Separable first order ODE involving tangent

  • #1
psie
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Homework Statement
Consider the DE ##\frac{\mathrm{d}x}{\mathrm{d}t}=t\tan x## with initial value ##x(0)=\frac\pi{6}##.
1. Find the solution.
2. Describe the region in which the solutions are defined.
Relevant Equations
A separable first order ODE is of the form ##x'=g(x)h(t)##.
By inspection, we see that ##x=k\pi## is a solution for ##k\in\mathbb Z##. Moreover, the equation implicitly assumes ##x\neq n\pi/2## for odd ##n\in\mathbb Z##, since ##\tan x## isn't defined there. So suppose ##x\neq k\pi##, i.e. ##\tan x\neq 0##, then rearranging and writing ##\tan x=\frac{\sin x}{\cos x}## we have $$\frac{x'\cos x}{\sin x}=\frac{\mathrm{d}}{\mathrm{d}t}\log(|\sin x|)=t.$$ Integrating and exponentiating, we obtain, $$|\sin x|=Ae^{\frac{t^2}{2}}$$ The initial condition implies ##A=1/2##. Now, how do I get rid of the absolute values here? Does it somehow follow from the initial condition that ##\sin x## has to be positive?
 
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  • #2
[tex]\sin x = \pm A e^{t^2/2}[/tex]
where A >0. There are two cases of x>0 and x<0 during the time after. The initial condition says ##x(0)=\pi/6 >0 ## Our solution is the former one.
[tex]\sin x = \frac{e^{t^2/2}}{2}[/tex]
 
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  • #3
I think from [itex](\log |\sin x|)' = t[/itex] the next step is [tex]
\log |\sin x| = \log |A| + \tfrac12 t^2[/tex] and hence [tex]
|\sin x| = |A|e^{\frac12 t^2}.[/tex] It follows from this that [itex]\sin x[/itex] and [itex]A[/itex] have the same sign, so we can drop the absolute value signs.
 
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FAQ: Separable first order ODE involving tangent

What is a separable first-order ODE?

A separable first-order ordinary differential equation (ODE) is an equation that can be expressed in the form dy/dx = g(x)h(y), where the variables x and y can be separated on opposite sides of the equation, allowing it to be written as (1/h(y)) dy = g(x) dx.

How do you solve a separable first-order ODE involving the tangent function?

To solve a separable first-order ODE involving the tangent function, you first separate the variables. For example, if you have dy/dx = tan(x) * y, you can rewrite it as (1/y) dy = tan(x) dx. Then, integrate both sides: ∫(1/y) dy = ∫tan(x) dx. This gives ln|y| = -ln|cos(x)| + C, where C is the constant of integration. Finally, solve for y to get y = C * sec(x).

What are the common mistakes to avoid when solving separable first-order ODEs?

Common mistakes include not properly separating the variables, neglecting the constant of integration, and errors in the integration process. Additionally, one should be cautious about the domains of the functions involved, such as the tangent function, which has vertical asymptotes where it is undefined.

Can you provide an example of a separable first-order ODE involving the tangent function and solve it?

Sure! Consider the ODE dy/dx = y * tan(x). Separate the variables to get (1/y) dy = tan(x) dx. Integrate both sides: ∫(1/y) dy = ∫tan(x) dx. This yields ln|y| = -ln|cos(x)| + C. Exponentiating both sides, we get |y| = e^C * sec(x). Letting e^C = K, where K is a new constant, we obtain y = K * sec(x).

What are the applications of separable first-order ODEs involving the tangent function?

Separable first-order ODEs involving the tangent function can be found in various applications, including physics (e.g., modeling the motion of a pendulum for small angles), engineering (e.g., analyzing certain electrical circuits), and biology (e.g., modeling population dynamics with specific growth rates). They are useful in any context where the rate of change of a quantity can be expressed as a product of functions of two different variables.

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