<s>[INST] Substitution methods of first order linear eqautions [problem help] Okay so I am having some computational issues towards the end of the problems
Problem 1: $y' = (4x+y)^2$
$z = 4x + y$
$y = z - 4x$
$y' = \frac{dz}{dx} -4$
$z^2 = \frac{dz}{dx}-4$
$z^2 + 4 = \frac{dz}{dx}$
$\int \frac{dx}{x} = \int \frac{dz}{z^2+4}$ 
$ln \vert x \vert + c = \ln \vert z^2+4 \vert$
$ce^x = (4x+y)^2+4$
$y(x) = \sqrt{ce^x-4}-4x$ 
I don't know it seems like every single problem with this substitution is a nightmare computationally.
Second problem:
$x(x+y)y'+y(3x+y)=0$
$y' = \frac{-y(3x+y)}{x(x+y)}$
This is a homogeneous equation
$y' = -v \frac{(3x+xv)}{(x+xv)}$
$y' = -v \frac{(3+v)}{(1+v)}$
$v + \frac{dv}{dx} =  \frac{-v^2-3v}{1+v}$
$x \frac{dv}{dx} = \frac{-v^2 -3v -v -v^2}{1+v}$
$x \frac{dv}{dx} = \frac{-2v^2-4v}{1+v}$
$\int \frac{1+v}{-2v(v-2)}dv = \int \frac{dx}{x}$
This a partial fractions, an ugly one at that:
$1+ v = \frac{A}{-2v}+\frac{B}{v-2}= A(v-2) -2Bv$
Now the only values of $v$ where I get any where is $v=2$ and $v = 0$ but I do not think these are viable because we get zeros in the fractions at this point. 
 [/INST] 1.)
Your integration is not correct.
$$
\int\frac{dz}{z^2+4}=\frac12\arctan(\frac z2)+C
$$
so that $z=2\tan(2(x-C))$, $y=2\tan(2(x-C))-4x$.
2.)
You want to compute the coefficients in the partial fraction decomposition
$$
-\frac{1+v}{2v(v-2)}=\frac{A}{v}+\frac{B}{v-2}
$$
To that end you multiply with the denominator to get
$$
1+v=-2A(v-2)-2Bv.
$$
You need to keep these two forms of the equation separate, do not mix them.
Indeed then inserting $v=0$ and $v=2$ gives $4A=1$ and $4B=-3$.
But, carefully check again the signs in your formulas
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