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Question

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

Solution

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

100D2y160Dy+64Iy=100D(Dy)160y+64y=100y160y+64y\begin{align*} 100D^2y - 160Dy + 64Iy &= 100D(Dy) -160y' + 64 y \\ & = 100y'' -160 y' + 64y \end{align*}

Let's solve the equation:

100y160y+64y=0100y'' -160 y' + 64y = 0

The characteristic equation of the homogeneous ODE is:

100λ2160λ+64=0100\lambda ^2 -160 \lambda + 64 = 0

100λ2160λ+64=0    λ1/2=160±1602410064200    λ1/2=160±25  60025  600200    λ1/2=160200\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:

λ=45=0.8\lambda = \frac{4}{5} = 0.8

A basis is:

{ϕ1(x)=eλ1x=e0.8xϕ2(x)=xeλ2x=xe0.8x\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=c1ϕ1(x)+c2ϕ2(x)=c1e0.8x+c2xe0.8xy = 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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