Question

How do you find the solution of $\dfrac{dy}{dx}=\dfrac{{{x}^{2}}+{{y}^{2}}-xy}{{{x}^{2}}}$ with $y\left( 1 \right)=0$?

Answer 1

Hint: We first try to form the equation in the form of $\dfrac{y}{x}$ as we take the variable $\dfrac{y}{x}=v$. We change the differential form from $\dfrac{dy}{dx}$ to $\dfrac{dv}{dx}$. We replace the variables and take integral to find the solution of the differential equation.

Complete step-by-step solution:
 We first simplify the right hand -side equation by dividing with ${{x}^{2}}$.
Therefore, $\dfrac{dy}{dx}=\dfrac{{{x}^{2}}+{{y}^{2}}-xy}{{{x}^{2}}}=1+{{\left( \dfrac{y}{x} \right)}^{2}}-\dfrac{y}{x}$.
Now we assume that $y=vx$ which gives $\dfrac{y}{x}=v$.
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
$\begin{align}
  & \dfrac{dy}{dx}=1+{{\left( \dfrac{y}{x} \right)}^{2}}-\dfrac{y}{x} \\
 & \Rightarrow v+x\dfrac{dv}{dx}=1+{{v}^{2}}-v \\
\end{align}$
We now simplify the equation to take respective variables in one side
\[\begin{align}
  & v+x\dfrac{dv}{dx}=1+{{v}^{2}}-v \\
 & \Rightarrow x\dfrac{dv}{dx}=1+{{v}^{2}}-2v={{\left( 1-v \right)}^{2}} \\
 & \Rightarrow \dfrac{dv}{{{\left( 1-v \right)}^{2}}}=\dfrac{dx}{x} \\
\end{align}\]
Now we take the integration on both sides of the equation and take $k$ as the integral constant.
So, \[\int{\dfrac{dv}{{{\left( 1-v \right)}^{2}}}}=\int{\dfrac{dx}{x}}+k\].
We know that $d\left( 1-v \right)=-dv$. We now change the differential from.
we get \[-\int{\dfrac{d\left( 1-v \right)}{{{\left( 1-v \right)}^{2}}}}=\int{\dfrac{dx}{x}}+c\].
We know that \[\int{\dfrac{dy}{y}}=\log \left| y \right|\]. We use that to get \[-\log \left| 1-v \right|=\log \left| x \right|+k\].
At $y\left( 1 \right)=0$ and $\dfrac{y}{x}=v$, we get $v=\dfrac{y}{x}=0$.
So, putting the values we get \[-\log 1=\log 1+k\Rightarrow k=0\]. We get \[-\log \left| 1-v \right|=\log \left| x \right|\]
We take all the logarithms in one side and get \[\log \left| x \right|+\log \left| 1-v \right|=0\].
We know the formula of logarithm as \[\log m+\log n=\log \left( mn \right)\].
Therefore, \[\log \left| x \right|+\log \left| 1-v \right|=\log \left| x\left( 1-v \right) \right|\].
We get \[\log \left| x\left( 1-v \right) \right|=0\]. To remove the modulus, we multiply 2 and get \[2\log \left| x\left( 1-v \right) \right|=0\].
We also know that \[{{\log }_{n}}m=l\Rightarrow {{n}^{l}}=m\] and \[a\log m=\log {{m}^{a}}\].
\[\begin{align}
  & 2\log \left| \left( x-vx \right) \right|=0 \\
 & \Rightarrow \log \left\{ {{\left( x-vx \right)}^{2}} \right\}=0 \\
\end{align}\]
We know that $y=vx$.
\[\begin{align}
  & \log \left\{ {{\left( x-vx \right)}^{2}} \right\}=0 \\
 & \Rightarrow {{\left( x-y \right)}^{2}}=1 \\
\end{align}\]
The solution of the differential equation $\dfrac{dy}{dx}=\dfrac{{{x}^{2}}+{{y}^{2}}-xy}{{{x}^{2}}}$ is \[{{\left( x-y \right)}^{2}}=1\].

Note: We need to be careful about the change of variable where we have to just withdraw the variable $y$. The constant will be calculated before we multiply with 2. The solution we got is a general solution which we get in the form of x,y And we can also cross check the solution by putting the given value y(1)=0 in solution and this will satisfy the solution expression.