Question
Solution of the differential equation $$ydx + \left( {x + {x^2}y} \right)dy = 0$$ is-
A.
$$\log \,y = Cx$$
B.
$$ - \frac{1}{{xy}} + \log \,y = C$$
C.
$$\frac{1}{{xy}} + \log \,y = C$$
D.
$$ - \frac{1}{{xy}} = C$$
Answer :
$$ - \frac{1}{{xy}} + \log \,y = C$$
Solution :
$$\eqalign{
& ydx + \left( {x + {x^2}y} \right)dy = 0 \cr
& \Rightarrow \frac{{dx}}{{dy}} = - \frac{x}{y} - {x^2}\,\,\,\, \Rightarrow \frac{{dx}}{{dy}} + \frac{x}{y} = - {x^2} \cr} $$
It is Bernoullis form. Divide by $${x^2}$$
$$\eqalign{
& {x^{ - 2}}\frac{{dx}}{{dy}} + {x^{ - \,1}}\left( {\frac{1}{y}} \right) = - 1 \cr
& {\text{Put }}{x^{ - \,1}} = t,\,\, - {x^{ - \,2}}\frac{{dx}}{{dy}} = \frac{{dt}}{{dy}} \cr
& {\text{We get,}} \cr
& - \frac{{dt}}{{dy}} + t\left( {\frac{1}{y}} \right) = - 1\,\,\,\,\,\, \Rightarrow \frac{{dt}}{{dy}} - \left( {\frac{1}{y}} \right)t = 1 \cr} $$
It is linear in $$t.$$
Integrating factor $$ = {e^{\int { - \,\frac{1}{y}dy} }} = {e^{ - \,\log \,y}} = {y^{ - 1}}$$
$$\therefore $$ Solution is $$t\left( {{y^{ - \,1}}} \right) = \int {\left( {{y^{ - \,1}}} \right)} dy + c$$
$$ \Rightarrow \frac{1}{x}.\frac{1}{y} = \log \,y + c \Rightarrow \log \,y - \frac{1}{{xy}} = c$$