The solution of the differential equation $(2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0$ is ( $C$ is an arbitrary…
The solution of the differential equation $(2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0$ is
( $C$ is an arbitrary constant)
- $\log [(2 x-4 y)+3]=x-2 y+C$
- $\log [2(2 x-4 y)+3]=2(x-2 y)+C$
- $\log [2(x-2 y)+5]=2(x+y)+C$
- $\log [4(x-2 y)+5]=4(x+2 y)+C$
Solution
Given differential equation,
$\begin{aligned}
& (2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0 \\
& \Rightarrow \quad \frac{d y}{d x}=\frac{(x-2 y+1)}{2(x-2 y+3)} \quad \ldots \text { (i) }
\end{aligned}$
Let, $x-2 y=v \Rightarrow 1-2 \frac{d y}{d x}=\frac{d y}{d x}$
$\therefore \frac{1}{2}\left(1-\frac{d y}{d x}\right)=\frac{d y}{d x}$
Now, substitute $v=x-2 y$ and $\frac{d y}{d x}$ in eq. (i) integration both the sides, we get
$\begin{aligned}
& \therefore \quad \frac{1}{2}\left(1-\frac{d v}{d x}\right)=-\left(\frac{v+1}{2 v+3}\right) \\
& \Rightarrow \quad \frac{d v}{d x}=1+\frac{2 v+2}{2 v+3} \\
& \Rightarrow \frac{2 v+2}{4 v+5} d v=d x \\
& \frac{1}{2} \int \frac{4 v+6}{4 v+5} d v=\int d x \\
& \Rightarrow \frac{1}{2} \int\left(\frac{4 v+6}{4 v+5}+\frac{1}{4 v+5}\right) d v=\int d x \\
& \Rightarrow \frac{1}{2} v+\frac{1}{2 \times 4} \log [4 v+5]=x+C \\
& \Rightarrow 4 v+\log [4 v+5]=8 x+C \\
& \Rightarrow 4(x-2 y)+\log [4(x-2 y)+5]=8 x+C \\
& \Rightarrow \log [4(x-2 y)+5]=4(x+2 y)+C
\end{aligned}$
Asked in: AP EAMCET 2016
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