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# Solve the IVP. Check that your answer satisfies the ODE as well as the initial conditions. Show the details of your work. 8y''-2y'-y=0, y(0)=-0.2, y'(0)=-0.325

Solution

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We have

$8y''-2y' - y = 0$

The characteristic equation,

$\begin{gather*} 8\lambda^{2} - 2\lambda - 1 = 0 \\ \text{its roots:} \quad \lambda_{1} =\dfrac{1}{2} \quad \text{and} \quad \lambda_{2}=-\dfrac{1}{4} \end{gather*}$

where we see that we have a case of two distinct real roots, which means that the general solution of this Second-Order Linear ODE would be of the form

$\mathbf{y=C_{1}e^{\lambda_{1} x} + C_{2} e^{\lambda_{2} x}}$

Therefore,

$y=C_{1}e^{\frac{1}{2} x} + C_{2}e^{-\frac{1}{4} x}$

and

$y'= \dfrac{1}{2} C_{1} e^{\frac{1}{2} x} - \dfrac{1}{4} C_{2} e^{-\frac{1}{4} x}$

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