Differential equation dy/dx=4y

Step

Given the equation:
$$\frac{d}{d x} y{\left(x \right)} = 4 y{\left(x \right)}$$
This differential equation has the form:
$$y' + P(x)y = 0$$,
where
$$P{\left(x \right)} = -4$$
and
and it is called linear homogeneous differential first-order equation:
It's the equation with separable variables.
The equation is solved using following steps:
From $y' + P(x)y = 0$ you get
$$\frac{dy}{y} = - P{\left(x \right)} dx$$, if y is not equal to 0
$$\int \frac{1}{y}\, dy = - \int P{\left(x \right)}\, dx$$
$$\log{\left(\left|{y}\right| \right)} = - \int P{\left(x \right)}\, dx$$
Or,
$$\left|{y}\right| = e^{- \int P{\left(x \right)}\, dx}$$
Therefore,
$$y_{1} = e^{- \int P{\left(x \right)}\, dx}$$
$$y_{2} = - e^{- \int P{\left(x \right)}\, dx}$$
The expression indicates that it is necessary to find the integral:
$$\int P{\left(x \right)}\, dx$$
Because
$$P{\left(x \right)} = -4$$, then
$$\int P{\left(x \right)}\, dx = \int \left(-4\right)\, dx = - 4 x + Const$$
Detailed solution of the integral
So, solution of the homogeneous linear equation:
$$y_{1} = e^{C_{1} + 4 x}$$
$$y_{2} = - e^{C_{2} + 4 x}$$
that leads to the correspondent solution for any constant C, not equal to zero:
$$y = C e^{4 x}$$