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Question: If \(xdy = ydx + {y^2}dy\) , \(y > 0\) and \(y\left( 1 \right) = 1\) , then what is \(y\left( { - 3}...

If xdy=ydx+y2dyxdy = ydx + {y^2}dy , y>0y > 0 and y(1)=1y\left( 1 \right) = 1 , then what is y(−3)y\left( { - 3} \right) equal to?
A.3 only
B.-1 only
C.Both -1 and 3
D.Neither -1 nor 3

Explanation

Solution

We can rearrange the differential equation and simplify it using quotient rule of differentiation. Then we can find the solution of the differential equation by integration. We can find the value of the constant of integration using the initial conditions. Then we can substitute the value of x and solve for y. We get the required answer by applying the condition y>0y > 0 .

Complete step-by-step answer:
We have the differential equation xdy=ydx+y2dyxdy = ydx + {y^2}dy
On rearranging, we get,
xdy−ydx=y2dyxdy - ydx = {y^2}dy
Now we can divide both sides of the equation with y2dx{y^2}dx .
⇒xdydx−yy2=dydx\Rightarrow \dfrac{{x\dfrac{{dy}}{{dx}} - y}}{{{y^2}}} = \dfrac{{dy}}{{dx}} .
Now we can multiply both the sides with -1.
⇒y−xdydxy2=−dydx\Rightarrow \dfrac{{y - x\dfrac{{dy}}{{dx}}}}{{{y^2}}} = - \dfrac{{dy}}{{dx}}
We know that by quotient rule of derivatives, d(xy)dx=y−xdydxy2\dfrac{{d\left( {\dfrac{x}{y}} \right)}}{{dx}} = \dfrac{{y - x\dfrac{{dy}}{{dx}}}}{{{y^2}}} . So, the differential equation will become
⇒d(xy)dx=−dydx\Rightarrow \dfrac{{d\left( {\dfrac{x}{y}} \right)}}{{dx}} = - \dfrac{{dy}}{{dx}}
We can cancel the term dxdx from both sides.
⇒d(xy)=−dy\Rightarrow d\left( {\dfrac{x}{y}} \right) = - dy
On integrating on both sides, we get,
⇒∫d(xy)=−∫dy+C\Rightarrow \int {d\left( {\dfrac{x}{y}} \right)} = - \int {dy} + C
⇒xy=−y+C\Rightarrow \dfrac{x}{y} = - y + C … (1)
Now we can apply the initial condition y(1)=1y\left( 1 \right) = 1 .
⇒11=−1+C\Rightarrow \dfrac{1}{1} = - 1 + C
On rearranging, we get,
⇒C=1+1\Rightarrow C = 1 + 1
⇒C=2\Rightarrow C = 2
On substituting the value of C in the equation, we get,
⇒xy=−y+2\Rightarrow \dfrac{x}{y} = - y + 2
We can multiply throughout with y.
⇒x=−y2+2y\Rightarrow x = - {y^2} + 2y
Now we need to find the value of y when x=−3x = - 3 . On substituting, we get,
⇒−3=−y2+2y\Rightarrow - 3 = - {y^2} + 2y
On rearranging, we get,
⇒y2−2y−3=0\Rightarrow {y^2} - 2y - 3 = 0
Now we have a quadratic equation. We can solve it for getting the value of y. For solving, we can split the middle term such that its product gives -3
⇒y2−3y+y−3=0\Rightarrow {y^2} - 3y + y - 3 = 0
Now we can take the common factors from the 1st 2 and last 2 terms.
⇒y(y−3)+(y−3)=0\Rightarrow y\left( {y - 3} \right) + \left( {y - 3} \right) = 0
On simplification, we get,
⇒(y−3)(y+1)=0\Rightarrow \left( {y - 3} \right)\left( {y + 1} \right) = 0
For the equation to be 0, either (y+1)\left( {y + 1} \right) or (y−3)\left( {y - 3} \right) must be equal to 0.
y+1=0⇒y=−1y + 1 = 0 \Rightarrow y = - 1
y−3=0⇒y=3y - 3 = 0 \Rightarrow y = 3
It is given in the question that y>0y > 0 . So, we can reject y=−1y = - 1 .
Therefore, the required solution is y=3y = 3
So, the correct answer is option A.

Note: An alternate solution to this problem is given by,
It is given that y>0y > 0 . So, -1 cannot be a solution. Therefore, we can reject options B and C.
We have the differential equation xdy=ydx+y2dyxdy = ydx + {y^2}dy
On rearranging, we get,
xdy−ydx=y2dyxdy - ydx = {y^2}dy
Now we can divide both sides of the equation with y2dx{y^2}dx.
⇒xdydx−yy2=dydx\Rightarrow \dfrac{{x\dfrac{{dy}}{{dx}} - y}}{{{y^2}}} = \dfrac{{dy}}{{dx}} .
Now we can multiply both the sides with -1.
⇒y−xdydxy2=−dydx\Rightarrow \dfrac{{y - x\dfrac{{dy}}{{dx}}}}{{{y^2}}} = - \dfrac{{dy}}{{dx}}
We know that by quotient rule of derivatives, d(xy)dx=y−xdydxy2\dfrac{{d\left( {\dfrac{x}{y}} \right)}}{{dx}} = \dfrac{{y - x\dfrac{{dy}}{{dx}}}}{{{y^2}}} .
⇒d(xy)dx=−dydx\Rightarrow \dfrac{{d\left( {\dfrac{x}{y}} \right)}}{{dx}} = - \dfrac{{dy}}{{dx}}
We can cancel the term dxdx from both sides.
⇒d(xy)=−dy\Rightarrow d\left( {\dfrac{x}{y}} \right) = - dy
On integrating on both sides, we get,
⇒∫d(xy)=−∫dy+C\Rightarrow \int {d\left( {\dfrac{x}{y}} \right)} = - \int {dy} + C
⇒xy=−y+C\Rightarrow \dfrac{x}{y} = - y + C … (1)
Now we can apply the initial condition y(1)=1y\left( 1 \right) = 1.
⇒11=−1+C\Rightarrow \dfrac{1}{1} = - 1 + C
On rearranging, we get,
⇒C=1+1\Rightarrow C = 1 + 1
⇒C=2\Rightarrow C = 2
On substituting the value of C in the equation, we get,
⇒xy=−y+2\Rightarrow \dfrac{x}{y} = - y + 2
⇒x=−y2+2y\Rightarrow x = - {y^2} + 2y
Now we can check whether y=3,x=−3y = 3,x = - 3 is a solution.
On substituting, we get,
⇒x=−(3)2+2(3)\Rightarrow x = - {\left( 3 \right)^2} + 2\left( 3 \right)
⇒x=−9+6=−3\Rightarrow x = - 9 + 6 = - 3
Therefore, y=3y = 3 is the only option which is correct.