(3)

$\text{We have,} \frac{dy}{dx}=\frac{x^2+y^2}{2xy}$

Putting $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$, we get

$v+x\frac{dv}{dx}=\frac{1}{2}\left(\frac{{\displaystyle 1+v^2}}{{\displaystyle v}}\right)$

$\Rightarrow x\frac{dv}{dx}=\frac{1}{2}(\frac{{\displaystyle 1}}{{\displaystyle v}}-v)=\frac{1}{2}\left(\frac{{\displaystyle 1-v^2}}{{\displaystyle v}}\right)$              $\therefore$ $\int \frac{2v}{1-v^2} dv=\int \frac{dx}{x}$

$\Rightarrow \log c-\log$$|1-v^2|$$=\log$$\left|x\right|$

           $c=x(1-\frac{{\displaystyle y^2}}{{\displaystyle x^2}})=\frac{x^2-y^2}{x}$

At $y\left(2\right)=0, c=2$

$\therefore$  $\frac{x^2-y^2}{x}=2$

Now, at $x=8,$ $2=\frac{8^2-y^2}{8}\Rightarrow y^2=64-16=48\Rightarrow y=4\sqrt{3}$

(3)

We have, $x^{2}f\left(x\right)-x=4{\int }_0^{x}tf\left(t\right) dt$

$\Rightarrow 2xf\left(x\right)+x^{2}f\text{'}\left(x\right)-1=4xf\left(x\right)$

$\Rightarrow x^2\frac{dy}{dx}-2xy=1 [ f(x)=y, f\text{'}(x)=\frac{dy}{dx}]$

$\Rightarrow \frac{dy}{dx}-\frac{2y}{x}=\frac{1}{x^2}, \text{which is a linear differential equation.}$

$\text{I.F.}=e^{\int -\frac{2}{x} dx}=e^{-2\log x}=\frac{1}{x^2}$

$\therefore$Solution is given by

$\frac{y}{x^2}=\int \frac{1}{x^4} dx+C\Rightarrow \frac{y}{x^2}=-\frac{1}{3x^3}+C$                    ...(i)

$\text{At }x=1, y=\frac{2}{3}$     $[\because f(1)=\frac{2}{3}]$

     $\frac{2}{3}=\frac{-1}{3}+C \Rightarrow C=1$

$\therefore$ $y=\frac{-1}{3x}+x^2\Rightarrow f\left(x\right)=\frac{-1}{3x}+x^2$

$\text{Now, } 18f\left(3\right)=18[\frac{{\displaystyle -1}}{{\displaystyle 9}}+9]=160$

(4)

$(1-x^2{y}^2) dx=ydx+xdy$

$\Rightarrow dx=\frac{d\left(xy\right)}{1-{\left(xy\right)}^2}$

$\int dx=\int \frac{d\left(xy\right)}{1-{\left(xy\right)}^2}$ $x=\frac{1}{2}\ln$$\left|\frac{1+xy}{1-xy}\right|$$+C$

$\text{We are given }y\left(1\right)=2\text{ and }y\left(2\right)=\alpha$

$\text{Put }x=1, y=2:1=\frac{1}{2}\ln$$\left|\frac{{\displaystyle 1+2}}{{\displaystyle 1-2}}\right|$$+C$

$C=1-\frac{1}{2}\ln 3$

$\text{Now put }x=2, y=\alpha$

$2=\frac{1}{2}\ln$$\left|\frac{{\displaystyle 1+2\alpha }}{{\displaystyle 1-2\alpha }}\right|$$+1-\frac{1}{2}\ln 3$$;$ $1+\frac{1}{2}\ln 3=\frac{1}{2}\ln$$\left|\frac{{\displaystyle 1+2\alpha }}{{\displaystyle 1-2\alpha }}\right|$

$2+\ln 3=\ln \left(\frac{{\displaystyle 1+2\alpha }}{{\displaystyle 1-2\alpha }}\right)$

$\Rightarrow$$\left|\frac{{\displaystyle 1+2\alpha }}{{\displaystyle 1-2\alpha }}\right|$$=3e^2$$\Rightarrow \frac{1+2\alpha }{1-2\alpha }=3e^2,\frac{1+2\alpha }{1-2\alpha }=-3e^2$

$\frac{1+2\alpha }{1-2\alpha }=3e^2\Rightarrow \alpha =\frac{3e^2-1}{2(3e^2+1)}$

and $\frac{1+2\alpha }{1-2\alpha }=-3e^2\Rightarrow \alpha =\frac{3e^2+1}{2(3e^2-1)}$

(3)

$\text{I.F.}=e^{\int \frac{5dx}{x(x^5+1)}} =e^{\int \frac{5x^{-6}}{x^{-5}+1}dx}$

Put $1+x^{-5}=t\Rightarrow -5x^{-6}dx=dt$

$\therefore$$\text{I.F.}=e^{\int \frac{-dt}{t}}=\frac{1}{t}=\frac{x^5}{1+x^5}$

$\text{Solution is given by }\frac{y{x}^5}{1+x^5}=\int \frac{x^5}{(1+x^5)}\times \frac{{(1+x^5)}^2}{x^7}dx+C$

$=\int x^3 dx+\int x^{-2}dx+C$               $\therefore$$\frac{y{x}^5}{1+x^5}=\frac{x^4}{4}-\frac{1}{x}+C$

At $x=1,y=2$

$\frac{2}{2}=\frac{1}{4}-1+C\Rightarrow C=\frac{7}{4}$              $\therefore$  $\frac{y{x}^5}{1+x^5}=\frac{x^4}{4}-\frac{1}{x}+\frac{7}{4}$

$\text{At }x=2$

$y\left(\frac{32}{33}\right)=\frac{21}{4}=\frac{693}{128}$

(2)

$\text{Given differential equation is}$

$(1+x^2) dy=y(x-y) dx$

$\Rightarrow \frac{dy}{dx}=\frac{xy-y^2}{1+x^2}\Rightarrow \frac{dy}{dx}=\frac{xy}{1+x^2}-\frac{y^2}{1+x^2}$

        $\frac{dy}{dx}-\frac{xy}{1+x^2}=\frac{-y^2}{1+x^2}$

$\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}\frac{x}{1+x^2}=\frac{-1}{1+x^2}$                       ...(i)

Put  $\frac{1}{y}=z \Rightarrow \frac{-1}{y^2}\frac{dy}{dx}=\frac{dz}{dx}$

So, (i) becomes

$\frac{dz}{dx}+\frac{xz}{1+x^2}=\frac{1}{1+x^2}$

Which is a linear differential equation.

$\text{I.F.}=e^{\int \frac{x}{1+x^2}dx}=e^{\int \frac{x}{1+x^2}dx}=e^{\ln \sqrt{1+x^2}}=\sqrt{1+x^2}$

$\therefore$ $\text{Solution is} z\sqrt{1+x^2}=\int \frac{\sqrt{1+x^2}}{1+x^2}dx$

$z\sqrt{1+x^2}=\int \frac{dx}{\sqrt{1+x^2}}$

$z\sqrt{1+x^2} =\log$$|x+\sqrt{1+x^2}|$$+C$

$\frac{\sqrt{1+x^2}}{y}=\log$$|x+\sqrt{1+x^2}|$$+C$                       ...(ii)

$\text{We have, }y\left(0\right)=1$

So, from (ii), $1=\log 1+C\Rightarrow C=1$

$\therefore$ $\text{(ii) becomes}$

$\Rightarrow \frac{\sqrt{1+x^2}}{y}=\log$$|x+\sqrt{1+x^2}|$$+1$

$\Rightarrow$$\frac{\sqrt{1+x^2}}{y}=\log$$|x+\sqrt{1+x^2}|$$+\log e$

$\Rightarrow$$\frac{\sqrt{1+x^2}}{y}=\log \left(e\right(x+\sqrt{1+x^2}\left)\right)$

$\Rightarrow$Also, $y\left(2\sqrt{2}\right)=\beta$

$\frac{3}{\beta }=\log \left(e\right(2\sqrt{2}+3\left)\right)\Rightarrow e^{\frac{3}{\beta }}=\left(e\right(3+2\sqrt{2}\left)\right)$

$\Rightarrow e^{3{\beta }^{-1}}=\left(e\right(3+2\sqrt{2}\left)\right)$

(3)

We have, $\frac{dy}{dx}=y+7, y_1\left(0\right)=0, y_2\left(0\right)=1$

$\int \frac{dy}{y+7}=\int dx$

$\ln$$|y+7|$$=x+k \Rightarrow y+7=e^{x+k}=e^x·e^k=C{e}^x [\because e^k=C]$

$\Rightarrow y=-7+C{e}^x$

$y_1\left(0\right)=0 \Rightarrow 0=-7+C \Rightarrow C=7$

$y_2\left(0\right)=1 \Rightarrow 1=-7+C \Rightarrow C=8$

$\therefore$ $y_1\left(x\right)=-7+7e^x \text{and} y_2\left(x\right)=-7+8e^x$

If $y_1\left(x\right)$ and $y_2\left(x\right)$ intersect at any point, then the values of both curves will be the same at that point.

$\therefore$   $-7+7e^x=-7+8e^x \Rightarrow e^x=0 \Rightarrow \text{Not possible}$

$\therefore$ $\text{The curves }{y}_1\left(x\right)\text{ and }{y}_2\left(x\right)\text{ will not intersect at any point.}$

(2)

$2(y+2)\ln (y+2)dx+(x+4-2\ln (y+2\left)\right)dy=0$

$\Rightarrow 2\ln (y+2)+(x+4-2\ln (y+2\left)\right)\frac{1}{y+2}·\frac{dy}{dx}=0$               ...(i)

Let $\ln (y+2)=t$   $\Rightarrow$   $\frac{1}{y+2}·\frac{dy}{dx}=\frac{dt}{dx}$

Equation (i) becomes,

       $2t+(x+4-2t)·\frac{dt}{dx}=0 \Rightarrow (x+4-2t)\frac{dt}{dx}=-2t$

$\Rightarrow \frac{dx}{dt}=\frac{2t-4-x}{2t} \Rightarrow \frac{dx}{dt}+\frac{x}{2t}=\frac{2t-4}{2t}$

which is a linear differential equation in $x$.

$I.F.=e^{\int \frac{1}{2t} dt} =e^{\frac{1}{2}\ln t} =t^{1/2}$

Now, $x{t}^{1/2}=\int \frac{2t-4}{2t} t^{1/2} dt+C$

$\Rightarrow x{t}^{1/2}=\int (t^{1/2}-\frac{{\displaystyle 2}}{{\displaystyle t^{1/2}}})dt+C$

$\Rightarrow x·t^{1/2}=\frac{t^{3/2}}{3/2}-2·\frac{t^{1/2}}{1/2}+C$

$\Rightarrow x·t^{1/2}=\frac{2t^{3/2}}{3}-4t^{1/2}+C \Rightarrow x=\frac{2}{3}·t-4+C· t^{-1/2}$

Put $t=\ln (y+2)$

        $x=\frac{2}{3}\ln (y+2)-4+C·(\ln {(y+2))}^{-1/2}$                   ...(ii)

Put $y=e^4-2$ and $x=1$

     $1=\frac{2}{3}\ln (e^4-2+2)-4+C(\ln {(e^4-2+2))}^{-1/2}$

$\Rightarrow 1=\frac{2}{3}\times 4-4+C\times \frac{1}{2} \Rightarrow \frac{C}{2}=5-\frac{8}{3}=\frac{7}{3} \Rightarrow C=\frac{14}{3}$

Equation (ii) becomes,

$x=\frac{2}{3}\ln (y+2)-4+\frac{14}{3}(\ln {(y+2))}^{-1/2}$

Put $y=e^9-2$

     $x(e^9-2)=\frac{2}{3}\ln (e^9-2+2)-4+\frac{14}{3}(\ln {(e^9-2+2))}^{-1/2}$

$\Rightarrow x(e^9-2)=\frac{2}{3}\times 9-4+\frac{14}{3}\times \frac{1}{3}=2+\frac{14}{9}=\frac{32}{9}$

(1)

$\frac{dy}{dx}+y\tan x=x\sec x, 0\le x\le \frac{\pi }{3}, y\left(0\right)=1$

$\text{I.F.}=e^{\int \tan x dx}=e^{{\log }_e\sec x}=\sec x$$\Rightarrow y·\sec x=\int x{\sec }^{2}x dx$

$\Rightarrow y\sec x=x\tan x-\int \tan x dx$

$\Rightarrow y·\sec x=x\tan x-{\log }_e\left(\sec x\right)+c$

$\text{Now }y\left(0\right)=1\Rightarrow 1=c$

$\text{Now, at }x=\frac{\pi }{6}, \text{we have}$

$\frac{2y}{\sqrt{3}}=\frac{\pi }{6}·\frac{1}{\sqrt{3}}-{\log }_e\frac{2}{\sqrt{3}}+1;$     $y=\frac{\pi }{12}-\frac{\sqrt{3}}{2}{\log }_e\frac{2}{\sqrt{3}}+\frac{\sqrt{3}}{2}$

$=\frac{\pi }{12}-\frac{\sqrt{3}}{2}[{\log }_e\frac{{\displaystyle 2}}{{\displaystyle \sqrt{3}}} -{\log }_{e}e]$$=\frac{\pi }{12}-\frac{\sqrt{3}}{2}{\log }_e\left(\frac{{\displaystyle 2}}{{\displaystyle e\sqrt{3}}}\right)$

(3)

$\alpha x=e^{x^{\beta }·y^{\gamma }}$ and $2x^{2}y\frac{dy}{dx}=1-x·y^2$

Let $y^2=t\Rightarrow 2y\frac{dy}{dx}=\frac{dt}{dx}$$\Rightarrow x^2\frac{dt}{dx}=1-xt$

$\Rightarrow \frac{dt}{dx}+\frac{t}{x}=\frac{1}{x^2}$

$\text{I.F.}=e^{{\log }_{e}x}=x$

$t\left(x\right)=\int \frac{1}{x^2}·xdx \Rightarrow y^2·x={\log }_{e}x+C$

$\Rightarrow 2{\log }_{e}2={\log }_{e}2+C \Rightarrow C={\log }_{e}2$

$\text{Hence, }x{y}^2={\log }_{e}2x$    $\therefore$$2x=e^{x·y^2}$

$\text{Hence, }\alpha =2, \beta =1, \gamma =2$      $\therefore$ $\alpha +\beta -\gamma =1$

(3)

$\text{Given: }{x}^{3}dy+(xy-1)dx=0$

$\Rightarrow x^{3}dy=(1-xy)dx$

$\Rightarrow \frac{dy}{dx}=\frac{1-xy}{x^3}\Rightarrow \frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3},$ which is a linear differential equation.

$\text{Now, (I.F.) }=e^{\int P dx}=e^{\int \frac{1}{x^2} dx}=e^{-1/x}$

Solution is given by, $y{e}^{-1/x}=\int \frac{1}{x^3}{e}^{-1/x}dx+C$

Putting $u=\frac{-1}{x} \Rightarrow du=\frac{1}{x^2}dx,$we get

$y{e}^{-1/x}=\int -e^{u}udu+c=-[u{e}^u-\int e^u du]+c$

     $=-[u{e}^u-e^u]+c=(1-u)e^u+c$

$\Rightarrow y{e}^{-1/x}=(1+\frac{{\displaystyle 1}}{{\displaystyle x}})e^{-1/x}+c$                           ...(i)

Put $x=\frac{1}{2}$ and $y=3-e$

$3e^{-2}-e^{-1}=3e^{-2}+c \Rightarrow c=-e^{-1}$

Equation (i) becomes,  $y{e}^{-1/x}=e^{-1/x}+\frac{e^{-1/x}}{x}-e^{-1}$

Put $x=1$ to get $y\left(1\right)$,

$\Rightarrow y{e}^{-1}=e^{-1}+e^{-1}-e^{-1}=e^{-1} \Rightarrow y=1$