Differential Equations

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

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

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    

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

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

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

 $\frac{dy}{dx}-y=2-e^{-x}$

$I.F.=e^{-{\displaystyle \int dx}}=e^{-x}$

$\therefore$ soln

$y{e}^{-x}={\displaystyle \int \left(2e^{-x}-e^{-2x}\right)dx}$

$y=-2+\frac{e^{-x}}{2}+C{e}^x$

as for x $\infty$ y finite c = 0

$\therefore y=\frac{e^{-x}}{2}-2$

$\begin{array}{l}\Rightarrow x+2y=-3\\ \Rightarrow a=-3b=\frac{-3}{2}\end{array}$

$\therefore a=4b=-3+6=3$

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

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

 $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt[]{1-x^2}},$ $I.F.e^{{\displaystyle \int \frac{xdx}{x^2-1}}}=\sqrt[]{\left|x^2-1\right|}=\sqrt[]{1-x^2}$ $\left(?x\in \left(-1,1\right)\right)$

Solution of differential equation is $y\sqrt[]{1-x^2}={\displaystyle \int \left(x^4+2x\right)dx=\frac{x^5}{5}+x^2+c}$

Curve is passing through origin, c = 0 $y=\frac{x^5+5x^2}{5\sqrt[]{1-x^2}}$

$\therefore {\displaystyle \underset{-\frac{\sqrt[]{3}}{2}}{\overset{\frac{\sqrt[]{3}}{2}}{\int }}\frac{x^5+5x^2}{5\sqrt[]{1-x^2}}dx=\frac{\pi }{3}-\frac{\sqrt[]{3}}{4}}$