First order Ordinary Differential Equation

[PAR]

Homework Statement

Solve the differential equation:

$$y' = 8sin(4yt) ; y(1) = 4$$

Homework Equations

The Attempt at a Solution

The integrating doesn't apply because I can't get the equation into:
y' + p*y = f(x) form

Also, I have tried separating variables, but I can't get the y inside of the sin(4yt) outside of the sin(). Basically, I need some help to get started.

Thanks in advance

 

[Avodyne]

Try changing variables: y(t)=z(t)/t.

 

[PAR]

Correction! The differential equation should be y' = 8sin(4yt) instead of y'=8y$$^{3}$$sin(4yt)

I've tried substitution where y(t) = z(t)/t

This gave me:

$$y' = z'/t - z/t^{2} = 8sin(4z)$$

The problem now is that I still have a function of z on the right hand side of the equation, and I don't know how to combine it with the left hand side to give me an a nice integrating factor or separation of variables. I was thinking I could divide the whole thing by sin(4z) anyway, and then integrate with respect to t. The z' would be integrable and so would the 8, but the resulting -z/(t$$^{2}$$sin(4z)) I don't know how to integrate with respect to t. Great help so far, but could still use more :).

 

[Gregg]

$$\frac{dy}{dt} = 8\sin (4yt)$$

$$y=\frac{z}{t} \Rightarrow 8 \sin (4yt) = 8 \sin (4z) = \frac{dy}{dt}$$


$$y=\frac{z}{t}$$

$$\frac{dy}{dt}=-\frac{dz}{t^2 dt}$$

$$\frac{dz}{dt} = -8t^2\sin (4z)$$

$$\int \frac{dz}{\sin (4z)} = \int -8t^2 dt$$

Maybe that will work?

 

[PAR]

$$\frac{dy}{dt} = 8\sin (4yt)$$

$$y=\frac{z}{t} \Rightarrow 8 \sin (4yt) = 8 \sin (4z) = \frac{dy}{dt}$$


$$y=\frac{z}{t}$$

$$\frac{dy}{dt}=-\frac{dz}{t^2 dt}$$

Don't you have to apply the product rule when you differentiate y?

Since y = z/t = zt$$^{-1}$$ and z is a function of t by definition shouldn't

y' = z'/t - z/t$$^{-2}$$ instead of y' = $$-\frac{dz}{t^2 dt}$$

 

[Gregg]

Don't you have to apply the product rule when you differentiate y?

Since y = z/t = zt$$^{-1}$$ and z is a function of t by definition shouldn't

y' = z'/t - z/t$$^{-2}$$ instead of y' = $$-\frac{dz}{t^2 dt}$$

Oh yeah god, use the product rule :$.

 

[PAR]

Thank you Gregg and Avodyne for the help, but I don't have a solution to this problem. Is there a different way to solve first order ODE other than separation of variables and the integration factor? Please, I need some help.