We need to solve the differential equation $(4x+y)^2 \frac{dx}{dy} = 1$.

1. Problem Description

We need to solve the differential equation $(4x+y)^2 \frac{dx}{dy} = 1$.

2. Solution Steps

First, rewrite the equation as:

$\frac{dx}{dy} = \frac{1}{(4x+y)^2}$

Now, let $v = 4x+y$. Then, differentiating with respect to $y$, we get:

$\frac{dv}{dy} = 4\frac{dx}{dy} + 1$

Substituting $\frac{dx}{dy} = \frac{1}{v^2}$ into the equation above, we have:

$\frac{dv}{dy} = \frac{4}{v^2} + 1 = \frac{4+v^2}{v^2}$

$\frac{v^2}{v^2+4} dv = dy$

Now, integrate both sides with respect to their respective variables:

$\int \frac{v^2}{v^2+4} dv = \int dy$

We can rewrite the integrand as:

$\frac{v^2}{v^2+4} = \frac{v^2+4-4}{v^2+4} = 1 - \frac{4}{v^2+4}$

So, the integral becomes:

$\int (1 - \frac{4}{v^2+4}) dv = \int dy$

$\int 1 dv - \int \frac{4}{v^2+4} dv = \int dy$

$v - 4 \int \frac{1}{v^2+2^2} dv = y + C$

Recall that $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \arctan(\frac{x}{a}) + C$. In our case, $a=2$, so:

$v - 4 (\frac{1}{2} \arctan(\frac{v}{2})) = y + C$

$v - 2 \arctan(\frac{v}{2}) = y + C$

Substituting $v = 4x+y$ back into the equation:

$4x+y - 2 \arctan(\frac{4x+y}{2}) = y + C$

$4x - 2 \arctan(\frac{4x+y}{2}) = C$

3. Final Answer

$4x - 2 \arctan(\frac{4x+y}{2}) = C$