Find a general solution. Check your answer by substitution. 10y''-32y'+25.6y=0

Solutions

Verified${\color{#c34632} {10y''-32y'+25.6y=0}}\;\;\;\; \Rightarrow \;\; \textbf{Homogeneous 2nd Order D.E , With Constant Coeff.}$

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

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

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

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

$\textbf{Then Characterestic Eq :}\;\; \Rightarrow \;\;{\color{#c34632} {10m^{2}-32m+25.6=0}}$

$m_{1,2}=\dfrac{-b \;\pm \sqrt{b^{2}-4ac}}{2a}=\dfrac{32\;\pm \sqrt{1024-1024}}{20}$

$m_{1,2}=\dfrac{32\;\pm 0}{20}=1.6\;\;\;,\;\;\; \Rightarrow \;\; \boxed{\;m=1.6\;}$

$\therefore \;\; \textbf{Solution is : }\; \boxed{\color{#c34632} {y=C_{1}e^{m x}+C_{2}xe^{m x}}}\;\;\; \Rightarrow \;\; \textbf{For Repeated Roots}$

$\therefore \boxed{\;\color{#4257b2} {y=C_{1}e^{1.6x}+C_{2}xe^{1.6x}}\;}$

The characteristic equation is

$10\lambda ^2 - 32\lambda + 25.6 = 0.$

Let's find the roots of the characteristic equation.

$\begin{align*} 10\lambda ^2 - 32\lambda + 25.6 = 0 &\iff \lambda _{1/2} = \frac{32 \pm \sqrt{32^2 -4 \cdot 256}}{20} \\ & \iff \lambda _{1/2} =\frac{32 \pm \sqrt{1024 - 1024}}{20} \\ & \iff \lambda _{1/2} = \frac{8}{5} = 1.6 \end{align*}$

So, it has the real double root:

$\lambda = 1.6$

A basis is:

$\begin{align*} \begin{cases} & \phi _1 (x) = e^{\lambda x} = e^{1.6 x} \\ & \phi _2 (x)= xe^{\lambda x} = xe^{1.6 x} \end{cases} \end{align*}$

The general solution is:

$\begin{align*} y &= c_1 \phi _1 (x) + c_2 \phi _2 (x) \\ &= \boxed{{\color{#4257b2}{ c_1e^{1.6 x} + c_2xe^{1.6 x} }}} \end{align*}$

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