Differential equation dx/dt+3x/20+t=4

Given the equation:
$$t + \frac{3 x{\left(t \right)}}{20} + \frac{d}{d t} x{\left(t \right)} = 4$$
This differential equation has the form:

y' + P(x)y = Q(x)

where
$$P{\left(t \right)} = \frac{3}{20}$$
and
$$Q{\left(t \right)} = - t + 4$$
and it is called linear inhomogeneous
differential 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
$$P{\left(t \right)} = \frac{3}{20}$$, then
$$\int P{\left(x \right)}\, dx = \int \frac{3}{20}\, dt = \frac{3 t}{20} + Const$$
Detailed solution of the integral
So, solution of the homogeneous linear equation:
$$y_{1} = e^{C_{1} - \frac{3 t}{20}}$$
$$y_{2} = - e^{C_{2} - \frac{3 t}{20}}$$
that leads to the correspondent solution
for any constant C, not equal to zero:
$$y = C e^{- \frac{3 t}{20}}$$
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

$$y = C{\left(t \right)} e^{- \frac{3 t}{20}}$$
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):
$$\frac{d}{d t} C{\left(t \right)} = \left(- t + 4\right) e^{\frac{3 t}{20}}$$
So,
$$C{\left(t \right)} = \int \left(- t + 4\right) e^{\frac{3 t}{20}}\, dt = - \frac{\left(60 t - 400\right) e^{\frac{3 t}{20}}}{9} + \frac{80 e^{\frac{3 t}{20}}}{3} + Const$$
Detailed solution of the integral
use C(x) at
$$y = C{\left(t \right)} e^{- \frac{3 t}{20}}$$
and we get a definitive solution for y(x):
$$e^{- \frac{3 t}{20}} \left(- \frac{\left(60 t - 400\right) e^{\frac{3 t}{20}}}{9} + \frac{80 e^{\frac{3 t}{20}}}{3} + Const\right)$$