Question

The solution of the differential equation $\left(\frac{dy}{dx}\right)=\frac{\left[x-2y+1\right]}{\left[2x-4y\right]}$ is

• A. ${\left(x-2y\right)}^2+2x=c$
• B. ${\left(x-2y\right)}^2+x=c$
• C. ${\left(x-2y\right)}^2+2x^2=c$
• D. ${\left(x-2y\right)}^2+x^2=c$

Answer 1

The correct option is A

${\left(x-2y\right)}^2+2x=c$

Explanation for correct option

Given: $\left(\frac{dy}{dx}\right)=\frac{\left[x-2y+1\right]}{2\left[x-y\right]}$

Let $x-2y=v$

Differentiating with respect to $x$ we get

$$\begin{gathered} \Rightarrow 1-2\frac{dy}{dx}=\frac{dv}{dx} \\ \Rightarrow \frac{1-\frac{dv}{dx}}{2}=\frac{dy}{dx} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{1-{\displaystyle \frac{dv}{dx}}}{2}=\frac{\left[v+1\right]}{2\left[v\right]} \\ \Rightarrow 1-\frac{dv}{dx}=\frac{v+1}{v} \\ \Rightarrow 1-\frac{v+1}{v}=\frac{dv}{dx} \\ \Rightarrow \frac{v-v-1}{v}=\frac{dv}{dx} \\ \Rightarrow -\frac{1}{v}=\frac{dv}{dx} \\ \Rightarrow -dx=vdv \end{gathered}$$

integrating both sides

$$\begin{gathered} \Rightarrow -\int dx=\int vdv \\ \Rightarrow -x=\frac{v^2}{2} \end{gathered}$$

putting $v=x-2y$

$$\begin{gathered} \Rightarrow -2x={\left(x-2y\right)}^2+C \\ \Rightarrow {\left(x-2y\right)}^2+2x=C \end{gathered}$$

Hence, option A is correct.