Differential Equations

$xdy=\left(\sqrt[]{x^2+y^2}+y\right)dx$

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

$\frac{d\left(\frac{y}{x}\right)}{\sqrt[]{1+{\left(\frac{y}{x}\right)}^2}}=\frac{dx}{x}\Rightarrow \mathcal{l}n\left(\frac{y}{x}+\sqrt[]{{\left(\frac{y}{x}\right)}^2+1}\right)=\mathcal{l}nx+C$

$\alpha =\frac{3}{2}$

$f\left(x\right)=\frac{2}{\sqrt[]{3}}{\displaystyle {\int }_0^{\sqrt[]{3}}f\left(\frac{{\lambda }^{2}x}{3}\right)d\lambda }$

Substitute $\frac{{\lambda }^{2}x}{3}=t$ $\frac{{}^{}}{}$

$f\left(x\right)=\frac{1}{\sqrt[]{x}}{\displaystyle {\int }_0^x\frac{f\left(t\right)}{\sqrt[]{t}}dt}$

differentiate using leibneitz rule

$\sqrt[]{x}.f\text{'}\left(x\right)+f\left(x\right).\frac{1}{2\sqrt[]{x}}=\frac{f\left(x\right)}{\sqrt[]{x}}$

f (1) = $\sqrt[]{3}\Rightarrow f\left(x\right)=\sqrt[]{3x}$

$sin\left(2x^2\right).109_e\left(tan{x}^2\right)dy+4xydx=4\sqrt[]{2}.x.\left(sin{x}^{2}cos\frac{\pi }{4}-cos{x}^{2}sin\frac{\pi }{4}\right)dx$

$\Rightarrow \mathcal{l}n\left(tan{x}^2\right)dy+\frac{4x}{sin\left(2x^2\right)}ydx=\frac{4x\left(sin{x}^2-cos{x}^2\right)}{sin\left(2x^2\right)}dx$

Integrate

$\Rightarrow y.\mathcal{l}n\left(tan{x}^2\right)=2.\mathcal{l}n\left(\frac{sin{x}^2+cos{x}^2-1}{sin{x}^2+cos{x}^2+1}\right)+C$

$x=\sqrt[]{\frac{\pi }{6}},y=1$ calculate C.

$y^{2}dx+\left(x^2?xy+y^2\right)dy=0$

$\frac{dx}{dy}+\frac{x^2?xy+y^2}{y^2}=0?\frac{dx}{dy}+{\left(\frac{x}{y}\right)}^2?\left(\frac{x}{y}\right)+1=0$

Put x = vy $?v+y\frac{dv}{dy}+v^2?v+1=0$ $\frac{}{}{}^{}$

$?\frac{ydv}{dy}+v^2+1=0$

$?\frac{?}{6}+\mathcal{l}n\left|y\right|=\frac{?}{4}$

$\mathcal{l}n\left|y\right|=\frac{?}{12}$

$\frac{x\left(cosx-sinx\right)}{e^x+1}+\frac{g\left(x\right)\left(e^x+1-x{e}^x\right)}{{\left(e^x+1\right)}^2}$

$=\sqrt[]{2}sin\left(x+\frac{\pi }{4}\right)+c,x>0$

$g\text{'}\left(x\right)=\sqrt[]{2}cos\left(x+\frac{\pi }{4}\right),x>0$

g + g' = 2cos x + c, x > 0

g – g' = 2 sin x + c, x > 0

  $f\left(x\right)+f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)=x^5+64$

$\underset{x\to 1}{lim}\frac{f\left(x\right)}{x-1}\left(\frac{f\left(1\right)}{0}\right)f\left(1\right)=0$

Let f(x) = x⁵ + ax⁴ + bx³ + cx² + dx + e

f(1) = 0 Þ 1 + a + b + c + d + e = 0

  $\Rightarrow x^5+\left(a+5\right)x^4+\left(b+4a+20\right)x^3+\left(c+3b+12a\right)x^2+\left(d+2c+6b\right)x+e+d+2c=x^5+64$

Þ e + d + 2c = 64

b + 4a + 20 = 0                 c – 1 – a – b = 64

c + 3b + 12a = 0

d + 2c + 6b = 0                 a + b – c = -65

1 + a + b + c + d + e = 0   13a + 4b = -65

b + 4c = -20 * 4

-3a = 15 a = -5

$\begin{array}{l}f\left(x\right)=x^5-5x^4+60x^2-120x+64\\ f\text{'}\left(x\right)=5x^4-20x^3+12x-120\\ f\text{'}\left(1\right)=5-20+120-120=-15\end{array}$

  $y\frac{dx}{dy}=2x+y^3\left(y+1\right)e^y$               

    $\frac{dx}{dy}-\frac{2}{y}x=y^3\left(1+y\right)e^y..........(i)$

Equation (i) is in linear form, so I.F. = y^-2

$\begin{array}{l}\therefore x{y}^{-2}=y{e}^y+c\Rightarrow 0=e+c\Rightarrow c=-e\\ \therefore x=e^3\left(e^e-1\right)\end{array}$                

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

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

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

Put x = 1, y = 1 $\Rightarrow ta{n}^{-1}=c\Rightarrow c=\frac{\pi }{4}$

$\therefore ta{n}^{-1}\left(xy\right)=\mathcal{l}nx=\frac{\pi }{4}$

$e\left(y\left(e\right)\right)=tan\left(1+\frac{\pi }{4}\right)=\frac{tan1+1}{1-tan1}$

2cos⁴ x – cos²x = $2{\left(\frac{1+cos2x}{2}\right)}^2-cos2x$

$=\frac{1+co{s}^{2}2x}{2}$

$\sqrt[]{2}{\displaystyle \int \frac{2dx}{1+co{s}^{2}2x}=\sqrt[]{2}{\displaystyle \int 2}\cdot \frac{se{c}^{2}2xdx}{2+ta{n}^{2}2x}}$

$=\sqrt[]{2}\cdot \frac{1}{\sqrt[]{2}}\cdot ta{n}^{-1}\left(\frac{tanx}{\sqrt[]{2}}\right)$

now IF = $e^{{\displaystyle \int Pdx}}$

$e^{\sqrt[]{2}{\displaystyle \int 2\frac{se{c}^{2}2xdx}{2+ta{n}^{2}2x}}}=e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}$

Solution: $y\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}={\displaystyle \int x.e^{ta{n}^{-1}\left(\sqrt[]{2}\cdot cot2x\right)}\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}dx}$

$y\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}=\frac{x^2}{2}\cdot e^{\pi /2}+C$

at $x=\frac{\pi }{4},y=\frac{{\pi }^2}{32},C=0$

at $x=\frac{\pi }{3},y=\frac{{\pi }^2}{18}{e}^{ta{n}^{-1}\alpha }$

$y\cdot e^{ta{n}^{-1}}\frac{\left(tan2\frac{\pi }{3}\right)}{\sqrt[]{2}}=\frac{{\pi }^2}{18}\cdot e^{\pi /2}$

$\frac{{\pi }^2}{18}\cdot e^{ta{n}^{-1}\alpha }\cdot e^{ta{n}^{-1}}\left(-\frac{\sqrt[]{3}}{2}\right)=\frac{{\pi }^2}{18}\cdot e^{\pi /2}$

tan^-1 a + tan^-1 $\left(-\sqrt[]{3}/2\right)=\frac{\pi }{2}$

cot^-1a = tan^-1 $\left(-\frac{\sqrt[]{3}}{2}\right)$

tan^-1 $\frac{1}{\alpha }=ta{n}^{-1}\left(-\frac{\sqrt[]{3}}{2}\right)$

$\frac{1}{\alpha }=-\sqrt[]{3}/2$

2 = $\frac{2}{3}$

 $\frac{dy}{dx}=\frac{y}{x}+\frac{\sqrt[]{y^2+16x^2}}{x}$

Put y = $V\cdot x$

differentiable worst = x

$\frac{dy}{dx}=V+x\cdot \frac{dV}{dx}$

V + $x\frac{dV}{dx}=V+\sqrt[]{V^2+16}$

$x\frac{dV}{dx}=\sqrt[]{V^2+16}$

apply variable separable method

${\displaystyle \int \frac{dV}{\sqrt[]{V^2+16}}={\displaystyle \int \frac{dx}{x}+\mathcal{l}nC}}$

$\Rightarrow \mathcal{l}n\left|V+\sqrt[]{V^2+16}\right|=\mathcal{l}nCx$

$\frac{y}{x}+\frac{\sqrt[]{y^2+16x^2}}{x}=Cx$

Given y(1) = 3 C = 8

Now at x = 2

$\frac{y}{2}+\frac{\sqrt[]{y^2+16.4}}{2}=8.2$

y2 + 64 = (32 – y)2

y = 15