⇒Defferential Equation

Correct Option is (C)

$\begin{array}{rl}& \cos (x+y)\mathrm{d}y=\mathrm{d}x\\ & \therefore \frac{\mathrm{d}x}{\mathrm{d}y}=\cos (x+y)\text{..... (i)}\end{array}$

Put $x+y=\mathrm{u}$ .... (ii)

Differentiating w.r.t. $y$, we get

$\frac{\mathrm{d}x}{\mathrm{d}y}+1=\frac{\mathrm{du}}{\mathrm{d}y}$

$\therefore \frac{\mathrm{d}x}{\mathrm{d}y}=\frac{\mathrm{du}}{\mathrm{d}y}-1$ ..... (iii)

Substituting (ii) and (iii) in (i), we get

$\begin{array}{rl}& \frac{\mathrm{du}}{\mathrm{d}y}-1=\cos \mathrm{u}\\ & \therefore \frac{\mathrm{du}}{1+\mathrm{cosu}}=\mathrm{d}y\\ & \therefore \frac{\mathrm{du}}{2{\cos }^2\left(\frac{\mathrm{u}}{2}\right)}=\mathrm{d}y\end{array}$

Integrating on both sides, we get

$\begin{array}{rl}& \frac{1}{2}\int {\sec }^2\left(\frac{\mathrm{u}}{2}\right)\mathrm{du}=\int \mathrm{d}y\\ \therefore y& =\tan \left(\frac{x+y}{2}\right)+\mathrm{c}\end{array}$

Correct Option is (B)

$\begin{array}{rl}& \frac{\mathrm{d}x}{\mathrm{d}y}+\frac{x}{y}=x^2\\ & \frac{1}{x^2}\frac{\mathrm{d}x}{\mathrm{d}y}+\frac{1}{xy}=1\text{.... (i)}\end{array}$

Let $\frac{1}{x}=\mathrm{t}$

Differentiating w.r.t. $y$, we get

$\begin{array}{ll}& \frac{-1}{x^2}\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{\mathrm{dt}}{\mathrm{d}y}\Rightarrow \frac{1}{x^2}\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{-\mathrm{dt}}{\mathrm{d}y}\\ \therefore & \text{ (i) }\Rightarrow \frac{-\mathrm{dt}}{\mathrm{d}y}+\frac{\mathrm{t}}{y}=1\\ \therefore & \frac{\mathrm{dt}}{\mathrm{d}y}-\frac{\mathrm{t}}{y}=-1\\ \therefore & \text{ I.F. }={\mathrm{e}}^{\int \frac{-1}{y}\mathrm{d}y}={\mathrm{e}}^{\log \left(\frac{1}{y}\right)}=\frac{1}{y}\end{array}$

$\therefore$ The solution of the given equation is

$\begin{array}{rl}& \mathrm{t}(\text{ I.F })=\int (-1)(\text{ I.F. })\mathrm{d}y+\mathrm{c}\\ & \mathrm{t}\left(\frac{1}{y}\right)=\int \frac{-1}{y}\mathrm{d}y+\mathrm{c}\end{array}$

$\begin{array}{rl}& \therefore \frac{t}{y}=-\log y+c\\ & \therefore \frac{1}{xy}=-\log y+c\\ & \therefore \frac{1}{x}=cy-y\log y\end{array}$

Correct Option is (A)

$\begin{array}{rl}& \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right)\text{.... (i)}\\ & \text{ Put }y=\mathrm{v}x\\ \therefore & \frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{v}+x\frac{\mathrm{dv}}{\mathrm{d}x}\\ \therefore & \mathrm{v}+x\frac{\mathrm{dv}}{\mathrm{d}x}=\mathrm{v}+\sec \mathrm{v}\text{... [From (i)]}\\ & \Rightarrow \cos \mathrm{v}d\mathrm{v}=\frac{1}{x}\mathrm{d}x\end{array}$

Integrating both sides, we get

$\begin{array}{rl}& \sin \mathrm{v}=\log (x)+\mathrm{c}\\ & \Rightarrow \sin \left(\frac{y}{x}\right)=\log (x)+\mathrm{c}\text{.... (ii)}\end{array}$

The curve passes through $(1,\frac{\pi }{6})$.

$\sin \left(\frac{\pi }{6}\right)=\log (1)+c\Rightarrow c=\frac{1}{2}$

$\therefore$ Equation (ii) becomes

$\sin \left(\frac{y}{x}\right)=\log (x)+\frac{1}{2}$

Correct Option is (D)

We have $\left(\frac{y}{x}\right)\cos \left(\frac{y}{x}\right)dx-[\left(\frac{x}{y}\right)\sin \left(\frac{y}{x}\right)+\cos \left(\frac{y}{x}\right)]dy=0$

$\begin{array}{rl}& \therefore \frac{dy}{dx}=\frac{\left(\frac{y}{x}\right)\cos \left(\frac{y}{x}\right)}{\left(\frac{x}{y}\right)\sin \left(\frac{y}{x}\right)+\cos \left(\frac{y}{x}\right)}\\ & \text{Put}\frac{y}{x}=v\Rightarrow y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ & \therefore v+x\frac{dv}{dx}=\frac{v\cos v}{\frac{1}{v}\sin v+\cos v}=\frac{v^2\cos v}{\sin v+v\cos v}\Rightarrow x\frac{dv}{dx}=\frac{-v\sin v}{\sin v+v\cos v}\\ & \therefore \int \frac{\sin v+v\cos v}{v\sin v}dv=\int \frac{-dx}{x}\Rightarrow \int \frac{1}{v}dv+\int \cot vdv=-\int \frac{dx}{x}\\ & \therefore \log |v|+\log |\sin x|=-\log |x|+\log k\Rightarrow \log |(v)(\sin v)(x)|=\log k\\ & \therefore y\sin \left(\frac{y}{x}\right)=k\end{array}$

Correct Option is (D)

$\begin{array}{rl}& (3xy+y^2)dx+(x^2+xy)dy=0\\ & \therefore \frac{dy}{dx}=-\frac{(3xy+y^2)}{(x^2+xy)}\end{array}$

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\begin{array}{rl}& \therefore v+x\frac{dv}{dx}=-\frac{(3v{x}^2+v^2{x}^2)}{(x^2+v^2)}=-\frac{3v+v^2}{1+v}\\ & \therefore x\frac{dv}{dx}=-\frac{3v+v^2}{1+v}-v=-\frac{(2v^2+4v)}{1+v}\\ & \therefore \int \frac{1+v}{(2v^2+4v)}dv=\int \frac{-1}{x}dx\\ & \therefore \frac{1}{2}\int \frac{v+1}{v^2+2v}dv=-\int \frac{dx}{x}\Rightarrow \frac{1}{4}\int \frac{2(v+1)}{v^2+2v}dv=-\log x+\log c_1\\ & \therefore \frac{1}{4}\log (v^2+2v)+\log x=\log c_1\Rightarrow \frac{1}{4}\log (\frac{y^2}{x^2}+\frac{2y}{x})+\log x=\log c_1\end{array}$

$\begin{array}{rl}& \therefore \frac{1}{4}\log \left(\frac{y^2+2xy}{x^2}\right)+\log x=\log c_1\Rightarrow \log \left(\frac{y^2+2xy}{x^2}\right)+\log x^4=4\log c_1\\ & \therefore \log \left[\left(\frac{y^2+2xy}{x^2}\right)\left(x^4\right)\right]=\log c_1^4\Rightarrow \log \left[x^2(y^2+2xy)\right]=\log c_1^4\\ & \therefore x^2(y^2+2xy)=c\end{array}$