Question
If $$xdy = ydx + {y^2}dy,\,y > 0$$ and $$y\left( 1 \right) = 1,$$ then what is $$y\left( { - 3} \right)$$ equal to?
A.
$$3$$ only
B.
$$ - 1$$ only
C.
Both $$ - 1$$ and $$3$$
D.
Neither $$ - 1$$ nor $$3$$
Answer :
$$3$$ only
Solution :
$$\eqalign{
& {\text{Given, }}xdy = ydx + {y^2}dy \cr
& \Rightarrow 1 = \frac{4}{x}\frac{{dx}}{{dy}} + \frac{{{y^2}}}{x} \cr
& \Rightarrow \frac{{dx}}{{dy}} + y = \frac{x}{y} \cr
& \Rightarrow \frac{{dx}}{{dy}} - \frac{x}{y} = - y......\left( 1 \right) \cr
& P = - \frac{1}{y},\,Q = - y \cr
& {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {Pdy} }} = {e^{\int { - \frac{1}{y}dy} }} = {e^{ - \log \,y}} = \frac{1}{y} \cr
& {\text{Multiplying Equation }}\left( 1 \right){\text{ by I}}{\text{.F}}{\text{.}} \cr
& \Rightarrow \frac{1}{y}\frac{{dx}}{{dy}} - \frac{x}{{{y^2}}} = - 1\,;\,\frac{x}{y} = \int {\frac{1}{y}\left( { - y} \right)dy + C} \cr
& \Rightarrow \frac{x}{y} = \int { - 1\,dy + C} \cr
& \Rightarrow \frac{x}{y} = - y + C \cr
& \left( {y\left( 1 \right) = 1\, \Rightarrow \frac{1}{1} = - 1 + C\, \Rightarrow C = 2} \right) \cr
& \Rightarrow \frac{x}{y} = - y + 2 \cr
& \Rightarrow x = - {y^2} + 2y \cr
& \Rightarrow y\left( { - 3} \right) \Rightarrow - 3 = - {y^2} + 2y \cr
& \Rightarrow {y^2} - 2y - 3 = 0 \cr
& \Rightarrow y = \frac{{ + 2 \pm \sqrt {4 + 12} }}{2} \cr
& \Rightarrow y = \frac{{2 \pm 4}}{2} \cr
& \Rightarrow y = 3,\,\, - 1 \cr
& {\text{Since }}y > 0{\text{ so }}y = 3. \cr} $$