Question

The solution of \[a(xdy+ydx)=xydy\] is:

Answer 1

Hint: In finding the solution of \[a(xdy+ydx)=xydy\] we have to write some basic derivative formulas.
We have to know \[\dfrac{d}{dy}\left( xy \right)=y\dfrac{dx}{dy}+x\dfrac{dy}{dy}\] and \[\int{\dfrac{1}{x}dx}=\log \left( \left| x \right| \right)+c\] , where \[x,y\] are constants and
 \[\dfrac{dy}{dy}=1\] .

Complete step-by-step answer:
From the question it is clear that we have to find the solution of \[a\left( xdy+ydx \right)=xydy\] .
Now, consider the given as equation (1).
 \[\Rightarrow a\left( xdy+ydx \right)=xydy\] ………………(1)
Now, on LHS open up the bracket by multiplying \[a\] with \[xdy\] and \[a\] with \[ydx\] . we get
 \[\Rightarrow a\times xdy+a\times ydx=xydy\]
consider this equation as equation (2)
 \[\Rightarrow a\times xdy+a\times ydx=xydy\] ……………….(2)
Now divide equation (2) with \[dy\] , we get
 \[\begin{align}
  & \Rightarrow \dfrac{a\times xdy}{dy}+\dfrac{a\times ydx}{dy}=\dfrac{xydy}{dy} \\
 & \\
\end{align}\]
After simplification, we get
 \[\Rightarrow ax\dfrac{dy}{dy}+ay\dfrac{dx}{dy}=xy\]
Now, take \[a\] common from LHS. We get
 \[\Rightarrow a\times \left( x\dfrac{dy}{dy}+y\dfrac{dx}{dy} \right)=xy\] ………………(3)
From the basic derivative formulas, we know that \[\dfrac{d}{dy}\left( xy \right)=y\dfrac{dx}{dy}+x\dfrac{dy}{dy}\] , where \[x,y\] are constants and \[\dfrac{dy}{dy}=1\] .
So, put \[y\dfrac{dx}{dy}+x\dfrac{dy}{dy}=\dfrac{d}{dy}\left( xy \right)\] in equation (3).
so, equation (2) becomes as
 \[\Rightarrow a\times \left( \dfrac{d}{dy}\left( xy \right) \right)=xy\]
 \[\Rightarrow a\times \left( \dfrac{d\left( xy \right)}{dy} \right)=xy\]
Now multiply with \[dy\] on both sides,
 \[a\times dy\left( \dfrac{d\left( xy \right)}{dy} \right)=xydy\]
After simplification we get,
  \[\Rightarrow a\times d\left( xy \right)=xydy\]
Now divide the whole equation as
 \[\Rightarrow a\times \dfrac{d\left( xy \right)}{xy}=\dfrac{xydy}{xy}\]
After simplification we get,
 \[\Rightarrow a\times \dfrac{d\left( xy \right)}{xy}=dy\] ………………..(4)
Now, apply integration on both sides
So, the equation (4) becomes as
 \[\Rightarrow \int{a\dfrac{d\left( xy \right)}{xy}}=\int{dy}\] …………..(5)
 Now we know that \[\int{kdx}=k\times \int{dx}\] , where \[k\] is constant.
So now we can write \[\int{a\dfrac{d\left( xy \right)}{xy}}=a\times \int{\dfrac{d\left( xy \right)}{xy}}\]
Equation (5) becomes as
 \[\Rightarrow a\times \int{\dfrac{d\left( xy \right)}{xy}}=\int{dy}\] …………….(6)
From the basic integration formulas, we know that \[\int{\dfrac{1}{x}dx}=\log \left( \left| x \right| \right)+c\] .
So, we can write \[\int{\dfrac{d\left( xy \right)}{xy}}=\log \left( \left| xy \right| \right)+c\]
So, equation (6) as
 \[\Rightarrow a\log \left( \left| xy \right| \right)+c=y+c\]
 \[\Rightarrow y=a\log \left( \left| xy \right| \right)+c\] .
So now, we can conclude that the solution of \[a\left( xdy+ydx \right)=xydy\] is \[y=a\log \left( \left| xy \right| \right)+c\] .

Note: Students should be careful while applying derivative formulas and integration formulas.
Wrong formulas may lead to this question being wrong. Students should do calculations right because small calculation errors can make finding the solution of \[a\left( xdy+ydx \right)=xydy\] wrong.