# Solve the following linear second order homogeneous differential

Solve the following linear second order homogeneous differential equation

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Gia Edwards
Solution:

The characteristic equation is given by
${\gamma }^{2}-\frac{3}{2}\gamma +\frac{9}{16}=0$
$⇒16{\gamma }^{2}-24\gamma +9=0$
Hence
$\gamma =\frac{24±\sqrt{{24}^{2}-4\cdot 16\cdot 9}}{2\cdot 16}$
$=\frac{24±\sqrt{576-576}}{32}$
$=\frac{24}{32}=\frac{3}{4}$
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
$={c}_{1}{e}^{\frac{3}{4}\cdot t}+{c}_{2}t{e}^{\frac{3}{4}\cdot t}$
$={e}^{\frac{3}{4}\cdot t}\left({c}_{1}+{c}_{2}\cdot t\right)$
given, $y\left(0\right)=1$, so

${y}^{\prime }\left(t\right)=\frac{3}{4}{e}^{\frac{3}{4}t}\left({c}_{1}+{c}_{2}t\right)+{e}^{\frac{3}{4}t}{c}_{2}$
${y}^{\prime }\left(0\right)=\frac{3}{4}{c}_{1}+{c}_{2}=-1$
${c}_{2}=-1-\frac{3}{4}{c}_{1}$
$=-1-\frac{3}{4}=-\frac{7}{4}$
Hence
$y\left(t\right)={e}^{\frac{3}{4}t}-\frac{7}{4}t{e}^{\frac{3}{4}t}$
$={e}^{\frac{3}{4}}\left(1-\frac{7}{4}t\right)$
Jeffrey Jordon