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7.1 x'=x+sin(t) solve de

karush

Well-known member
Jan 31, 2012
2,886
Solve the differential equation
$x'=x+\sin(t)$
ok this uses x rather than y which threw me off
so rewrite as
$x'-x=\sin(t)$
thus $u(t)=e^{-t}$
$e^{-t}x'-e^{-t}x=e^{-t}\sin{t}$
and
$(e^{-t}x)'=e^{-t}\sin{t}$
intergrate thru
$\displaystyle e^{-t}x=\int {e^{-t} \sin(t)} dt = -1/2 e^{-t}\sin(t) - 1/2 e^{-t} \cos(t) +e^{-t} c$
divide thru
$\displaystyle x(t)=-\frac{\sin t}{2}-\frac{\sin t}{2}+ce^t$


ok well typos probably




W|A returned $\displaystyle x(t) = c_1 e^t - \frac{\sin(t)}{2} - \frac{cos(t)}{2}$
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Let's back up to:

\(\displaystyle \frac{d}{dt}\left(e^{-t}x\right)=e^{-t}\sin(t)\)

Now, when we integrate, we get:

\(\displaystyle e^{-t}x=-\frac{1}{2}e^{-t}\left(\sin(t)+\cos(t)\right)+c_1\)

Hence:

\(\displaystyle x(t)=c_1e^t-\frac{1}{2}\left(\sin(t)+\cos(t)\right)\)

This is equivalent to W|A returned. :)
 

karush

Well-known member
Jan 31, 2012
2,886
Isn't the 1/2 distributed?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775