Search any question & find its solution
Question:
Answered & Verified by Expert
The solution of the differential equation $\frac{d y}{d x}=\frac{x-2 y+1}{2 x-4 y}$ is
Options:
Solution:
1435 Upvotes
Verified Answer
The correct answer is:
$(x-2 y)^2+2 x=c$
Given that, $\frac{d y}{d x}=\frac{x-2 y+1}{2 x-4 y}$
Put $x-2 y=z \Rightarrow 1-2 \frac{d y}{d x}=\frac{d z}{d x}$
$$
\begin{aligned}
& \therefore \quad \frac{1}{2}\left[-\frac{d z}{d x}+1\right]=\frac{z+1}{2 z} \\
& \Rightarrow \quad-\frac{d z}{d x}+1=\frac{z+1}{z} \\
& \Rightarrow \quad \frac{d z}{d x}=-\frac{1}{z} \\
& \Rightarrow \quad z d z=-d x \\
& \Rightarrow \quad \frac{z^2}{2}=-x+c_1 \\
& \Rightarrow(x-2 y)^2+2 x=c \\
&
\end{aligned}
$$
Put $x-2 y=z \Rightarrow 1-2 \frac{d y}{d x}=\frac{d z}{d x}$
$$
\begin{aligned}
& \therefore \quad \frac{1}{2}\left[-\frac{d z}{d x}+1\right]=\frac{z+1}{2 z} \\
& \Rightarrow \quad-\frac{d z}{d x}+1=\frac{z+1}{z} \\
& \Rightarrow \quad \frac{d z}{d x}=-\frac{1}{z} \\
& \Rightarrow \quad z d z=-d x \\
& \Rightarrow \quad \frac{z^2}{2}=-x+c_1 \\
& \Rightarrow(x-2 y)^2+2 x=c \\
&
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.