Find a general solution. Show the details of your calculation. (100D²-160D+64l)y=0

Solution

VerifiedFirst, we need to apply the given operator to the given function:

$\begin{align*} 100D^2y - 160Dy + 64Iy &= 100D(Dy) -160y' + 64 y \\ & = 100y'' -160 y' + 64y \end{align*}$

Let's solve the equation:

$100y'' -160 y' + 64y = 0$

The characteristic equation of the homogeneous ODE is:

$100\lambda ^2 -160 \lambda + 64 = 0$

$\begin{align*} 100\lambda ^2 -160 \lambda + 64 = 0 &\iff \lambda _{1/2} = \frac{160 \pm \sqrt{160^2 - 4 \cdot 100 \cdot 64}}{200} \\ &\iff \lambda _{1/2} = \frac{160 \pm \sqrt{25 \; 600 - 25 \; 600}}{200} \\ &\iff \lambda _{1/2} = \frac{160}{200} \\ \end{align*}$

So, it has the real double root:

$\lambda = \frac{4}{5} = 0.8$

A basis is:

$\begin{align*} \begin{cases} & \phi _1 (x) = e^{\lambda _1 x} = e^{0.8 x} \\ & \phi _2 (x)= xe^{\lambda _2 x} = xe^{0.8 x} \end{cases} \end{align*}$

The general solution is:

$y = c_1 \phi _1 (x) + c_2 \phi _2 (x) = \boxed{{\color{#4257b2}{ c_1e^{0.8 x} + c_2xe^{0.8 x} }}}$

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