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?
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 $x^2+y^2=c$.