Mathematics - Differential Equation Question with Solution | TestHub

Let $y=f\left(x\right)$
$\frac{dy}{dx}+\frac{y}{x}=2$ (linear first order differential equation)

$\mathrm{I}.\mathrm{F}.=e^{\mathrm{\int }\frac{dx}{x}}=e^{{\log }_e\left(x\right)}=x$, (I.F.=Integrating factor)

$\therefore y·x=2\int xdx+c$
$\therefore yx=x^2+c$

$\Rightarrow f\left(x\right)=x+\frac{c}{x};$
As $f\left(1\right)\ne 1 \Rightarrow c\ne 0$
$\Rightarrow f^{\text{'}}\left(x\right)=1-\frac{c}{x^2}, c \ne 0$
$\underset{x\to 0^{+}}{\lim }{f}^{\text{'}}\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{\lim }\left(1-c{x}^2\right)=1$
$\underset{x\to 0^{+}}{\lim }xf\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{\lim }x\left(\frac{1}{x}+cx\right)=\underset{x\to 0^{+}}{\lim }\left(1+c{x}^2\right)=1$
$\underset{x\to 0^{+}}{\lim }{x}^2{f}^{\mathit{\text{'}}}\left(x\right)=\underset{x\to 0^{+}}{\lim }{x}^2\left(1-\frac{c}{x^2}\right)=\underset{x\to 0^{+}}{\lim }\left(x^2-c\right)=-c\ne 0$ as $c\ne 0$
$f\left(x\right)=x+\frac{c}{x}, c\ne 0$
$\therefore \underset{x\to 0^{+}}{\lim }f\left(x\right)=\mathrm{\infty }$ $\Rightarrow$function is not bounded in $\left(0, 2\right)$.