## Related questions with answers

Find a general solution. Show the details of your calculation. 4y''+32y'+63y=0

Solutions

VerifiedThe characteristic equation is

$4\lambda ^2 + 32 \lambda + 63 = 0$

$\begin{align*} 4\lambda ^2 + 32 \lambda + 63 = 0 &\iff \lambda _{1/2} = \frac{-32 \pm \sqrt{32^2 - 4 \cdot 4 \cdot 63}}{2\cdot 4} \\ & \iff \lambda _{1/2} =\frac{-32 \pm \sqrt{1204 - 1008}}{8} \\ & \iff \lambda _{1/2} = \frac{-32 \pm 4}{8} \end{align*}$

So, it has the distinct real roots:

$\lambda _1 = -\frac{28}{8} = -\frac{7}{2} = -3.5\quad \wedge \quad \lambda _2 = -\frac{36}{8} = -\frac{9}{2} = -4.5$

A basis is:

$\begin{align*} \begin{cases} & \phi _1 (x) = e^{\lambda _1 x} = e^{-3.5x} \\ & \phi _2 (x)= e^{\lambda _2 x}= e^{-4.5x} \end{cases} \end{align*}$

The general solution of the homogeneous ODE is:

${\color{#4257b2}{ y}} = c_1 \phi _1 (x) + c_2 \phi _2 (x) = {\color{#4257b2}{ c_1e^{-3.5x} + c_2e^{-4.5x} }}$

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