Solve the following linear second order homogeneous differential equation

$y{}^{\u2033}-\frac{3}{2}{y}^{\prime}+\frac{9}{16}y=0,$ $y\left(0\right)=1,y\text{'}\left(0\right)=-1$

Blackettyl2j
2022-03-25
Answered

Solve the following linear second order homogeneous differential equation

$y{}^{\u2033}-\frac{3}{2}{y}^{\prime}+\frac{9}{16}y=0,$ $y\left(0\right)=1,y\text{'}\left(0\right)=-1$

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Gia Edwards

Answered 2022-03-26
Author has **12** answers

Solution:

$y{}^{\u2033}-\frac{3}{2}{y}^{\prime}+\frac{9}{16}y=0,\text{}y\left(0\right),\text{}{y}^{\prime}\left(0\right)=-1$

The characteristic equation is given by

${\gamma}^{2}-\frac{3}{2}\gamma +\frac{9}{16}=0$

$\Rightarrow 16{\gamma}^{2}-24\gamma +9=0$

Hence

$\gamma =\frac{24\pm \sqrt{{24}^{2}-4\cdot 16\cdot 9}}{2\cdot 16}$

$=\frac{24\pm \sqrt{576-576}}{32}$

$=\frac{24}{32}=\frac{3}{4}$

Since the root are repeated therefore the solution are

$y}_{1}\left(t\right)={e}^{\gamma \cdot t$ and $y}_{2}\left(t\right)=t{e}^{\gamma \cdot t$

$={e}^{\frac{3}{4}}$ and $y}_{2}\left(t\right)=t{e}^{\frac{3}{4}\cdot t$

Let$y}_{c$ be the complimentary solt

Then

$y}_{c}\left(t\right)={c}_{1}{y}_{1}+{c}_{2}{y}_{2$ , where $c}_{1$ and $c}_{2$ constant

$={c}_{1}{e}^{\frac{3}{4}\cdot t}+{c}_{2}t{e}^{\frac{3}{4}\cdot t}$

$={e}^{\frac{3}{4}\cdot t}({c}_{1}+{c}_{2}\cdot t)$

given,$y\left(0\right)=1$ , so

$y\left(0\right)=1\text{}{c}_{1}=1$

$y}^{\prime}\left(t\right)=\frac{3}{4}{e}^{\frac{3}{4}t}({c}_{1}+{c}_{2}t)+{e}^{\frac{3}{4}t}{c}_{2$

${y}^{\prime}\left(0\right)=\frac{3}{4}{c}_{1}+{c}_{2}=-1$

$c}_{2}=-1-\frac{3}{4}{c}_{1$

$=-1-\frac{3}{4}=-\frac{7}{4}$

Hence

$y\left(t\right)={e}^{\frac{3}{4}t}-\frac{7}{4}t{e}^{\frac{3}{4}t}$

$={e}^{\frac{3}{4}}(1-\frac{7}{4}t)$

The characteristic equation is given by

Hence

Since the root are repeated therefore the solution are

Let

Then

given,

Hence

Jeffrey Jordon

Answered 2022-03-31
Author has **2262** answers

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