Question

Find the solution of the differential equation $\dfrac{dy}{dx}=\dfrac{xy}{{{x}^{2}}+{{y}^{2}}}$.

Answer 1

Hint: We first cross multiplies the given expression $\dfrac{dy}{dx}=\dfrac{xy}{{{x}^{2}}+{{y}^{2}}}$. Then we interchange the terms and divide with ${{y}^{3}}$ to find the differential form of \[\dfrac{x}{y}\]. We then integrate the expression of \[\dfrac{dy}{y}=\dfrac{x}{y}d\left( \dfrac{x}{y} \right)\]. We simplify the integration to find the solution.

Complete step by step answer:
We first cross multiplies the given expression $\dfrac{dy}{dx}=\dfrac{xy}{{{x}^{2}}+{{y}^{2}}}$.
We get ${{x}^{2}}dy+{{y}^{2}}dy=xydx$. Changing sides, we get \[{{y}^{2}}dy=xydx-{{x}^{2}}dy\].
We take $x$ common on the right side and get
\[{{y}^{2}}dy=xydx-{{x}^{2}}dy \\
\Rightarrow {{y}^{2}}dy=x\left( ydx-xdy \right) \\ \]
We now divide with ${{y}^{3}}$ both sides to get
\[\dfrac{{{y}^{2}}dy}{{{y}^{3}}}=\dfrac{x\left( ydx-xdy \right)}{{{y}^{3}}} \\
\Rightarrow \dfrac{dy}{y}=\dfrac{x}{y}\times \dfrac{ydx-xdy}{{{y}^{2}}} \\ \]
We now take the term \[\dfrac{x}{y}\] and find its differential form.
$\dfrac{d\left( \dfrac{x}{y} \right)}{dx}=\dfrac{y-x\dfrac{dy}{dx}}{{{y}^{2}}} \\
\Rightarrow d\left( \dfrac{x}{y} \right)=\dfrac{ydx-xdy}{{{y}^{2}}} \\ $
We can write
\[\dfrac{dy}{y}=\dfrac{x}{y}\times \dfrac{ydx-xdy}{{{y}^{2}}} \\
\Rightarrow \dfrac{dy}{y}=\dfrac{x}{y}\times d\left( \dfrac{x}{y} \right) \\ \]
Now we take the integration non both sides
\[\int{\dfrac{dy}{y}}=\int{\dfrac{x}{y}d\left( \dfrac{x}{y} \right)}+c\]
We know the integration form \[\int{\dfrac{dm}{m}}=\log \left| m \right|+c\] and \[\int{{{m}^{n}}dm}=\dfrac{{{m}^{n+1}}}{n+1}+c\].
\[\int{\dfrac{dy}{y}}=\int{\dfrac{x}{y}d\left( \dfrac{x}{y} \right)}+c \\
\Rightarrow \log \left| y \right|=\dfrac{{{x}^{2}}}{2{{y}^{2}}}+c \\
\Rightarrow 2\log \left| y \right|=\dfrac{{{x}^{2}}}{{{y}^{2}}}+k \\ \]
We took $k=2c$. Both k and c are constants.
We now apply the logarithmic formula of \[a\log b=\log {{b}^{a}}\].
\[2\log \left| y \right|=\dfrac{{{x}^{2}}}{{{y}^{2}}}+k \\
\Rightarrow \log {{y}^{2}}=\dfrac{{{x}^{2}}}{{{y}^{2}}}+k \\
\Rightarrow {{y}^{2}}={{e}^{\dfrac{{{x}^{2}}}{{{y}^{2}}}+k}}={{e}^{\dfrac{{{x}^{2}}}{{{y}^{2}}}}}{{e}^{k}} \\
\Rightarrow {{y}^{2}}=K{{e}^{\dfrac{{{x}^{2}}}{{{y}^{2}}}}} \\ \]
We again took $K={{e}^{k}}$. K is constant.
The simplified solution of the differential equation $\dfrac{dy}{dx}=\dfrac{xy}{{{x}^{2}}+{{y}^{2}}}$ is \[{{y}^{2}}=K{{e}^{\dfrac{{{x}^{2}}}{{{y}^{2}}}}}\].

Note:We can also take the equation in the form of $\dfrac{y}{x}$ as we take the variable $\dfrac{y}{x}=v$. Differentiating with respect to the term $x$, we get $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.
We replace the values in the differential form and get
$\dfrac{dy}{dx}=\dfrac{xy}{{{x}^{2}}+{{y}^{2}}} \\
\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}=\dfrac{\dfrac{y}{x}}{1+{{\left( \dfrac{y}{x} \right)}^{2}}}=\dfrac{v}{1+{{v}^{2}}} \\
\Rightarrow x\dfrac{dv}{dx}=\dfrac{v}{1+{{v}^{2}}}-v=\dfrac{-{{v}^{3}}}{1+{{v}^{2}}} \\
\Rightarrow \dfrac{1+{{v}^{2}}}{{{v}^{3}}}dv=\dfrac{dx}{x} \\
$
We integrate the equation to get the same solution of \[{{y}^{2}}=K{{e}^{\dfrac{{{x}^{2}}}{{{y}^{2}}}}}\].