# Help with solving for a flow curve:

1,005

You cannot solve a second-order differential equation by separating variables as you did. You're supposed to know (or have been shown) that the general solution of $\dfrac{d^2x}{dt^2}+x=0$ is $x(t)=c_1 \cos t + c_2\sin t$.

An alternative solution fitting multivariable calculus is to note that $G(x,y)=(x,y)$ is orthogonal to the flow line at the point $(x,y)$. But $G(x,y)=\frac12\nabla(x^2+y^2)$, and the gradient is normal to level curves. So the flow lines are the level curves $x^2+y^2=c$.

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### Sidharth Ghoshal

Updated on December 15, 2020

• Sidharth Ghoshal almost 3 years

So I'm preparing for a final exam in multivariable and our textbook posed the following question:

find the flow lines of F(x,y) = (-y, x)

Which I can't seem to solve correctly.

We are told that a vector function: $g(t)$ is a flow line if: $g'(t) = F(g(t))$

I thus let: $g(t) = (x(t),y(t))$

and then receive the system of differential equations:

$$\frac{dx}{dt} = -y$$ $$\frac{dy}{dt} = x$$

I observe:

$$\frac{dx}{dt} = -y \rightarrow -\frac{dx}{dt} = y \rightarrow - \frac{d^2x}{dt^2} = \frac{dy}{dt} = x$$

Now I solve the equation:

$$\frac{d^2x}{dt^2} = -x$$ $$-\frac{1}{x} d^2 x = 1 dt$$ $$-\ln(x) dx = [t + C_1]dt$$ $$x - x\ln(x) = \frac{1}{2}t^2 + C_1t + C_2$$ $$\ln(\frac{e^x}{x^x}) = \frac{1}{2}t^2 + C_1t + C_2$$ $$\frac{e^x}{x^x} = e^{\frac{1}{2}t^2 + C_1t + C_2}$$ $$(\frac{e}{x})^x = e^{\frac{1}{2}t^2 + C_1t + C_2}$$

let $u = \frac{x}{e}$ $$(\frac{1}{u})^{eu} = e^{\frac{1}{2}t^2 + C_1t + C_2}$$ $$u^{-u} = e^{\frac{1}{2e}t^2 + C_1t + C_2}$$ $$u^{u} = e^{-\frac{1}{2e}t^2 + C_1t + C_2}$$ $$x = e*(e^{-\frac{1}{2e}t^2 + C_1t + C_2})_{1/2}$$

where $s_{b} = s^{s^{s^{...}}}$ "b times" (tetration)

But the book claims the x component is: $C_1\cos(t) - C_2\sin(t)$

What exactly am I missing here?

• Sidharth Ghoshal almost 10 years
why is that true? I mean I could prove it now that you have given it but how did you arrive at it?
• Ted Shifrin almost 10 years
Well, one way is to assume a solution of the form $x(t)=e^{kt}$ (this works generally for any constant-coefficient linear differential equation). Substituting, you find $k^2+1=0$, so $k=\pm i$. But $e^{\pm it}=\cos t \pm i\sin t$. More to learn :)
• Stephen Montgomery-Smith almost 10 years
The equation $x''(t) + x(t) = 0$ is an example of a constant coefficient second order differential equation. This should be covered by any first course on differential equations.
• Stephen Montgomery-Smith almost 10 years
Also, most differential equations are solved by guessing what the solution is. There is no systematic way to solve ODE. And a first course on ODE should cover the more well known examples of ODEs for which successful guesses have been made.
• Sidharth Ghoshal almost 10 years
@Ted Shifrin, what exactly whas the intuition behind using $e^kt$ for your substitution?
• Ted Shifrin almost 10 years
Because exponentials are the functions whose derivatives are constant multiples of the original function. This will make more sense as you study differential equations and linear algebra.
• Sidharth Ghoshal almost 10 years
Can I ask you a follow up question? Is there some way to simplify my fractional tetration into complex exponentials? Like a formula or Identity of some sort?
• Ted Shifrin almost 10 years
I don't know. You might want to post that specifically as a separate question.