Question

The solution of the initial value problem $\frac{dy}{dx}=-2xy;y\left(0\right)=2$ is

• A. $1+e^{-x^2}$
• B. $2e^{-x^2}$
• C. $1+e^{x^2}$
• D. $2e^{x^2}$

Answer 1

The correct option is B $2e^{-x^2}$

$\frac{dy}{dx}=-2xy;y\left(0\right)=2$

$\frac{dy}{y}=-2xdx$

$lny=-x^2+c$

Condition : y(0) = 2 $\Rightarrow c=ln2$

$ln\left(\frac{y}{2}\right)=-x^2$

$y=2e^{-x^2}$