Differential Equations

^{$y.\frac{dy}{dx}=x\left[\frac{y^2}{x^2}+\frac{?\left(y^2/x^2\right)}{?\text{'}\left(y^2/x^2\right)}\right],x>0,?>0$}

$Let\frac{y}{x}=t$

$\frac{dy}{dx}=t+x.\frac{dt}{dx}$

$ln\left(?\left(\frac{y^2}{4}\right)\right)=ln4+ln\left(?\left(1\right)\right)$

$=ln4\left(?\left(1\right)\right)$

$?\left(\frac{y^2}{4}\right)=4.?\left(1\right)$

$\underset{x\to 0}{lim}{\left(2-cosx.\sqrt[]{cos2x}\right)}^{\frac{x+2}{x^2}}$ $\left(1^{\infty }\right)$

= $\underset{x\to 0}{lim}{e}^{2\left(\frac{sinx\sqrt[]{cos2x}-cosx.\frac{1\left(-2sin2x\right)}{2\sqrt[]{cos2x}}}{2x}\right)}$

$=\underset{x\to 0}{lim}{e}^{2\left(\frac{1}{2}+1\right)}=e^3=e^a$

a = 3

$\frac{dy}{dx}=\frac{2^{x+y}-2^x}{2^y}\Rightarrow {\displaystyle \int \frac{2^{y}dy}{2^y-1}={\displaystyle \int 2^{x}dx\Rightarrow put{2}^y-1=t\Rightarrow 2^{y}ln2dy=dt}}$

$\frac{1}{ln2}{\displaystyle \int \frac{dt}{t}={\displaystyle \int 2^{x}dx\Rightarrow \frac{1}{ln2}lnt=\frac{2^x}{ln2}+\frac{C}{ln2}}}$    put x = 0 and y = 1 we get C = -1

$\therefore ln\left(2^y-1\right)=2^x-1$    put x = 1 and we get y = log₂ (1 + e)

Let $l={\displaystyle \int \frac{2e^x+3e^{-x}}{4e^x+7e^{-x}}}dx={\displaystyle \int \frac{2e^{2x}+3}{4e^{2x}+7}dx}$  

$={\displaystyle \int \frac{2e^{2x}}{4e^{2x}+7}dx+{\displaystyle \int \frac{3e^{-2x}}{4+7e^{-2x}}dx}}$               

Put $4e^{2x}+7=t4+7e^{-2x}=\lambda$  

$8e^{2x}dx=dt-14e^{-2x}dx=d\lambda$              

$=\frac{1}{4}{\displaystyle \int \frac{dt}{t}-\frac{3}{14}{\displaystyle \int \frac{d\lambda }{\lambda }=\frac{1}{4}lnt-\frac{3}{14}ln\lambda +c}}$

$u=\frac{13}{2},v=\frac{1}{2}$              

->so, u + v = 7

$\frac{dy}{dx}=2\left(y+2sinx-5\right)x-2cosx$

where P = -2x, Q = 4x sin x – 10x – 2 cosx

Solution be $y.e^{-x^2}={\displaystyle \int \left(4xsinx-10x-2cosx\right).e^{-x^2}dx+c}$

$y.e^{-x^2}{\displaystyle \int \left(4xsinx-10x-2cosx\right).e^{-x^2}dx+c}$     

$y.e^{-x^2}=e^{-x^2}\left(5-2sinx\right)+c........\left(i\right)$  

Put x = 0, y = 7 then 7 = 5 + c i.e. c = 2

Put x = p then $y.e^{-{\pi }^2}=5.e^{-{\pi }^2}+2$

y = 5 + $2e^{x^2}$

$\left(2x-10y^3\right)dy+ydx=0$              

$\frac{dx}{dy}+\frac{2x}{y}-10y^2=0$              

$\frac{dx}{dy}+\frac{2}{y}x=10y^2\to$ Linear differential equation

$P=\frac{2}{y},Q=10y^2$              

$Nowputx=2,y=\beta then2{\beta }^2=2{\beta }^5-2$              

or ${\beta }^5-{\beta }^2-1=0$  

So B will be roots of $y^5-y^2-1=0$  

Given curve is f(x) + xf’(x) = x² i.e. y + x $\frac{dy}{dx}=x^2$

where P = $\frac{1}{x},Q=x$

$I.F.=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{1}{x}dx}}=e^{lnx}=x$

Solution be y.x = ${\displaystyle \int x.xdx}$

$xy=\frac{x^3}{3}+c.........\left(i\right)$

(i) passes through (-2, 2) then $-4=\frac{-8}{3}+c$

$c=\frac{-4}{3}$        

(i) -> 3xy = x³ – 4

or x³ – 3xf(x) – 4 = 0

Let $l={\displaystyle \int \frac{dx}{{\left(x^2+x+1\right)}^2}.............\left(i\right)}$

Now ${\displaystyle \int \frac{dx}{x^2+x+1}={\displaystyle \int \left(\frac{1}{x^2+x+1}\right)dx}}$

by integration by parts

=> ${\displaystyle \int \frac{dx}{{\left(x^2+x+1\right)}^2}=\frac{4}{3\sqrt[]{2}}ta{n}^{-1}\frac{2x+1}{\sqrt[]{3}}+\frac{1}{3}\frac{\left(2x+1\right)}{\left(x^2+x+1\right)}+c}$

$\Rightarrow a=\frac{4}{3\sqrt[]{3}},b=\frac{1}{3}$

Now, $9\left(\sqrt[]{3}a+b\right)=9\left(\frac{4}{3}+\frac{1}{3}\right)=15$

$L.R.=\frac{\left|3\times 2+4x-3-5\right|}{\sqrt[]{3^2+4^2}}=\frac{11}{5}$              

Equation of family of parabolas

${\left(x-h\right)}^2=\frac{11}{5}\left(y-k\right)$              

Differentiate 2 (x – h) = $\frac{11}{5}\frac{dy}{dx}$  

Again differentiate 2 = $\frac{11}{5}\frac{d^{2}y}{d{x}^2}$  

$11\frac{d^{2}y}{d{x}^2}=10$              

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

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

Put $e^{-y}=t\Rightarrow -e^{-y}\frac{dy}{dx}=\frac{dt}{dx}$  

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$              

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$

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

Soln. tx = ${\displaystyle \int \frac{x}{2x^2}dx=\frac{1}{2}lnx+c}$  

$x=1\Rightarrow e^{-y}=\frac{1}{2}\Rightarrow y=ln2$