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

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
$\begin{aligned} & \dfrac{dy}{dx}=1+{{\left( \dfrac{y}{x} \right)}^{2}}-\dfrac{y}{x} \\\ & \Rightarrow v+x\dfrac{dv}{dx}=1+{{v}^{2}}-v \\\ \end{aligned}$
We now simplify the equation to take respective variables in one side

& 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{aligned}$$ Now we take the integration on both sides of the equation and take $k$ as the integral constant. So, $$\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{aligned} & 2\log \left| \left( x-vx \right) \right|=0 \\\ & \Rightarrow \log \left\\{ {{\left( x-vx \right)}^{2}} \right\\}=0 \\\ \end{aligned}$$ We know that $y=vx$. $$\begin{aligned} & \log \left\\{ {{\left( x-vx \right)}^{2}} \right\\}=0 \\\ & \Rightarrow {{\left( x-y \right)}^{2}}=1 \\\ \end{aligned}$$ **The solution of the differential equation $\dfrac{dy}{dx}=\dfrac{{{x}^{2}}+{{y}^{2}}-xy}{{{x}^{2}}}$ is $${{\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.