## Related questions with answers

Find a general solution. Check your answer by substitution. 100y''+240y'+(196π²+144)y=0

Solutions

Verified${\color{#c34632} {100y''+240y'+(196\pi^{2} +144)y=0}}\;\;\;\; \Rightarrow \;\; \textbf{Homogeneous 2nd Order D.E , With Constant Coeff.}$

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

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

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

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

$\textbf{Then Characterestic Eq :}\;\; \Rightarrow \;\;{\color{#c34632} {100m^{2}+240m+(196\pi^{2} +144)=0}}$

$m_{1,2}=\dfrac{-b \;\pm \sqrt{b^{2}-4ac}}{2a}=\dfrac{-240\;\pm \sqrt{57600-(78400\pi^{2} +57600)}}{200}$

$m_{1,2}=\dfrac{-240\;\pm 280\pi i}{200}=\dfrac{-240-280\pi i}{200}\;\;,\;\; \dfrac{-240+280\pi i}{200}$

$m_{1,2}=\dfrac{-6}{5}\pm \dfrac{7}{5}\pi i\;\;\;\; \Rightarrow \;\; \boxed{\;m_{1,2}=\alpha \pm \beta i\;}$

$\boxed{\;\alpha =\dfrac{-6}{5}\;}\;\;\;.\;\;\; \boxed{\;\beta =\dfrac{7}{5}\pi \;}$

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

$\therefore \boxed{\;\color{#4257b2} {y=e^{\dfrac{-6}{5} x}(C_{1}\cos(\dfrac{7}{5}\pi x)+C_{2}\sin(\dfrac{7}{5}\pi x))}\;}$

The characteristic equation is

$100\lambda ^2 + 240\lambda + 196\pi ^2 +144= 0.$

Let's find the roots of the characteristic equation.

$\begin{align*} 100\lambda ^2 + 240\lambda + 196\pi ^2 +144= 0 &\iff \lambda _{1/2} = \frac{-240 \pm \sqrt{240^2 -400(196\pi ^2 +144)}}{2 \cdot 100} \\ & \iff \lambda _{1/2} = \frac{-240 \pm 4\sqrt{60^2 -100(49\pi ^2 +36)}}{200}\\ & \iff \lambda _{1/2} = \frac{-60 \pm 10 \sqrt{ \cancel{36} - 49 \pi ^2 - \cancel{36} }}{50}\\ & \iff \lambda _{1/2} = \frac{-6 \pm \sqrt{ -49 \pi ^2}}{5}\\ & \iff \lambda _1 =\frac{-6 + 7\pi i}{5} \quad \lambda _2 = \frac{-6 - 7\pi i}{5} \end{align*}$

So, it has the complex conjugate roots:

$\lambda _1 = -1.2 + 1.4\pi i \; (=a + ib)\quad \wedge \quad \lambda _2 = -1.2 - 1.4 \pi i \; (=a-ib)$

A basis is:

$\begin{align*} \begin{cases} & \phi _1 (x) = e^{ax} \cos (bx) = e^{-1.2x} \cos (1.4 \pi x) \\ &\phi _2 (x) =e^{ax} \sin (bx) = e^{-1.2x}\sin (1.4 \pi x) \end{cases} \end{align*}$

The general solution is:

$\begin{align*} y & = c_1 \phi _1 (x) + c_2 \phi _2 (x) \\ & = c_1 e^{-1.2x} \cos (1.4 \pi x)+ c_2e^{-1.2x}\sin (1.4 \pi x)\\ & = \boxed{{\color{#4257b2}{ e^{-1.2x} [c_1 \cos (1.4 \pi x)+ c_2\sin (1.4 \pi x)]}}} \end{align*}$

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