Considering that $a$ and $b$ are odd functions, prove that the solution of the differential equation $y' + ay = b$ is even

Question

$a$ and $b$ are two odd functions, continuous on R.

How can I prove that the solution of the differential equation $y' + ay = b$ is even.

I tried to find the solutions of the differential equation using the method of "variation of constants", but I have a sign problem.

Here is what I tried :

The homogeneous solution of the equation is $$y_{h} = \lambda.e^{-A(x)}$$ Where $A(x) = \int a(x)dx$

Furthermore, $A(x)$ is a even function, considering that $a(x)$ is odd.

Using the "variation of the constants" method, we get a particular solution of the equation:

$y_{p} =\lambda (x).e^{-A(x)}$

$y'_{p} =\lambda '(x).e^{-A(x)} - a(x).y_{p}$

Therefore : $\lambda '(x).e^{-A(x)} = b(x) $

$\lambda (x) = \int b(x).e^{A(x)} dx $ $$y_{p} = e^{-A(x)}. \int b(x).e^{A(x)} dx$$

Therefore, the solution of the differential equation is : $$ y(x) = \lambda.e^{-A(x)} + e^{-A(x)}. \int b(x).e^{A(x)} dx$$

Considering that $b$ is odd and $A$ is even , we get: $$y(-x) = \lambda.e^{-A(x)} - e^{-A(x)}. \int b(x).e^{A(x)} dx$$

But $$y(x) \neq y(-x)$$

Answer 1

I think it is simpler than that - you can use the definitions of odd and even functions.

You have $y'(x)+a(x)y(x)=b(x)$

If you set $x=-z$ you get $y'(-z)+a(-z)y(-z)=b(-z)$

Now from the chain rule you have $y'(-x)=-y'(x)$, so rename $z$ as $x$ in the second equation and add to eliminate $y'$ and see what you get.

Answer 2

Let $y_1: \mathbb R \to \mathbb R$ a solution of $y' + ay = b$ and put $c:=y_1(0)$. Then $y_1$ is the unique (!) solution of the initial value problem

$(*) \quad y' + ay = b, y(0)=c$ .

Next define $y_2(x):=y_1(-x)$ and show that $y_2$ is also a solution of $(*)$.

Hence we have $y_2=y_1$

Answer 3

Let us decompose the solution in its even and odd parts $y=y_e+y_o$, and let us take the even part of the equation. As the derivative of an odd function is even and conversely, we have

$$y'_o+ay_o=0.$$

But this can be written

$$(\log y_o)'+a=0$$

which is not possible because the two terms have different parities, and $y_o=0$.