MHB How to Solve the Differential Equation x'=x+sin(t)?

  • Thread starter Thread starter karush
  • Start date Start date
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
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:
Physics news on Phys.org
Let's back up to:

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

Now, when we integrate, we get:

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

Hence:

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

This is equivalent to W|A returned. :)
 
Isn't the 1/2 distributed?
 
karush said:
Isn't the 1/2 distributed?

You could if you want. I chose not to.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Replies
7
Views
3K
Replies
2
Views
2K
Replies
6
Views
1K
Replies
5
Views
3K
Replies
5
Views
3K
Replies
3
Views
2K
Back
Top