# Make and solve the given equation x′′ − x′ = 6 + e^{2}t

Make and solve the given equation $x\prime \prime -x\prime =6+{e}^{2}t$
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

pattererX
Given: $x\prime \prime -x\prime =6+{e}^{2}t$
Recall: The general solution to $a\left(x\right)y\prime \prime +b\left(x\right)y\prime +c\left(x\right)y=g\left(x\right)$ is given by
$y={y}_{h}+{y}_{p},$
where y_{h}, the complimentary function is the solution to the homogenens
system $a\left(x\right)y\prime \prime +b\left(x\right)y\prime +c\left(x\right)y=0$
and ${y}_{p}$, the particular solution, is any
function that satisfies the non-homogeneous equation.
Here it is given that $y”-{y}^{\prime }=6+{e}^{2}t.$ Now for the complimentary function,
put $y={e}^{m}x$ in the homogeneous form
$y”-{y}^{\prime }=0,$ which is given by
${m}^{2}-m=0$. Therefore
${y}_{n}=a+b{e}^{x},$
where a, b are arbitrary constant.
Now leeting $D=\frac{d}{dx}$ we have
${y}_{p}=\frac{1}{{D}^{2}-D}\left(6+{e}^{2}t\right)$
$=\frac{1}{{D}^{2}-D}6+\frac{1}{{D}^{2}-D}{e}^{2}t$
$=6\frac{1}{\left(-D\right)}\left(1-D{\right)}^{-1}+{e}^{2}t\frac{1}{{2}^{2}-2}$
$=6\frac{1}{\left(-D\right)}\left(1+D+{D}^{2}+\cdots \right)+{e}^{2}t\frac{1}{2}$
$=6\frac{1}{\left(-D\right)}+\frac{{e}^{2}t}{2}$
$=-6t+\frac{{e}^{2}t}{2}$
Therefore the general solution is given by
$y=a+b{e}^{x}+-6t+\frac{{e}^{2}t}{2}$
where a, b are arbitrary constant.
Jeffrey Jordon