Give the correct answer and solve the given equation: displaystyle{x}{y}{left.{d}{x}right.}-{left({y}+{2}right)}{left.{d}{y}right.}={0}

Use Laplace transforms to solve the following initial value problem.
$x^{″}+x=3\cos 3t,x(0)=1,x^{\prime }(0)=0$
The solution is $x(t)=?$

How to solve this first-order differential equation,I don't have any idea which type this is..
${\displaystyle y^{\prime }+{(x-y)}^2=0}$
I have ${\displaystyle y^{\prime }+x^2-2xy+y^2=0}$, and then I divide by ${\displaystyle x^2}$
${\displaystyle t^{\prime }+t(\frac{1}{x}-2x)+t^{2}x=-x}$
and replace ${\displaystyle \frac{y}{x}=t}$, so I have then
It seems like it's a Riccati's diff. equation,but I don't know how to do it beacuse I don't have any particular solution..
On Wolfram Alpha solution is
$y=\frac{1}{c_1{e}^{2ix}+\frac{i}{2}}+x+i$

${\displaystyle d^2\frac{y}{{dx}^2}-2\frac{dy}{dx}+10y=0}$ where $x=0;y=0$ and ${\displaystyle \frac{dy}{dx}=4?x=0;y=0}$ and ${\displaystyle \frac{dy}{dx}=4}$?