Answer
1) We show that really $y(0)=4$ i.e. that $y$ satisfies the given initial condition.
2) We find $y'$ and put $y$ and $y'$ into the given differential equation to show that $y$ really is its' solution.
by 1) and 2) we verified the particular solution.
Work Step by Step
1) Show that $y(0) = 4$.
Indeed
$$y(0)=4e^{-6\times0^2} = 4e^0 = 4\times 1 = 4.$$
2) Show that $y$ really is a solution to this equation.
First we will find $y'$:
$$y'=(4e^{-6x^2})'=4(e^{-6x^2})' = 4e^{-6x^2}(-6x^2)' = -24e^{-6x^2}2x = -48xe^{-6x^2}.$$
Now we put $y$ and $y'$ into the given differential equation
The Left side is:
$$y'=-48xe^{-6x^2}.$$
The Right side is:
$$-12xy = -12x(4e^{-6x^2}) = -48xe^{-6x^2}.$$
Since both sides are equal to each other we see that $y$ really is a solution to this differential equation.
1) and 2) say that $y$ is verified as a particular solution to this differential equation.
