Divide both sides of the equation by the multiplier of the derivative of y':
$$x$$
We get the equation:
$$\frac{x \frac{d}{d x} y{\left(x \right)} - 3 y{\left(x \right)}}{x} = \frac{6}{x}$$
This differential equation has the form:
y' + P(x)y = Q(x)
where
$$P{\left(x \right)} = - \frac{3}{x}$$
and
$$Q{\left(x \right)} = \frac{6}{x}$$
and it is called
linear inhomogeneousdifferential first-order equation:First of all, we should solve the correspondent linear homogeneous equation
y' + P(x)y = 0
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
$$P{\left(x \right)} = - \frac{3}{x}$$, then
$$\int P{\left(x \right)}\, dx$$ =
= $$\int \left(- \frac{3}{x}\right)\, dx = - 3 \log{\left(x \right)} + Const$$
Detailed solution of the integralSo, solution of the homogeneous linear equation:
$$y_{1} = x^{3} e^{C_{1}}$$
$$y_{2} = - x^{3} e^{C_{2}}$$
that leads to the correspondent solution
for any constant C, not equal to zero:
$$y = C x^{3}$$
We get a solution for the correspondent homogeneous equation
Now we should solve the inhomogeneous equation
y' + P(x)y = Q(x)
Use variation of parameters method
Now, consider C a function of x
$$y = x^{3} C{\left(x \right)}$$
And apply it in the original equation.
Using the rules
- for product differentiation;
- of composite functions derivative,
we find that
$$\frac{d}{d x} C{\left(x \right)} = Q{\left(x \right)} e^{\int P{\left(x \right)}\, dx}$$
Let use Q(x) and P(x) for this equation.
We get the first-order differential equation for C(x):
$$\frac{d}{d x} C{\left(x \right)} = \frac{6}{x^{4}}$$
So, C(x) =
$$\int \frac{6}{x^{4}}\, dx = Const - \frac{2}{x^{3}}$$
Detailed solution of the integraluse C(x) at
$$y = x^{3} C{\left(x \right)}$$
and we get a definitive solution for y(x):
$$x^{3} \left(Const - \frac{2}{x^{3}}\right)$$