Differential Equations

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

$C=+\frac{\pi }{2}$

now at x = $\mathcal{l}n\sqrt[]{3}$

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$  

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

-> $C=+\frac{\pi }{2}$

now at x =  $\mathcal{l}n\sqrt[]{3}$  

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$  

$\left[\frac{x}{\sqrt[]{x^2-y^2}}+e^{y/x}\right]x\frac{dy}{dx}=x+\left[\frac{x}{\sqrt[]{x^2-y^2}}+e^{y/x}\right]y$

$\Rightarrow e^{y/x}\left[xdy-ydx\right]=xdx+\frac{x}{\sqrt[]{x^2-y^2}}\left(ydx-xdy\right)$

$\Rightarrow e^{y/x}d\left(y/x\right)=\frac{dx}{x}-\frac{d\left(y/x\right)}{\sqrt[]{1-{\left(y/x\right)}^2}}$

Integrating

$e^{y/x}=\mathcal{l}nx-si{n}^{-1}\left(\frac{y}{x}\right)+c$

Passes (1, 0)

⇒ 1 = c

$\Rightarrow \alpha =\frac{1}{2}exp\left(\sqrt[]{e}-1+\frac{\pi }{6}\right)$

$\frac{dy}{dx}+\frac{2x}{x-1}\cdot y=\frac{1}{{\left(x-1\right)}^2}$

IF  $=e^{{\displaystyle \int \frac{2x}{x-1}dx}}$  

=  $e^{2x}\cdot {\left(x-1\right)}^2$  

$y\cdot e^{2x}{\left(x-1\right)}^2=\{e^{2x}\frac{{\left(x-1\right)}^2}{{\left(x-1\right)}^2}dx+C$

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

y(2) =  $\frac{1+e^4}{2e^4},\Rightarrow C=\frac{1}{2}$  

y(3) =  $\frac{e^{\alpha }+1}{\beta e^{\alpha }}=\frac{e^6+1}{8e^6}$  

$\frac{dy}{dx}+2ytanx=sinx,\therefore I.F.e^{{\displaystyle \int 2tanxdx}}=se{c}^{2}x$  

= cos x – 2 cos² x = $-2{\left(cosx-\frac{1}{4}\right)}^2+\frac{1}{8}\therefore y_{max}=\frac{1}{8}$

${\displaystyle \int e^{sinx}}\left(\frac{5+co{s}^{2}x}{\left(2-sinx\right)\sqrt[]{3+co{s}^{2}x}}\right)cosxdx$

put sin x = t

${\displaystyle {}^{}}\frac{{}^{}}{\text{exception 25:}}$  ${\displaystyle \int e^t}\left(\frac{6-t^2}{\left(2-t\right)\sqrt[]{4-t^2}}\right)$ dt

${\displaystyle \int e^t}\left(\sqrt[]{\frac{2+t}{2-t}}+\frac{2}{{\left(2-t\right)}^{3/2}{\left(2+t\right)}^{1/2}}\right)dt$

If $g\left(t\right)=\sqrt[]{\frac{2+t}{2-t}},g\text{'}\left(t\right)=\frac{2}{{\left(2-t\right)}^{3/2}{\left(2+t^2\right)}^{1/2}}$ $\text{exception 25:}\frac{}{{}^{}{{}^{}}^{}}$

$e^t\sqrt[]{\frac{2+t}{2-t}}+c$

$f\left(\frac{\pi }{2}\right)=\sqrt[]{3}e$

$xdy+ydx=\frac{xydx}{\sqrt[]{1-x^2}}$

$\frac{d\left(xy\right)}{xy}=\frac{dx}{\sqrt[]{1-x^2}}$

Integrate both sides we get

logexy = sin1 x + C

$y(ydx+xdy)=x^5\left(\frac{xdy-ydx}{x^2}\right)$

$\Rightarrow d\left(xy\right)=x^5{y}^{-1}\left(\frac{xdy-ydx}{x^2}\right)\Rightarrow d\left(xy\right)=(xy{)}^{\alpha }{\left(\frac{y}{x}\right)}^{\beta }d\left(\frac{y}{x}\right)$

$\alpha -\beta =5\mathrm{}\alpha +\beta =-1$

$2\alpha =4\mathrm{}\alpha =2\mathrm{}\beta =-3$

$\Rightarrow \mathrm{}\left(xy{)}^{-2}d\right(xy)={\left(\frac{y}{x}\right)}^{-3}d\left(\frac{y}{x}\right)\Rightarrow -(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}+c$

$\Rightarrow \mathrm{}-(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}+c\mathrm{}$ Passing $\left(\mathrm{1,1}\right)$

$\Rightarrow \mathrm{}-1=-\frac{1}{2}+c\Rightarrow c=-\frac{1}{2};-(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}-\frac{1}{2}$

The equation of the line through P (x, y) making an angle with the x-axis which is supplementary to the angle made by the tangent at P (x, y) is

$Y-y=-\frac{dy}{dx}\left(X-x\right)$                    …. (1)

At the point where it meets the x-axis

Y = 0, X = x    $+\frac{y}{\frac{dy}{dx}}\Rightarrow OA=x+\frac{y}{\frac{dy}{dx}}$    …. (2)

The line through P (x, y) and perpendicular to (1) is

$Y-y=\frac{dx}{dy}\left(X-x\right)$           

At the point where it meets the y-axis

$X=0,Y=y=\frac{x}{\frac{dy}{dx}}\Rightarrow OB=y-\frac{x}{\frac{dy}{dx}}$           .

$\boxed{x^2+2xy-y^2=2}$