Solving ode with complex numbers

cragar
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I want to solve y''+y'+y=(sin(x))^2 and try to use
y=Ae^{ix} but then when I square it I get A^2 e^{2ix}
I found y' and y'' and solved for A and it didn't work I guess I could use the formula for reducing powers but I would like to try and get around that.
 
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hint
$$(D^3+4D)(\sin(x))^2=0$$
or
(sin(x))^2 is a solution of y'''+4y

chose the particular solution from those solutions
 
Thread 'Direction Fields and Isoclines'

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...
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