Find a real general solution. Show the details of your work. 4x²y''+5y=0

Solution

VerifiedLet's substitute:

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

into the given ODE. This gives:

$4x^2m(m-1)x^{m-2} + 5 x^m = 0$

$4\cancel{x^2}m(m-1)x^m \cdot \cancel{x^{-2}} + 5 x^m = 0$

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

$4m(m-1) + 5 = 0 \iff 4m^2 - 4m + 5 = 0 \quad \textcolor{Fuchsia}{(*)}$

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

$\begin{align*} 4m^2 - 4m + 5 = 0 &\iff m_{1/2} = \frac{4 \pm \sqrt{(-4)^2 - 4\cdot 4 \cdot 5 }}{2 \cdot 4} \\ &\iff m_{1/2} = \frac{4 \pm \sqrt{-64}}{8} \\ &\iff m_{1/2} = \frac{4 \pm 8i}{8} \end{align*}$

So, it has the complex conjugate roots:

$m _1 = \frac{1}{2} + i \quad \wedge \quad m _2 = \frac{1}{2} - i$

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