$(x+y)\frac{dy}{dx} -(4x+y)=0, y(1)=2$ Hi I am stuck solving this problem: 
'Employ a change of variables $z=\frac{y}{x}$ to solve the differential equation: $$(x+y)\frac{dy}{dx} -(4x+y)=0, y(1)=2$$
Enter the result $y(3)$. 
I have done the substitution and then integrated by I cannot get an explicit expression for y, which leaves me confused as to how I can find $y(3)$. 
Please help!
 A: Starting from
$$(x+y)\frac{dy}{dx} -(4x+y)=0$$
rewrite as
$$\frac{dy}{dx}=\frac{4x+4}{x+y}=\frac{4+\frac xy}{1+\frac xy}$$
then with the substitution $z=\frac{y}{x}$ we have $\frac{dy}{dx}=z+x\frac{dz}{dx}$ so
$$z+x\frac{dz}{dx}=\frac{4+z}{1+z} \implies x\frac{dz}{dx}=\frac{4-z^2}{1+z}$$
which we can rewrite as
$$\frac{1+z}{4-z^2}dz=\frac{1}{x}dx$$
where a partial fraction decomposition of the LHS forms
$$\frac{1+z}{4-z^2}=\frac{-1}{4(z+2)}-\frac{3}{4(z-2)}$$
hence
$$-\frac{1}{4}\left(\frac{1}{z+2}+\frac{3}{z-2}\right)dz=\frac{1}{x}dx$$
whereby integrating both sides forms
$$-\frac{1}{4}\Big(\ln|z+2|+3\ln|z-2|\Big)=\ln|x|+C$$
or
$$\ln|z+2|+3\ln|z-2|=-4\ln|x|+C$$
which can be rewritten as
$$(z+2)(z-2)^3=Cx^{-4} \implies C=x^4(z+2)(z-2)^3$$
therefore substituting $z=\frac{y}{x}$
$$C=x^4\left(\frac{y}{x}+2\right)\left(\frac{y}{x}-2\right)^3$$
the initial/boundary condition of $y(1)=2$ forms
$$C=1(4)(0)=0$$
hence since $x\neq 0$ we see that
$$(y+2x)(y-2x)^3=0$$
which implies
$$y=-2x,\quad y=2x$$
where $y=-2x$ cannot be the solution since $y(1)=2$. Therefore, $y=2x$ is the solution and $y(3)=6$.
A: Taking $z = \frac{y}{x}$ yields $y=xz$, and hence $\frac{dy}{dx} = x\frac{dz}{dx}+z$.
From this, we find by substitution
\begin{align*}
(x+xz)(x\frac{dz}{dx}+z) - (4x+xz) &= 0 \\
\text{ divide by } x, \text{ and simplify}\\
x \frac{dz}{dx} +z +zx \frac{dz}{dx} +z^2 -4-z &= 0 \\
\frac{dz}{dx}x(1+z) &= 4-z^2 \\
\frac{dz(1+z)}{4-z^2} &= \frac{dx}{x} \\
dz \left( \frac{3}{4(2-z)} - \frac{1}{4(2+z)} \right) &= \frac{dx}{x}
\end{align*}
Solving this differential equation, we get for some constant $K$,
$$\frac{1}{|(z+2)(z-2)^3|} = Kx^4.$$
Substituting $z = \frac{y}{x}$ again, we get after simplifying
$$ 1 = K|y^2-4x^2|(y-2x)^2 $$
Using the information that $y(1)=2$, we should be able to find $K$, but we get the following.
\begin{align*}
1 &= K|2^2-4 \cdot 1^2|(2-2 \cdot 1)^2 \\
1 &= 0
\end{align*}
Are you sure you copied the condition $y(1)=2$ correctly? There is definetily some mistake in your formulation of the problem anyway.
The next step would be to solve the polynomial equation
$$1 = K|y^2-4 \cdot 3^2|(y-2 \cdot 3)^2$$
Finding $y$ that fits this will then be the solution to your problem.
