$\frac{d y}{d x}=y+3>0 \quad y(0)=2, y(\log 2)=?$

$\int \frac{d y}{y+3}=\int d x$

$\log |y+3|=x+c$

$\mathrm{y}(0)=2$

$\log |2+3|=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 5$

y. $(\log 2)=?$

$\log |y+3|=\log 2+\log 5$

$\log |y+3|=\log 10$

$y+3=10$

$y=7$

$\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-\mathrm{t})$

$\int \mathrm{d} \mathrm{V}=\int-\mathrm{k}(\mathrm{T}-\mathrm{t}) \mathrm{dt}$

$V=-k\left[T t-\frac{t^{2}}{2}\right]+C$

At $\mathrm{t}=0 \quad \mathrm{V}=\mathrm{I} \Rightarrow \mathrm{C}=\mathrm{I}$

$\mathrm{V}=-\mathrm{kt}\left(\mathrm{T}-\frac{\mathrm{t}}{2}\right)+\mathrm{I}$

$V(T)=-k T\left(T-\frac{T}{2}\right)+I=\frac{-k T^{2}}{2}+I$

Equation of tangent at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ is

$y-y_{1}=\frac{d y_{1}}{d x_{1}}\left(x-x_{1}\right)$

$\mathrm{x}-$ intercept $=\mathrm{x}_{1}-\mathrm{y}_{1} \frac{\mathrm{d} \mathrm{x}_{1}}{\mathrm{d} y_{1}}$

According to question

$x_{1}=\frac{x_{1}-y_{1} \frac{d x_{1}}{d y_{1}}}{2}$

$\Rightarrow \mathrm{x}_{1}=-\mathrm{y}_{1} \frac{\mathrm{d} \mathrm{x}_{1}}{\mathrm{d} y_{1}}$

$\int \frac{d y}{y}=\int-\frac{d x}{x}$

$\Rightarrow \quad \ell n y=-\ell n x+\ell n c$

$\Rightarrow \quad \mathrm{y}=\frac{\mathrm{c}}{\mathrm{x}} \quad \Rightarrow \quad \mathrm{xy}=\mathrm{c}$

Now at $x=2, y=3$

$\Rightarrow \quad c=6$

$\therefore \quad x y=6 \quad \Rightarrow \quad y=\frac{6}{x}$

$\mathrm{y}^{2} \mathrm{d} \mathrm{x}+\left(\mathrm{x}-\frac{1}{\mathrm{y}}\right) \mathrm{d} \mathrm{y}=0$

$\Rightarrow \quad \mathrm{y}^{2} \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} y}+\mathrm{x}=\frac{1}{\mathrm{y}} \quad \Rightarrow \quad \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} y}+\frac{\mathrm{x}}{\mathrm{y}^{2}}=\frac{1}{\mathrm{y}^{3}}$

Integrating factor (I.F.) $=\mathrm{e}^{\int \frac{1}{\mathrm{y}^{2}} \mathrm{d} \mathrm{y}}=\mathrm{e}^{-1 / \mathrm{y}}$

General solution is –

\[ \mathbf{x} \cdot \mathrm{e}^{-1 / \mathrm{y}}=\int \frac{1}{y^{3}} e^{-1 / y} \mathrm{d} y+c \]

Let $\mathrm{I}_{1}=\int \frac{1}{y^{3}} e^{-1 / y} \mathrm{d} y$

put $\quad \frac{-1}{y}=t$

$y^{-2} d y=d t$

$\therefore \quad \mathrm{I}_{1}=-\int \mathrm{te}^{\mathrm{t}} \mathrm{dt}$

$=-e^{t}(t-1)$

$=e^{t}(1-t)$

General solution is

$\mathrm{xe}^{-1 / \mathrm{y}}=\mathrm{e}^{-1 / \mathrm{y}}\left(1+\frac{1}{\mathrm{y}}\right)+\mathrm{C}$

$\Rightarrow \quad \mathrm{x}=1+\frac{1}{y}+\mathrm{Ce}^{1 / \mathrm{y}}$

Put x = 1, y = 1

$\therefore \quad 1=1+\frac{1}{1}+\mathrm{Ce}^{1 / 1}$

$\Rightarrow \quad C=-1 / \mathrm{e}$

$\therefore \quad x=1+\frac{1}{y}-\frac{e^{1 / y}}{e}$

$\frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=\frac{1}{2} \mathrm{p}(\mathrm{t})-450$

integrate

$\int \frac{\mathrm{d} \mathrm{p}}{\mathrm{p}-900}=\int \frac{1}{2} \mathrm{d} \mathrm{t}$

$\ell \mathrm{n}|(\mathrm{p}-900)|=\frac{1}{2} \mathrm{t}+\mathrm{C} \quad \ldots .(1)$

given $\mathrm{t}=0 \rightarrow \mathrm{p}=850 \quad \therefore \mathrm{C}=\ell \mathrm{n} 50$

from ( 1)

$\mathrm{P}=100 \mathrm{x}-12 \mathrm{x}^{3 / 2} \cdot \frac{2}{3}+\mathrm{C}$

x = 0,

P = 2000

C = 2000

$\mathrm{P}_{(\mathrm{x}=25)}=2500-1000+2000=3500$

$\mathrm{S}=4 \pi \mathrm{r}^{2}$

$\frac{\mathrm{ds}}{\mathrm{dt}}=8=8 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \& \quad \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^{3}$

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}$

$\Rightarrow \frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=4 \mathrm{r}$

put $z=y^{2}$

$\Rightarrow \quad \frac{\mathrm{d} z}{\mathrm{dx}}=2 \mathrm{y} \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}$

$\Rightarrow \quad$ D.E. is $\frac{\mathrm{d} z}{\mathrm{dx}}=\frac{2 \mathrm{z}^{2}}{2\left(\mathrm{xz}-\mathrm{x}^{2}\right)}$

$\Rightarrow \quad z(x d z-z d x)=x^{2} d z$

$\Rightarrow \quad \int d\left(\frac{z}{x}\right)=\int \frac{d z}{z}$

$\Rightarrow \quad \frac{\mathrm{z}}{\mathrm{x}}=\ell \mathrm{nz}+\ell \mathrm{nC}$

$\frac{\mathrm{d} y}{\mathrm{dx}}=1-\frac{1}{\mathrm{x}^{2}}$

$\Rightarrow \quad \mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}+1$

$\Rightarrow \quad \mathrm{y}=-2-\frac{1}{2}+1=-\frac{3}{2}$

$\int \mathrm{d}\left(\left(1+\mathrm{x}^{2}\right) \mathrm{y}\right)=\int 4 \mathrm{x}^{2} \mathrm{d} \mathrm{x}$

$\Rightarrow \quad \mathrm{y}\left(1+\mathrm{x}^{2}\right)=\frac{4 \mathrm{x}^{3}}{3}+\mathrm{C}$ passes through $(0,0)$

$\Rightarrow \quad 3 \mathrm{y}\left(1+\mathrm{x}^{2}\right)=4 \mathrm{x}^{3}$

$\frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=\frac{1}{2} \mathrm{p}(\mathrm{t})-200$

$\int_{10}^{p(t)} \frac{d p(t)}{p(t)-400}=\int_{0}^{t} \frac{d t}{2} \Rightarrow \log \left|\frac{p(t)-400}{-300}\right|=\frac{t}{2}$

$|p(t)-400|=300 e^{t / 2}$

$400-p(t)=300 e^{t / 2}$

$\therefore \mathrm{p}(\mathrm{t})=400-300 \mathrm{e}^{\mathrm{t} / 2}$

$(x \log x) \frac{d y}{d x}+y=2 x \log x,(x \geq 1)$

$\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=2$

I.F. $=e^{\int \frac{d x}{x \log x}}=e^{\log (\log x)}=\log x$

Solution is $y \log x=\int 2 \log x d x$

Given differential equation

$\mathrm{ydx}+\mathrm{xy}^{2} \mathrm{dx}=\mathrm{xdy}$

$\Rightarrow \frac{\mathrm{xdy}-\mathrm{ydx}}{\mathrm{y}^{2}}=\mathrm{xdx}$

$\Rightarrow-\mathrm{d}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\mathrm{d}\left(\frac{\mathrm{x}^{2}}{2}\right)$

Integrating we get

$-\frac{x}{y}=\frac{x^{2}}{2}+C$

$\because$ It passes through $(1,-1)$

$\therefore \quad 1=\frac{1}{2}+\mathrm{C} \Rightarrow \mathrm{C}=\frac{1}{2}$

$\therefore \quad x^{2}+1+\frac{2 x}{y}=0 \Rightarrow y=\frac{-2 x}{x^{2}+1}$

$\therefore \mathrm{f}\left(-\frac{1}{2}\right)=\frac{4}{5}$

$(2+\sin x) \frac{d y}{d x}+(y+1) \cos x=0$

$\frac{d}{d x}(2+\sin x)(y+1)=0$

$(2+\sin x)(y+1)=c$

$x=0, y=1 \Rightarrow c=4$

$\mathrm{y}+1=\frac{4}{2+\sin \mathrm{x}}$

$y\left(\frac{\pi}{2}\right)=\frac{4}{3}-1=\frac{1}{3}$