## Related questions with answers

Find a general solution. Check your answer by substitution. y''+y'+3.25y=0

Solutions

Verified${\color{#c34632} {y''+y'+3.25y=0}}\;\;\;\; \Rightarrow \;\; \textbf{Homogeneous 2nd Order D.E , With Constant Coeff.}$

$\Rightarrow \textbf{Form :}\;\; \boxed{\;ay''+by'+cy=0\;}$

$Where\; : \;a\;,\;b\;,\;c \;\; \Rightarrow Constants$

$In\; order\; to\; solve\; this\; D.E \;, \;we\; need \;to\; get\; the\; "characterestic\; equation\;" by\; replacing$

$\boxed{y'' \;\; \Rightarrow \;\; m^{2}\;\;\;,\;\;\; y'\;\; \Rightarrow \;\; m\;\;\;,\;\;\;y\;\; \Rightarrow \;\; 1}$

$\textbf{Then Characterestic Eq :}\;\; \Rightarrow \;\;{\color{#c34632} {m^{2}+m+3.25=0}}$

$m_{1,2}=\dfrac{-b \;\pm \sqrt{b^{2}-4ac}}{2a}=\dfrac{-1\;\pm \sqrt{1-13}}{2}$

$m_{1,2}=\dfrac{-1\;\pm 2\sqrt{3}i}{2}$

$m_{1,2}=\dfrac{-1}{2}\pm \sqrt{3}\;i\;\;\;\; \Rightarrow \;\; \boxed{\;m=\alpha \pm \beta i\;}$

$\therefore \alpha = \dfrac{-1}{2} \;\;\; , \;\;\; \beta =\sqrt{3}$

$\therefore \;\; \textbf{Solution is : }\; \boxed{\color{#c34632} {y=e^{\alpha x}[C_{1}\cos \beta +C_{2}\sin \beta ]}}\;\;\; \Rightarrow \;\; \textbf{For Complix Roots}$

$\therefore \boxed{\;\color{#4257b2} {y=e^{(-1/2)x}[C_{1}\cos \sqrt{3} +C_{2}\sin \sqrt{3} ]}\;}$

The characteristic equation is

$\lambda ^2 + \lambda + 3.25= 0.$

Let's find the roots of the characteristic equation.

$\begin{align*} \lambda ^2 + \lambda + 3.25= 0 &\iff \lambda _{1/2} = \frac{-1 \pm \sqrt{1^2 -4\cdot 3.25}}{2} \\ & \iff \lambda _{1/2} =\frac{-1 \pm \sqrt{-12}}{2} = \frac{-1 \pm 2\sqrt{3}i}{2} \\ & \iff \lambda _1 =\frac{-1 + 2\sqrt{3}i}{2}, \quad \lambda _2 = \frac{-1 - 2\sqrt{3}i}{2} \end{align*}$

So, it has the complex conjugate roots:

$\lambda _1 = -\frac{1}{2} - i\sqrt{3} (=a + ib)\quad \wedge \quad \lambda _2 = -\frac{1}{2} + i\sqrt{3} (=a-ib)$

A basis is:

$\begin{align*} \begin{cases} & \phi _1 (x) = e^{ax} \cos (bx) = e^{-\frac{x}{2}} \cos (\sqrt{3} x) \\ &\phi _2 (x) =e^{ax} \sin (bx) = e^{-\frac{x}{2}}\sin (\sqrt{3} x) \end{cases} \end{align*}$

The general solution is:

$\begin{align*} y & = c_1 \phi _1 (x) + c_2 \phi _2 (x) \\ & = c_1e^{-\frac{x}{2}} \cos (\sqrt{3} x)+ c_2e^{-\frac{x}{2}}\sin (\sqrt{3} x) \\ & = \boxed{{\color{#4257b2}{ e^{-0.5 x} \left(c_1\cos (\sqrt{3} x)+ c_2\sin (\sqrt{3} x) \right) }}} \end{align*}$

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