Given the equation:
$$\frac{d}{d t} x{\left(t \right)} = 3 t^{2} x{\left(t \right)}$$
This differential equation has the form:
y' + P(x)y = 0,
where
$$P{\left(t \right)} = - 3 t^{2}$$
and
and it is called
linear homogeneousdifferential first-order equation:It's an equation with multiple 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
$$P{\left(t \right)} = - 3 t^{2}$$, then
$$\int P{\left(x \right)}\, dx$$ =
= $$\int \left(- 3 t^{2}\right)\, dt = - t^{3} + Const$$
Detailed solution of the integralSo, solution of the homogeneous linear equation:
$$y_{1} = e^{C_{1} + t^{3}}$$
$$y_{2} = - e^{C_{2} + t^{3}}$$
that leads to the correspondent solution
for any constant C, not equal to zero:
$$y = C e^{t^{3}}$$