Thread: Linear , homogenous, separation and change of variables

(3,-1) is the intersection between two lines right?



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Originally Posted by romeobernard

(3,-1) is the intersection between two lines right?

You could look at it that way, we really just needed to solution to the system of linear equations in order to make a good substitution.

dv/du = {u+(2v)} / {(2u)+v}

Next? I cant follow those terms..thanks

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I went further than that and wrote it in first order homogeneous form, which has a well-known theory for solving. You are familiar with such equations, right?

Originally Posted by MarkFL

I went further than that and wrote it in first order homogeneous form, which has a well-known theory for solving. You are familiar with such equations, right?

Is this right
V^2(z-1)^3 = c(z+1)

Turns to this

(vz-v)^3 = c(vz+v)

How?thanks

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Originally Posted by romeobernard

Is this right
V^2(z-1)^3 = c(z+1)

Turns to this

(vz-v)^3 = c(vz+v)

How?thanks

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I saw this problem on the book.

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Well, what I would do is let:

$\displaystyle w=\frac{v}{u}\implies v=uw\implies \frac{dv}{du}=w+u\frac{dw}{du}$

And so our ODE becomes:

$\displaystyle w+u\frac{dw}{du}=\frac{1+2w}{2+w}$

$\displaystyle u\frac{dw}{du}=\frac{1+2w-w(2+w)}{2+w}$

$\displaystyle u\frac{dw}{du}=\frac{1-w^2}{2+w}$

Can you proceed?

After those equation (u+2v)du - (2u+v)dv i used the homogeneous DE

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