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Problem 46: Prove that the set of all solutions to the DE $y\text{'}+9y=0$ is a vector space

Problem 46: Prove that the set of all solutions to the DE $y\text{'}+9y=0$ is a vector space

Solution: Given $\V = \{y:y'+9y=0\}$

I. Properties under addition
i. Closure Property
Let $u,v \in \V$ then $u'+9u=0$ and $v'+9v=0$ \begin{align*} u+v = (u+v)'+9(u+v) \\ = (u'+v')+ 9(u+v)\\ = u'+9u+v'+9v\\ = 0+0=0\\ \therefore u+v \in \V. \end{align*} $\Rightarrow \V$ is closed under addition.

ii. Associativity:
Let $u,v,w \in \V$ then $u'+9u=0$, $v'+9v=0$ and $w'+9w=0$ Now, \begin{align*} \left[u+v\right]+w = \left[(u'+9u)+(v'+9v)\right]+(w'+9w)\\ = \left[(u'+9u+v'+9v)\right]+(w'+9w)\\ = \left[(u'+9u+v'+9v+w'+9w)\right]\\ = \left[(u'+9u)+(v'+9v+w'+9w)\right]\\ = (u'+9u)+\left[(v'+9v+w'+9w)\right]\\ = (u'+9u)+\left[(v'+9v)+(w'+9w)\right]\\ = u+\left[(v+w\right]\\ \end{align*} $\Rightarrow$ Addition is associative in $\V$.

iii. Existence of additive identity:
For all $u \in \V, \exists 0 \in \V$
Now \begin{align*} u+0=(u'+9u)+0 = (u'+9u) = u\\ 0+u=0+(u'+9u) = (u'+9u) = u\\ \end{align*} $\Rightarrow 0$ is the additive identity in $\V$.

iv. Existence of additive inverse:
Let $u \in \V$, be any element $-u\in \V$
Now \begin{align*} u+-u= (u'+9u)+ (-u'-9u)=0\\ -u+u= (-u'-9u)+(u'+9u)=0\\ \end{align*} $\Rightarrow -u$ is the additive inverse of $u$ for each $u \in \V$.

v. Commutative Law:
Let $ u,v \in \V$, Now \begin{align*} u+v= (u'+9u)+(v'+9v)\\ = (v'+9v)+(u'+9u) = v+u \end{align*} $\Rightarrow$ addition is commutative in $\V$.

II. Properties under Scalar Multiplication
vi. Closure under Scalar Multiplication
Let $\alpha\in \R$ and $u \in \V$ \begin{align*} \alpha u = (\alpha u)'+9(\alpha u)\\ = \alpha u' + 9\alpha u\\ = \alpha (u'+9u) = \alpha \cdot 0 = 0\\ \therefore \alpha u \in \V \end{align*} $\Rightarrow \V$ is closed under Scalar Multiplication.

vii.Distributivity of Addition of Real Numbers
Let $\alpha, \beta\in \R$ and $u \in \V$ \begin{align*} (\alpha+\beta)u= ((\alpha+\beta) u)'+9(\alpha+\beta) u)\\ = \alpha u'+\beta u' + 9 \alpha u + 9 \beta u\\ = \alpha u' + 9 \alpha u + \beta u' + 9 \beta u\\ = \alpha (u'+9u) + \beta (u'+9u)\\ = \alpha u + \beta u \end{align*} $\Rightarrow (\alpha+\beta)u = \alpha u+\beta u$.
$\Rightarrow $ Distributivity of Addition of Real Numbers does exists.

viii. Distributivity of Addition of Vectors
Let $\alpha\in \R$ and $u, v \in \V$ \begin{align*} \alpha\left[u+v\right]= \alpha\left[(u'+9u)+(v'+9v)\right]\\ = \left[\alpha((u'+9u)+(v'+9v))\right]\\ = \left[\alpha(u'+9u)+\alpha(v'+9v)\right]\\ = \left[\alpha(u'+9u)+\alpha(v'+9v)\right]\\ =\alpha(u'+9u)+\alpha(v'+9v)\\ =\alpha u+\alpha v\\ \end{align*} $\Rightarrow $ Distributivity of Addition of Vectors exists.

ix.Associativity of multiplication
Let $\alpha, \beta\in \R$ and $u\in \V$ \begin{align*} (\alpha\beta)u= (\alpha\beta)(u'+9u)\\ = \alpha(\beta(u'+9u)\\ =\alpha(\beta u) \end{align*} $\Rightarrow$ Associativity of multiplication exists.

x. Unity element multiplication
Let 1 be unity element of field $\F$ and $u\in \V$
Now \begin{align*} 1.u= 1\cdot(u'+9u)\\ =(u'+9u)\\ = u \end{align*} $\Rightarrow 1.u = u \forall u \in \V$.

Hence, $\V$ is a vector space over $\R$.

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