Find a real general solution. Show the details of your work. xy''+2y'=0

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Put $y = x^m$ .

Let's substitute:

y' = mx^{m-1}, \quad y'' = m(m-1)x^{m-2}

into the given ODE. This gives:

xm(m-1)x^{m-2} + 2 mx^{m-1}= 0

Multiply the equation by $x \neq 0$ :

x^2m(m-1)x^{m-2} + 2 xmx^{m-1}= 0

\cancel{x^2}m(m-1)x^m \cdot \cancel{x^{-2}} + 2 \cancel{x} mx^{m}\cancel{ x^{-1}} = 0

We can see that $x^m$ is a common factor, dropping it gives:

m(m-1) + 2m= 0 \iff m^2 + m = 0 \quad \textcolor{Fuchsia}{(*)}

So, $y = x^m$ is a solution of the given ODE if $m$ is a root of the equation $\textcolor{Fuchsia}{(*)}$

Let's find the roots of the equation $\textcolor{Fuchsia}{(*)}$ .

m^2 + m = 0 \iff m(m+1) = 0 \iff m = 0 \quad \vee \quad m = -1

So, it has the distinct real roots:

m _1 = 0 \quad \wedge \quad m _2 = -1

Real different roots $m_1$ and $m_2$ provide two real solutions:

y_1 = x^{m_1} = x^{0} = 1 \quad \wedge \quad y_2 = x^{m_2} = x^{-1} = \frac{1}{x}

Their quotient is not constant, so the solutions $y_1$ and $y_2$ are linearly independent and constitute a basis of solutions for the given ODE, for all x for which $y_1,y_2 \in \mathbb{R}$ .

So, the general solution is:

{\color{#4257b2}{y}}= c_1 y_1 = c_2 y_2 = \boxed{{\color{#4257b2}{c_1 + \frac{c_2}{x} }}}

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