Differential Equations

l + m – n = 0

l + m = n . (i)

l² + m² = n²

Now from (i)

l² + m² = (l + m)²

=> 2lm = 0

=>lm = 0

l = 0 or m = 0

=> m = n Þ l = n

if we take direction consine of line

$0,\frac{1}{\sqrt[]{2}},\frac{1}{\sqrt[]{2}}and\frac{1}{\sqrt[]{2}},0,\frac{1}{\sqrt[]{2}}$

cos a = $\frac{1}{2}$              

$si{n}^4\alpha +co{s}^4\alpha ={\left(\frac{\sqrt[]{3}}{2}\right)}^4={\left(\frac{1}{2}\right)}^4=\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=\frac{5}{8}$  

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

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

x = 0, y = 0

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

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

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

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

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

$x_1=\underset{x\to \infty }{lim}\frac{2^{\frac{x^n}{e^x}}-3^{\frac{x^n}{e^x}}}{\frac{x^n}{e^x}},put\frac{x^n}{e^x}=t$

$x_2=\underset{x\to \infty }{lim}\frac{co{t}^{-1}\left(\sqrt[]{x+1}-\sqrt[]{x}\right)}{se{c}^{-1}\left({\left(\frac{2x+1}{x-1}\right)}^x\right)}$

$x_2=\frac{2}{\pi }\underset{x\to \infty }{lim}ta{n}^{-1}\left(\sqrt[]{x+1}+\sqrt[]{x}\right)$

$x_2=\frac{2}{\pi }.\frac{\pi }{2}=1$

Differentiating

y.  $\frac{-2x}{2\sqrt[]{1-x^2}}+\sqrt[]{1-x^2}\cdot y^{\mathrm{\text{'}}}=\frac{x\cdot 2y{y}^{\mathrm{\text{'}}}}{2\sqrt[]{1-y^2}}-\sqrt[]{1-y^2}$

Put $x=\frac{1}{2},y=-\frac{1}{4}$ and $x,y=-\frac{1}{8}$

$y^{\mathrm{\text{'}}}=\frac{-\sqrt[]{5}}{2}$

dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.

$af\left(x\right)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta }{x}............\left(i\right)$  

Replace x by   $\frac{1}{x}$

$af\left(\frac{1}{x}\right)+\alpha f\left(x\right)=\frac{b}{x}+\beta x..............\left(ii\right)$  

$a\left(f\left(x\right)+f\left(\frac{1}{x}\right)\right)+\alpha \left(f\left(x\right)+f\left(\frac{1}{x}\right)\right)=b\left(x+\frac{1}{x}\right)+\beta \left(x+\frac{1}{x}\right)$           

$\therefore \frac{f\left(x\right)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}=\frac{b+\beta }{a+\alpha }=\frac{2}{1}=2$           

$x\frac{dy}{dx}+y=b{x}^4$       

$\frac{dy}{dx}+\frac{y}{x}=b{x}^3$           

As per question

$\frac{32b}{25}+\left(10-b\right)ln2-\frac{b}{32}=\frac{62}{5}$           

$\left(10-b\right)ln2=\frac{62}{5}-\frac{31}{25}b=\frac{31}{5}\left(2-\frac{b}{5}\right)$           
$\therefore b=10$

$2x{y}^2-y)dx+xdy=0$

$\Rightarrow \frac{dy}{dx}+2y^2-\frac{y}{x}=0$

$Put\frac{1}{y}=z$

Then $-\frac{1}{y^2}\frac{dy}{dx}=\frac{dz}{dx}$

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

Point (2, 1) c = 2 – 4 = 2 y = $\frac{x}{x^2-2}$

$\left|y\left(1\right)\right|=1$

As per questions

$\frac{dy}{dx}=\frac{x^2-4x+y+8}{x-2}$           

$\frac{dy}{dx}=\frac{{\left(x-2\right)}^2+\left(y+4\right)}{\left(x-2\right)}$           

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

Let $\frac{y+4}{x-2}=t$  

(y + 4) = t(x – 2)

Putting in equation (i)

$\left(x-2\right)\frac{dt}{dx}+t=\left(x-2\right)+t$        

    $\frac{dt}{dx}=1$        

dt = dx

Integrating on both the sides t = x + c

$\frac{y+4}{x-2}=x+c$  

Passing through origin C = -2

  $\therefore$         equation of curve $\frac{y+4}{x-2}=x-2$

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

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

(1) -> $v+\left(x+2\right)\frac{dv}{dx}=e^v+v.$  

$\Rightarrow \frac{dv}{e^v}=\frac{dx}{x+2}$ Integrating both side

Let $e^{e-2/3}=a\Rightarrow 0<\left|\frac{x+2}{3}\right|<a$  

$\therefore a=-3a-2\&\beta =3a-2$              

$\Rightarrow \left|\alpha +\beta \right|=4$