Solving y’ + a(x) y = 0

Let’s consider the 1st-Order linear differential equation

y'(x)+ a(x) y(x) =0\quad\quad\quad(1)

We see y(x) \equiv 0 is a solution.

However, if (1) has a solution whose value at x is not zero, it must be true that

{1 \over y(x)} y(x)' = -a(x)

i.e.,

log(|y(x)|) = -\int a(x) dx + c_1\quad\quad\quad(2)

where c_1 is a constant.

From (2), we obtain

y(x) = c\;e^{-\int a(x) dx}\quad\quad\quad(3)

where c is a constant. It is either +e^{c_1} or -e^{c_1}.

We assert and prove that for any constant c, a function defined by (3)  is a solution of (1):

(c\;e^{-\int a(x)dx)})' +a(x)c\;e^{-\int a(x) dx}=-c\;a(x)e^{-\int a(x) dx} +a(x) c\;e^{-\int a(x) dx} =0.

Notice when c=0, (3) yields y(x) \equiv 0, the zero solution of (1).

Moreover, to see there are no other solutions, let  u(x) be any solution of (1), we have

(u(x) e^{\int a(x)dx})' = u'(x) e^{\int a(x) dx} + u(x) a(x) e^{\int a(x) dx}

= e^{\int a(x)dx } (u'(x) + a(x) u(x))

= 0.

Therefore,

(u(x) e^{\int a(x) dx})' = 0.

That is

u(x) e^{\int a(x)dx} = c

or,

u(x) = c\;e^{-\int a(x) dx}

where c is a constant. Hence, any solution of (1) belongs to the family of functions defined by (3)

2 thoughts on “Solving y’ + a(x) y = 0

  1. Pingback: Solving y’ + a(x) y = b(x), Part 1 | Vroom

  2. Pingback: Solving y’ + a(x) y = b(x), Part 1 | Vroom

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