Question

For the differential equation $\frac{dy}{dx}+\frac{1-2x}{x^2}y=1.$ The solution to the system is valid for which of the following interval(s)?

• A. $(0,1)$
• B. $(-1,1)$
• C. $(-\infty ,\infty )$
• D. $(-1,0)$

Answer 1

Answer: (A), (D)

The correct options are
A $(0,1)$
D $(-1,0)$

$\frac{dy}{dx}+\frac{1-2x}{x^2}y=1$ ...(1)
Here $P=\frac{1-2x}{x^2}\Rightarrow \int Pdx=\int \frac{1-2x}{x^2}dx$

$=\int (\frac{1}{x^2}-\frac{2}{x})dx=-\frac{1}{x}-2\log x$
$\therefore I.F.=e^{-1/x}.e^{-2\log x}=\frac{1}{x^2}{e}^{-1/x}$
Multiplying (1) by $I.F.$ we get
$\frac{1}{x^2}{e}^{-1/x}\frac{dy}{dx}+\frac{1-2x}{x^4}{e}^{-1/x}y=\frac{1}{x^2}{e}^{-1/x}$
Integrating both sides we get
$y.\left(\frac{1}{x^2}{e}^{-1/x}\right)=\int \frac{1}{x^2}{e}^{-1/x}dx+c=e^{-1/x}+c$

$\Rightarrow y=x^2(1+c{e}^{1/x})$

And this is not defined for $x=0$

Hence option (A) and (D) are true.