436. Solve the equation:

$$
3 x-4 y=5 \sqrt{x^{2}+y^{2}}
$$
$\triangle$ Let's introduce vectors $\bar{u}$ and $\bar{v}$, and choose their coordinates in such a way that the left side of the equation expresses the dot product of the vectors in coordinates, while the right side expresses the product of the lengths of the vectors. Suitable vectors are $\bar{u}(x ; y)$ and $\bar{v}(3 ;-4)$. We apply inequality (4):

$$
\bar{u} \cdot \bar{v}=3 x-4 y \leq|\bar{u}| \cdot|\bar{v}|=\sqrt{x^{2}+y^{2}} \cdot \sqrt{3^{2}+4^{2}}=5 \sqrt{x^{2}+y^{2}} .
$$

By the condition, the left and right sides of this inequality are equal. We use the proportion (3):

$$
\frac{x}{3}=\frac{y}{-4}, \quad y=-\frac{4}{3} x
$$

This, in fact, is the answer - with the addition of the restriction $x \geq 0$, since the ratio $\frac{x}{3}$ must be non-negative - due to the fact that the proportionality coefficient in equality (3) when using inequality (4) is non-negative.

Answer: $(x ;-4 x / 3)$, where $x$ is any non-negative number.