Question

Solution of the differential equation $x^2\frac{dy}{dx}+y=1$ is:
(where $c$ is integration constant)

• A. $y=1+c{e}^{\frac{1}{x}}$
• B. $y=1+c{e}^{-\frac{1}{x}}$
• C. $x=1+c{e}^{\frac{1}{y}}$
• D. $x=1+c{e}^{-\frac{1}{y}}$

Answer 1

The correct option is A $y=1+c{e}^{\frac{1}{x}}$

The given differential equation can be written as
$\frac{dy}{dx}+\frac{1}{x^2}y=\frac{1}{x^2},$ which is linear
On comparing with $\frac{dy}{dx}+Py=Q,$
We get $P=\frac{1}{x^2}$ and $Q=\frac{1}{x^2}$
$I.F.=e^{\int \left(\frac{1}{x^2}\right)dx}=e^{-\frac{1}{x}}$

Therefore the solution is
$y{e}^{-\frac{1}{x}}=\int e^{-\frac{1}{x}}\left(\frac{1}{x^2}\right)dx+c$
$=e^{-\frac{1}{x}}+c$
$\Rightarrow y=1+c{e}^{\frac{1}{x}}$