Given the equation:
$$\frac{d}{d t} x{\left(t \right)} = - k x{\left(t \right)} + r$$
This differential equation has the form:
y' + P(x)y = Q(x)
where
$$P{\left(t \right)} = k$$
and
$$Q{\left(t \right)} = r$$
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
$$P{\left(t \right)} = k$$, then
$$\int P{\left(x \right)}\, dx$$ =
= $$\int k\, dt = k t + Const$$
Detailed solution of the integralSo, solution of the homogeneous linear equation:
$$y_{1} = e^{C_{1} - k t}$$
$$y_{2} = - e^{C_{2} - k t}$$
that leads to the correspondent solution
for any constant C, not equal to zero:
$$y = C e^{- k t}$$
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
$$y = C{\left(t \right)} e^{- k t}$$
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):
$$\frac{d}{d t} C{\left(t \right)} = r e^{k t}$$
So, C(x) =
$$\int r e^{k t}\, dt = \begin{cases} \frac{r e^{k t}}{k} & \text{for}\: k \neq 0 \\r t & \text{otherwise} \end{cases} + Const$$
Detailed solution of the integraluse C(x) at
$$y = C{\left(t \right)} e^{- k t}$$
and we get a definitive solution for y(x):
$$e^{- k t} \left(\begin{cases} \frac{r e^{k t}}{k} & \text{for}\: k \neq 0 \\r t & \text{otherwise} \end{cases} + Const\right)$$