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

Question
Differential equations
Make and solve the given equation $$x′′ − x′ = 6 + e^{2}t$$

2020-11-09
Given: $$x′′ − x′ = 6 + e^{2}t$$
Recall: The general solution to $$a(x)y′′ + b(x)y′ + c(x)y = g(x)$$ is given by
$$y = y_{h} + y_{p},$$
where y_{h}, the complimentary function is the solution to the homogenens
system $$a(x)y′′ + b(x)y′ + c(x)y = 0$$
and $$y_{p}$$, the particular solution, is any
function that satisfies the non-homogeneous equation.
Here it is given that $$y” - y’ = 6 + e^{2}t.$$ Now for the complimentary function,
put $$y = e^{m}x$$ in the homogeneous form
$$y” - y’ = 0,$$ which is given by
$$m^{2} - m = 0$$. Therefore
$$y_{n} = a + be^{x},$$
where a, b are arbitrary constant.
Now leeting $$D = \frac{d}{dx}$$ we have
$$y_{p} =\frac{1}{D^{2}-D}(6+e^{2}t)$$
$$=\frac{1}{D^{2}-D}6+\frac{1}{D^{2}-D}e^{2}t$$
$$=6\frac{1}{(-D)}(1-D)^{-1}+e^{2}t\frac{1}{2^{2}-2}$$
$$=6\frac{1}{(-D)}(1+D+D^{2}+\cdots)+e^{2}t \frac{1}{2}$$
$$=6\frac{1}{(-D)}+\frac{e^{2}t}{2}$$
$$=-6t+\frac{e^{2}t}{2}$$
Therefore the general solution is given by
$$y = a + be^{x} + -6t + \frac{e^{2}t}{2}$$
where a, b are arbitrary constant.

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