Engineering First order differential equations (movement of a rotary solenoid)

[Ben_Walker1978]

Homework Statement

The movement of a rotary solenoid is given by the following differential equation

Relevant Equations

$${5} \frac{\text{d}\theta}{\text{d}t}+{6}\theta=0$$

My question i am trying to solve:

I have successfully done first order equations before but this one has got me a little stuck. My attempt at the general solution below:

$${5} \frac{\text{d}\theta}{\text{d}t}=-6\theta$$

$${5} \frac{\text{d}\theta}{\text{d}t} =\frac{\text{-6}\theta}{5}$$

$$d\theta = \frac{-6\theta}{5} dt $$

The integral should then = $$\theta = \frac{1}{5} in(5)$$

$$\theta = -6 \times (\frac{1}{5}) in (5) + c$$

This is my attempt an the general solution. Is this correct? Or could anyone help if i have gone wrong please?

Thank you

 

[anuttarasammyak]

¥frac{d¥theta}{¥theta}=-¥frac{6}{5} dt
You can integrate the both sides.

Edit to fix broken LaTeX:
$\frac{d \theta} \theta = -\frac 6 5 dt$

 

[Mark44]

Homework Statement:: The movement of a rotary solenoid is given by the following differential equation
Relevant Equations:: $${5} \frac{\text{d}\theta}{\text{d}t}+{6}\theta=0$$

My question i am trying to solve:

View attachment 300338

I have successfully done first order equations before but this one has got me a little stuck. My attempt at the general solution below:

$${5} \frac{\text{d}\theta}{\text{d}t}=-6\theta$$

$${5} \frac{\text{d}\theta}{\text{d}t} =\frac{\text{-6}\theta}{5}$$

$$d\theta = \frac{-6\theta}{5} dt $$

The integral should then = $$\theta = \frac{1}{5} in(5)$$

$$\theta = -6 \times (\frac{1}{5}) in (5) + c$$

There is no "in" function. If you're reading this off your calculator it is ln (lowercase ell n), short for logarithmus naturalis, or in English, natural logarithm.

This is my attempt an the general solution. Is this correct? Or could anyone help if i have gone wrong please?

No, your solution is not correct.
Going from $\frac{d\theta} \theta = -6/5 dt$, you should get $\ln|\theta| = -6/5t + C$
How do then solve for $\theta$ in that equation?

Thank you

 

[Babadag]

If θo it is θ(t) for t=0 then θ(t)=θo*e^(-6/5*t). How to get it?​

 

[MidgetDwarf]

The solution to these type of differential equations is called separation of variables. Where the idea is, to separate the infinitesimal dx and dy on opposite sides of the equal signs (if possible). Then integrate with respect to each infinitesimal on each side. Do not forget your constant of integration on the right.