If $x d y+\left(y+y^2 x\right) d x=0$ and $y=1$ at $x=1$. then
If $x d y+\left(y+y^2 x\right) d x=0$ and $y=1$ at $x=1$. then
- $y=\frac{x}{1+\log x}$
- $y=\frac{1+\log x}{x}$
- $y=x(1+\log x)$
- $y=\frac{1}{x(1+\log x)}$
Solution
Since, $x d y+\left(y+y^2 x\right) d x=0$
$\Rightarrow \frac{d y}{d x}+\frac{y}{x}+y^2=0...(i)$
Let $y=\frac{1}{t} \Rightarrow \frac{d y}{d x}=-\frac{1}{t^2} \frac{d t}{d x}$
$\begin{aligned}
& \Rightarrow-\frac{1}{t^2} \frac{d t}{d x}+\frac{1}{t x}+\frac{1}{t^2}=0 \Rightarrow \frac{d t}{d x}-\frac{t}{x}=1 \\
& \text { I.F. }=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}
\end{aligned}$
So, solution is given by
$\frac{t}{x}=\int \frac{1}{x} d x=\log x+c \Rightarrow \frac{1}{y x}=\log x+c$
Since, $y(1)=1 \Rightarrow \frac{1}{1}=\log 1+c \Rightarrow c=1$.
$\Rightarrow \frac{1}{y x}=\log x+1 \Rightarrow y=\frac{1}{x(1+\log x)}$.
Asked in: AP EAMCET 2024 (19 May Shift 2)
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