01 - Space of Solutions of a Homogenous Linear ODE is a Vector Space

Space of Solutions of an ODE is a a Vector Space

Let $y_1,y_2$ be solutions of a homogenous linear ODE
Let ${\alpha }_1,{\alpha }_2$ be real numbers

Then

$${\alpha }_1{y}_1+{\alpha }_2{y}_2\text{ is also a solution of the ODE}$$

Hence the solutions to the ODE form a Real Vector Space

Proof

We know that

$$a_k(x)\frac{d^k{y}_1}{d{x}^k}+a_{k-1}(x)\frac{d^{k-1}{y}_1}{d{x}^{k-1}}+\cdots +a_1(x)\frac{d{y}_1}{dx}+a_0(x)y_1=0$$ $$a_k(x)\frac{d^k{y}_2}{d{x}^k}+a_{k-1}(x)\frac{d^{k-1}{y}_2}{d{x}^{k-1}}+\cdots +a_1(x)\frac{d{y}_2}{dx}+a_0(x)y_2=0$$

Adding ${\alpha }_1$ times the first equation to ${\alpha }_2$ times the second equation gives

$$a_k(x)\frac{d^k({\alpha }_1{y}_1+{\alpha }_2{y}_2)}{d{x}^k}+a_{k-1}(x)\frac{d^{k-1}({\alpha }_1{y}_1+{\alpha }_2{y}_2)}{d{x}^{k-1}}+\cdots +a_1(x)\frac{d({\alpha }_1{y}_1+{\alpha }_2{y}_2)}{dx}+a_0(x)({\alpha }_1{y}_1+{\alpha }_2{y}_2)=0$$

Hence ${\alpha }_1{y}_1+{\alpha }_2{y}_2$ is also a solution to the ODE

Note that this holds as differentiation is a Linear Map