2}{2} \sin ^{-1} \frac{y}{a}\right]_0^a \)
\(=4\left[\left(\frac{a}{2} \times 0+\frac{a^2}{2} \sin ^{-1} 1\right)-0\right] \)
\(=4 \frac{a^2}{2} \frac{\pi}{2}=\pi a^2 \)

Since \(1+y^2 \neq 0\) therefore separating the variables, the given differential equation can be written as
\(\frac { dy }{ dx } =\frac { 1+{ y }^{ 2 } }{ 1+{ x }^{ 2 } } \)....(1)
Integrating both sides of equation (1), we get
\(\int \frac{d y}{1+y^2}=\int \frac{d x}{1+x^2}\)
or \(\tan ^{-1} y=\tan ^{-1} x+C\)
which is the general solution of equation (1).

The given ellipse is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Since (1) is symmetrical about both axes,
Therefore, area of the ellipse = 4 (Shaped area).

\(=4(area\ OAB)\)
\(But\ area\ OAB=\overset { 4 }{ \underset { 0 }{ \int { } } } ydx\quad [Taking\ vertical\ strips]\)
\(=\overset { 4 }{ \underset { 0 }{ \int { } } } \frac { 3 }{ 4 } \sqrt { 16-{ x }^{ 2 } } dx\)
\([\because \frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow \frac { { y }^{ 2 } }{ 9 } =1-\frac { { x }^{ 2 } }{ 16 } \Rightarrow y=\frac { 3 }{ 4 } \sqrt { 16-{ x }^{ 2 } } (\because y>0)]\)
\(=\frac { 3 }{ 4 } \left[ \frac { x\sqrt { 16-{ x }^{ 2 } } }{ 2 } +\frac { 16 }{ 2 } { sin }^{ -1 }\frac { x }{ 4 } \right] _{ 0 }^{ 4 }\)
\(=\frac { 3 }{ 4 } [[2(0)+8{ sin }^{ -1 }(1)]-[0-0]]\)
\(=\frac { 3 }{ 4 } \left[ 8\frac { \pi }{ 2 } \right] =3\pi \)
\(\therefore From\ (2),area\ of\ the\ ellipse =4(3\pi )=12\pi sq.units.\)

\(\left(x^{3}+x^{2}+x+1\right) \frac{d y}{d x}=2 x^{2}+x \)
\(\Rightarrow \frac{d y}{d x}=\frac{2 x^{2}+x}{\left(x^{3}+x^{2}+x+1\right)} \)
\(\Rightarrow d y=\frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)} d x \)
Integrating both sides, we get:
\(\int d y=\int \frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)} d x \)
\(\text { Let } \frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+1} . \)
\(\Rightarrow \frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{A x^{2}+A+(B x+C)(x+1)}{(x+1)\left(x^{2}+1\right)}\)
\(\Rightarrow 2 x^{2}+x=A x^{2}+A+B x^{2}+B x+C x+C \)
\(\Rightarrow 2 x^{2}+x=(A+B) x^{2}+(B+C) x+(A+C) \)
Comparing the coefficients of \(x^{2} \text { and } x \text { , we get: }\)
\(A+B=2 \)
\(B+C=1 \)
\(A+C=0 \)
Solving these equations, we get:
\(A=\frac{1}{2}, B=\frac{3}{2} \text { and } C=\frac{-1}{2}\)
Substituting the values of A,B, and C in equation (2), we get:
\(\frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{1}{2} \cdot \frac{1}{(x+1)}+\frac{1}{2} \frac{(3 x-1)}{\left(x^{2}+1\right)}\)
Therefore, equation (1) becomes:
\(\int d y=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{3 x-1}{x^{2}+1} d x \)
\(\Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{2} \int \frac{x}{x^{2}+1} d x-\frac{1}{2} \int \frac{1}{x^{2}+1} d x \)
\(\Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \cdot \int \frac{2 x}{x^{2}+1} d x-\frac{1}{2} \tan ^{-1} x+\mathrm{C} \)
\(\Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \log \left(x^{2}+1\right)-\frac{1}{2} \tan ^{-1} x+\mathrm{C} \)
\(\Rightarrow y=\frac{1}{4}\left[2 \log (x+1)+3 \log \left(x^{2}+1\right)\right]-\frac{1}{2} \tan ^{-1} x+\mathrm{C} \)
\(\Rightarrow y=\frac{1}{4}\left[(x+1)^{2}\left(x^{2}+1\right)^{3}\right]-\frac{1}{2} \tan ^{-1} x+\mathrm{C} \)
Now, y = 1 when x = 0
\(\Rightarrow 1=\frac{1}{4} \log (1)-\frac{1}{2} \tan ^{-1} 0+\mathrm{C}\)
\(\Rightarrow 1=\frac{1}{4} \times 0-\frac{1}{2} \times 0+\mathrm{C}\)
\(\Rightarrow \mathrm{C}=1 \)
 Substituting C = 1 in equation (3), we get:
\(y=\frac{1}{4}\left[\log (x+1)^{2}\left(x^{2}+1\right)^{3}\right]-\frac{1}{2} \tan ^{-1} x+1\)

As differential equation is
1+ \(3\frac { dy }{ dx } +3{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 3 }={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
Exponent of highest order derivative is 3.

Degree 1

\(y=\cos x+\mathrm{C}\)
Differentiating both sides of this equation with respect to, we get:
\(y^{\prime}=\frac{d}{d x}(\cos x+\mathrm{C}) \)
\(\Rightarrow y^{\prime}=-\sin x \)
Substituting the value of  y' in the given differential equation, we get: 
\(\text {L.H.S. }=y^{\prime}+\sin x=-\sin x+\sin x=0=\text { R.H.S. }\)
Hence, the given function is the solution of the corresponding differential equation.

y = e^x + 1
Differentiating both sides of this equation with respect to x, we get:
\(\frac{d y}{d x}=\frac{d}{d x}\left(e^{x}+1\right) \)
\(\Rightarrow y^{\prime}=e^{x} \)
Now, differentiating equation (1) with respect to x, we get:
\(\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(e^{x}\right) \)
\(\Rightarrow y^{\prime \prime}=e^{x} \)
Substituting the values of  y' and y'' in the given differential equation, we get the L.H.S. as: 
\(y^{\prime \prime}-y^{\prime}=e^{x}-e^{x}=0=\text { R.H.S. }\)
Thus, the given function is the solution of the corresponding differential equation.

Given curve is \(y=\sqrt{3 x+4} \)
On squaring both sides, we get
\(y^{2}=3 x+4 \)
\(\Rightarrow y^{2}=3\left(x+\frac{4}{3}\right) \Rightarrow y^{2}=3\left[x-\left(\frac{-4}{3}\right)\right] \)
which is the equation of the parabola of the form
y² = 4 a x. whose vertex is \(\left(-\frac{4}{3}, 0\right)\) and symmetrical about x-axis.
As  \(y=\sqrt{3 x+4} \) is a positive square root, so we take a upper part of the parabola \(y^{2}=3 x+4 \)
upper part of the parabola \(y^{2}=3 x+4 \)
The area of the region bounded by the curve
\(y=\sqrt{3 x+4}\) between  x = 0, x = 4 and the x-axis,is the area shown in the figure given below
 
∴ Required area \(= \int_{0}^{4} y d x=\int_{0}^{4}(\sqrt{3 x+4}) d x  \)
\(=\int_{0}^{4}(3 x+4)^{1 / 2} d x=\left[\frac{(3 x+4)^{3 / 2}}{\frac{3}{2} \cdot 3}\right]_{0}  \)
\(\begin{array}{l} =\frac{2}{9}\left[(12+4)^{3 / 2}-(4)^{3 / 2}\right] \quad\left[\because \int(a x+b)^{n} d x=\frac{(a x+b)^{n+1}}{a(n+1)}\right] \end{array}  \)
\(=\frac{2}{9}\left[(16)^{3 / 2}-(4)^{3 / 2}\right]  \)
\(=\frac{2}{9}\left[\left(2^{4}\right)^{3 / 2}-\left(2^{2}\right)^{3 / 2}\right]  \)
\(=\frac{2^{4}}{9}\left[(2)^{6}-(2)^{3}\right]=\frac{2}{9}(64-8)  \)
\(=\frac{2}{9} \times 56=\frac{112}{9} \text { sq units } \)
Hence, the required is \( \frac{112}{9} \text { sq units }.\)

Given, differential equation is \(\begin{aligned} \frac{d y}{d x} & =1+x^2+y^2+x^2 y^2 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =1\left(1+x^2\right)+y^2\left(1+x^2\right) \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =\left(1+x^2\right)\left(1+y^2\right) \end{aligned}\)
On separating the variables, we get \(\Rightarrow \frac{d y}{1+y^2}=\left(1+x^2\right) d x\)
On integrating both sides, we get \(\int \frac{d y}{1+y^2}=\int\left(1+x^2\right) d x\)
\(\Rightarrow \quad \tan ^{-1} y=x+\frac{x^3}{3}+C\)          ...(i)
Also, given that y = 1, when x = 0.
On putting x = 0 and y = 1 in Eq. (i), we get
tan^-1 1 = C
\(\Rightarrow \quad \tan ^{-1}(\tan \pi / 4)=C \quad\left[\because \tan \frac{\pi}{4}=1\right]\)
\(\Rightarrow \quad C=\pi / 4\)
On putting the value of C in Eq. (i), we get
\(\begin{aligned} \tan ^{-1} y & =x+\frac{x^3}{3}+\frac{\pi}{4} \end{aligned}\)
\(\begin{aligned} \therefore \quad y & =\tan \left(x+\frac{x^3}{3}+\frac{\pi}{4}\right) \end{aligned}\)
which is the required solution.

As shown in the Figure, the line y = 3x + 2 meets x-axis at x\(=\frac{-2}{3}\) and its graph lies below x-axis for \(x \in\left(-1, \frac{-2}{3}\right)\) and above x-axis for \(x \in\left(\frac{-2}{3}, 1\right)\)
The required area = Area of the region ACBA + Area of the region ADEA
\(=\left|\int_{-1}^{\frac{-2}{3}}(3 x+2) d x\right|+\int_{\frac{-2}{3}}^1(3 x+2) d x\)
\(=\left|\left[\frac{3 x^2}{2}+2 x\right]_{-1}^{\frac{-2}{3}}\right|+\left[\frac{3 x^2}{2}+2 x\right]_{\frac{-2}{3}}^1=\frac{1}{6}+\frac{25}{6}=\frac{13}{3}\)