Solve differential equation y'+4y=128x

Need your help!
1. Identify the substitution method used to solve the given differential equations. Support your answer with the appropriate calculations.
2. Solve for the solution to the given differential equation that satisfies the given condition
${\displaystyle y^{\prime }=2{(3x+y)}^2-1;y\left(0\right)=1}$

Solve the differential equatio
$\frac{dy}{dx}={\left(\frac{x+2y-3}{2x+y+3}\right)}^2$
The answer is given as:
$(x+3{)}^3-(y+3{)}^3=c(x-y+6{)}^4$
My attempt :I tried to expand the terms in the numerator and the denominator, arrange them into differentials and then integrate then, but I couldn't arrange all the terms in integrable format.

I have to solve following system of equations:
$x^{″}=-x-z+e^{-t},z^{\prime }=-2x-2z+3e^{-t}$
I would like to have formula for x' in order to find fundamental matrix and so on. So from above quations I obtain $z^{\prime }-2x^{″}=e^{-t}$, integrate sides and have $2x^{\prime }=z+e^{-t}$
So I have system of two first-order differential equation
$2x^{\prime }=z+e^{-t},z^{\prime }=-2x-2z+3e^{-t}$
Is it correct?

Solve differential equation $u^{\prime }-5u=v{e}^{-5v}$

For which nonnegative integers n is
${\displaystyle n^2\le n!?}$

Solve differential equation $xydy-y^{2}dx=(x+y{)}^2{e}^{(}-y/x)$

Im clueless on how to solve the following question...
$x{e}^y\frac{dy}{dx}=e^y+1$
What i've done is...
$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{x{e}^e};\frac{dy}{dx}-\frac{1}{x{e}^e}=\frac{1}{x}$
Find the integrating factor..
$$\begin{gathered} v(x)=e^{P(x)};whereP(x)=\int p(x)dx\Rightarrow P(x)=\int \frac{1}{x}dx=ln|x| \\ v(x)=e^{P(x)}=e^{ln|x|}=x; \\ y=\frac{1}{v(x)}\int v(x)q(x)dx=\frac{1}{x}\int x\frac{1}{x}dx=1+c \end{gathered}$$
I know I made a mistake somewhere. Would someone advice me on this?