Question

If the interval contained in the domain of definition of non-zero solution of the differential equation $(x-3{)}^2\cdot y^{{}^{\prime }}+y=0$ is $(-\infty ,\infty )-\left\{k\right\},$ then $k$ is

Answer 1

$(x-3{)}^2\cdot y^{{}^{\prime }}+y=0$
$\Rightarrow (x-3{)}^2\frac{dy}{dx}=-y$
$\Rightarrow \underset{}{\overset{}{\int }}\left(\frac{1}{y}\right)dy=\underset{}{\overset{}{\int }}\frac{-1}{(x-3{)}^2}dx$
$\Rightarrow \ln |y|=\frac{-(x-3{)}^{-2+1}}{-2+1}+\ln |c|,x\ne 3\Rightarrow \ln \left(\frac{|y|}{|c|}\right)=\frac{1}{x-3},x\ne 3\Rightarrow \frac{|y|}{|c|}=e^{\frac{1}{x-3}},x\ne 3\Rightarrow y=\pm c\cdot e^{\frac{1}{x-3}},x\ne 3$

$\therefore$ The domain for above solution is $(-\infty ,\infty )-\text{{}3\text{}}$
$\Rightarrow k=3$