Question

Let $y=f\left(x\right)$ be the solution of the differential equation $\frac{d y}{d x}=-2x\left(y - 1\right)$ with $f\left(0\right)=1,$ then $\underset{x \rightarrow \in fty}{l i m}f\left(x\right)$ is equal to

• A. $\frac{1}{2}$
• B. $0$
• C. $e$
• D. $1$

Answer 1

Answer: (D)

$\frac{d y}{d x}=-2x\left(y - 1\right)$
$\frac{d y}{d x}+2xy=2x$
I. $F .=e^{\int 2 x d x}=e^{x^{2}}$
$y\cdot e^{x^{2}}=\displaystyle \int e^{x^{2}} 2 x d x$
$y\cdot e^{x^{2}}=e^{x^{2}}+C$
$\because y\left(0\right)=1\Rightarrow 1=1+C\Rightarrow C=0$
Hence, the solution is $ye^{x^{2}}=e^{x^{2}}$
$\Rightarrow y=1=f\left(x\right)$
$\underset{x \rightarrow \in fty}{l i m}f\left(x\right)=1$