Question

The set of real (or complex) situations of equation y'(x) +P(x) y(x) - Q(x) , forms a real (or complex) vector space.

Answer 1

Answer on Question #40026 – Math - Differential Calculus

We have the equation $y'(x) + P(x)y(x) - Q(x) = 0$. Show that the set of real (or complex) situations of this equation does not form a real (or complex) vector space.

Solution.

Let $y_{1}$ and $y_{2}$ be any two solutions of the given differential equation. Then it is to be shown that $y_{1} + y_{2}$ and $a y_{1}$ are also solutions where $a$ is a scalar. We have

$$\begin{array}{l} (y_{1} + y_{2})' + P(x)(y_{1} + y_{2}) - Q(x) = y_{1}' + y_{2}' + P(x)y_{1} + P(x)y_{2} - Q(x) \\ = \left(y_{1}' + P(x)y_{1} - Q(x)\right) + \left(y_{2}' + P(x)y_{2} - Q(x)\right) + Q(x) = Q(x) \neq 0, \end{array}$$

as $y_{1}$ and $y_{2}$ are solutions.

Hence $(y_{1} + y_{2})$ is not a solution.

Further,

$$(a y_{1})' + P(x)(a y_{1}) - Q(x) = a y_{1}' + a P(x) y_{1} - a Q(x) + a Q(x) - Q(x) = Q(x)(a - 1) \neq 0,$$

as $y_{1}$ is a solution.

Thus $a y_{1}$ is not a solution.

So we see that our set does not form a real (or complex) vector space.