Given the equation:
$$\frac{d}{d x} y{\left(x \right)} = 4 y{\left(x \right)} - 8$$
This differential equation has the form:
y' + P(x)y = Q(x)
where
$$P{\left(x \right)} = -4$$
and
$$Q{\left(x \right)} = -8$$
and it is called
linear homogeneousdifferential 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
$$P{\left(x \right)} = -4$$, then
$$\int P{\left(x \right)}\, dx$$ =
= $$\int \left(-4\right)\, dx = - 4 x + Const$$
Detailed solution of the integralSo, 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:
$$y = C e^{4 x}$$
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
$$y = C{\left(x \right)} e^{4 x}$$
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):
$$\frac{d}{d x} C{\left(x \right)} = - 8 e^{- 4 x}$$
So, C(x) =
$$\int \left(- 8 e^{- 4 x}\right)\, dx = Const + 2 e^{- 4 x}$$
Detailed solution of the integraluse C(x) at
$$y = C{\left(x \right)} e^{4 x}$$
and we get a definitive solution for y(x):
$$e^{4 x} \left(Const + 2 e^{- 4 x}\right)$$