Differential equation question

Hurkyl
Staff Emeritus
Gold Member
P.S.

$$x(t)=k_1 e^t Cos[\sqrt{2}t]+k_2e^t Sin[\sqrt{2}t]-1/3$$
$$y(t)=k_1e^t\left(-\sqrt{2}Sin(\sqrt{2}t)\right)+k_2e^t\left(\sqrt{2} Cos(\sqrt{2}t)\right)+1/3$$

Can be simplified: remember your trigonometry! They're equivalent to:

$$x(t) = A e^t \sin(\sqrt{2} t + B) - 1/3$$
$$y(t) = A \sqrt{2} e^t \cos(\sqrt{2} t + B) + 1/3$$

where $A^2 = k_1^2 + k_2^2$ and $\tan B = k_1 / k_2$.

This gives a nice way to solve for t as -B + Log(f(x, y)) where f is a linear polynomial. You ought to be able to reduce it to a nonparametric equation, but I haven't worked out the details yet.

Last edited:
saltydog
Homework Helper
Ok, I'll look at your suggestions Hurkyl. Thanks.

However, continuing with the complex analysis, for:

$$v=\frac{i}{\alpha(\sqrt{2}-i)}-\frac{w}{\sqrt{2}(\alpha+1)}+\frac{K}{w^{\alpha}}$$

with:

$$v=\frac{x}{i},\quad w=i+(\sqrt{2}-i)u,\quad u=y-\frac{\sqrt{2}x}{i},\quad \alpha=\frac{\sqrt{2}+i}{\sqrt{2}-i}$$

I obtain:

\begin{align*} \frac{i}{\alpha(\sqrt{2}-i)} &= \frac{1}{3}+\frac{i\sqrt{2}}{3} \\ \\ -\frac{w}{\alpha(\sqrt{2}-i} &= \left(y-\frac{x}{2}-\frac{1}{2}\right)+ i\left(\frac{1}{\sqrt{8}}+\sqrt{2}x+\frac{y}{\sqrt{8}}\right) \\ \\ \frac{K}{w^{\alpha}} &= KExp\left(\frac{ln(r)}{3}+\frac{\sqrt{8}}{3}\theta\right)\left[Cos\left(\frac{\theta}{3}+\frac{\sqrt{8}}{3}ln(r)\right)-iSin\left(\frac{\theta}{3}+\frac{\sqrt{8}}{3}ln(r)\right)\right] \end{align}

With:

$$r=\sqrt{2(x+y)^2+(1+2x-y)^2}$$

$$\theta=ArcTan\left[\frac{1+2x-y}{x+y}\right]$$

It's very messy but can now be placed in a form:

$$f(x,y)+ig(x,y)=C$$

@@
my goodness~
that is one tough O.D.E!

saltydog
Homework Helper
asdf1 said:
@@
my goodness~
that is one tough O.D.E!

Asdf,

All that complex stuf, maybe not necessary and don't wish to give you the impression that the ODE was complex. It wasn't really. Hurkly and I just pursued a different approach which I just left at the point I did. However, the important part is that the solution is attainable via the substitutions suggested and/or the parametric route I chose. The parametric form provides a clean means of plotting the solution and is one I'd suggest using.

However, the other one you proposed up there allows for yet another approach which also can be solved parametrically via differential operators.

I'll make some postings.

Hurkyl
Staff Emeritus
Gold Member
Aha! I finally figured out the geometric picture I was searching for.

So, we have the equation:

$$\frac{dy}{dx} = \frac{L_1(x, y)}{L_2(x, y)}$$

where the L's are linear.

The L's are computing the distance from some fixed line. (Times some constant).

dy/dx is computing the slope of the tangent line at (x, y)... or, the tangent of the angle the tangent line makes with the horizontal.

So now we can get a coordinate free picture:

"The tangent of the angle between the tangent line at (x, y) with some fixed line is proportional to the ratio of the distances from (x, y) two these other two fixed lines"

And it suggests a good way to look at the problem: after shifting coordinates so that the two fixed lines intersect at the origin, you should convert to polar coordinates!

My initial stab at the algebra suggests that the computation is straightforward, although tedious and involving the complexes.

By the way, this is a good substitution for trig integrals:

sin θ = 2t / (1 + t²)
cos θ = (1 - t²) / (1 + t²)
dθ = (2 dt) / (1 + t²)

(I think I have dθ right)

Yes, you're right~
There's lots of different approaches to a question~
It's just that...
You guys are amazing!
I'd never of thought to come up with those ideas~
It makes me want to start jumping into mathematics books and brush up on my math~

Hurkyl
Staff Emeritus
Sigh, not all that enlightening. Everything except the "some constant" depend on the coefficients in the original problem.