Differential Equation Exercise 15
Let's consider the following first-order differential equation:
$$ y' + y = 1 $$
This is a linear differential equation of the form y' + a(x)y = b(x), where a(x) = 1 and b(x) = 1.
To solve it, we'll apply the method of variation of constants.
$$ y = e^{-\int a(x)\,dx} \cdot \left[ \int b(x) \cdot e^{\int a(x)\,dx} \,dx + c \right] $$
Substituting a(x) = 1 and b(x) = 1, we get:
$$ y = e^{-\int 1\,dx} \cdot \left[ \int 1 \cdot e^{\int 1\,dx} \,dx + c \right] $$
Since the integral of 1 is x, this becomes:
$$ y = e^{-x} \cdot \left[ \int e^x \,dx + c \right] $$
The antiderivative of ex is simply ex, so we have:
$$ y = e^{-x} \cdot \left[ e^x + c \right] $$
Multiplying through:
$$ y = e^{-x} \cdot e^x + e^{-x} \cdot c $$
$$ y = 1 + c \cdot e^{-x} $$
This is the general solution to the differential equation.
$$ y = c \cdot e^{-x} + 1 $$
And that wraps it up.
