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Question

# Find a real general solution. Show the details of your work. x²y''+0.7xy'-0.1y=0

Solution

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Let's substitute:

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

into the given ODE. This gives:

$x^2m(m-1)x^{m-2} + 0.7 x mx^{m-1} -0.1 x^m = 0$

$\cancel{x^2}m(m-1)x^m \cdot \cancel{x^{-2}} + 0.7 \cancel{x} mx^{m}\cancel{ x^{-1}} -0.1 x^m = 0$

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

$m(m-1) + 0.7m -0.1 = 0 \iff m^2 - 0.3m - 0.1 = 0 \quad \textcolor{Fuchsia}{(*)}$

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 $\textcolor{Fuchsia}{(*)}$.

\begin{align*} m^2 - 0.3m -0.1 = 0 &\iff m_{1/2} = \frac{0.3 \pm \sqrt{0.3^2 + 4\cdot 0.1 }}{2} \\ &\iff m_{1/2} = \frac{0.3 \pm \sqrt{0.09 + 0.4}}{2} \\ &\iff m_{1/2} = \frac{0.3 \pm 0.7}{2} \end{align*}

So, it has the distinct real roots:

$m _1 = 0.5 \quad \wedge \quad m _2 = -0.2$

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

$y_1 = x^{m_1} = x^{0.5} \quad \wedge \quad y_2 = x^{m_2} = x^{-0.2}$

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:

${\color{#4257b2}{y}}= c_1 y_1 + c_2 y_2 = \boxed{{\color{#4257b2}{c_1x^{0.5} + c_2 x^{-0.2} }}}$

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