Differential Equations

$cose{c}^{2}xdy+2dx=\left(1+ycos2x\right)cose{c}^{2}xdx.$

$\Rightarrow \frac{dy}{dx}+2sin2x=1+ycos2x.$

$I.F.=e^{-{\displaystyle \int cos2xdx}}=e^{-\frac{sin2x}{2}}$

$\therefore Solutiony{e}^{-\frac{sin2x}{2}}={\displaystyle \int e^{-\frac{sin2x}{2}}.cos2xdx.Put\frac{sin2x}{2}=t\Rightarrow cos2xdx=dt}$

$y\left(0\right)=-1+e^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\Rightarrow {\left(y\left(0\right)+1\right)}^2={\left(e^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2=e^{-1}$

${\displaystyle \int dy}={\displaystyle \int \frac{cos\left(\frac{1}{2}co{s}^{-1}\left(e^{-x}\right)\right)}{e^x\sqrt[]{1-{\left(e^{-x}\right)}^2}}}dxputcos2\theta =e^{-x}\Rightarrow -2sin2\theta d\theta =-e^{-x}dx$

$\therefore {\displaystyle \int dy}={\displaystyle \int \frac{2cos\theta sin2\theta d\theta }{\sqrt[]{1-co{s}^{2}2\theta }}\Rightarrow y=2sin\theta +c\Rightarrow y=-1,\theta =0,c=0}$

$\therefore y=2\sqrt[]{\frac{1-e^{-x}}{2}}-1at\left(\alpha ,0\right),e^{\alpha }=2$

Truth table

Hence according to option

(4) is most appropriate option

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

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

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

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

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

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

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

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

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

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