# Thread: Linear , homogenous, separation and change of variables

1. ## Linear , homogenous, separation and change of variables

Hi can you help me with this problem i dont know where do I start solving this one. Thanks

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2. ## Re: Linear , homogenous, separation and change of variables

I would let:

$\displaystyle u=2x-y+3\implies du=2\,dx-dy$

Can you now transform the given ODE into a linear equation?

3. ## Re: Linear , homogenous, separation and change of variables

y+c = -ln|2x+y+4|

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4. ## Re: Linear , homogenous, separation and change of variables

Well, I'm trying to help you derive the solution. And are you sure the solution given isn't:

$\displaystyle y+C=-\ln|2x-y+4|$ ?

5. ## Re: Linear , homogenous, separation and change of variables

But how to come up to a positive 4?

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6. ## Re: Linear , homogenous, separation and change of variables

This is what i got sir.

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7. ## Re: Linear , homogenous, separation and change of variables

Originally Posted by romeobernard

But how to come up to a positive 4?
You will see why when you make the substitution and solve the resulting linear equation. Can you make the substitution I suggested and write the resulting ODE in standard linear form?

8. ## Re: Linear , homogenous, separation and change of variables

Originally Posted by romeobernard
This is what i got sir.

When you found $\displaystyle du$ you made an error...check my first post to see what I wrote.

9. ## Re: Linear , homogenous, separation and change of variables

Ok mark now i get it. You will get 1 from dividing dy/dy.

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10. ## Re: Linear , homogenous, separation and change of variables

Originally Posted by romeobernard
Ok mark now i get it. You will get 1 from dividing dy/dy.
What I wound up with is:

$\displaystyle \frac{du}{dy}+u=-1$

I solved it as a linear equation and got the implicit solution, after back-substitution:

$\displaystyle 2x-y+4=c_1e^{-y}$

This is equivalent to the solution you cited from your book.

11. ## Re: Linear , homogenous, separation and change of variables

This is what i get mark base on youre substitutions.. correct me if im wrong thanks.

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12. ## Re: Linear , homogenous, separation and change of variables

This what i get.

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13. ## Re: Linear , homogenous, separation and change of variables

Yes, that looks good.

You chose to separate variables, which is fine.

14. ## Re: Linear , homogenous, separation and change of variables

And how about this one.? Its a linear stated in the book.can help me how to figure out this one. Thanks

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15. ## Re: Linear , homogenous, separation and change of variables

That pic is sideways for me, but I will try my best.

I would write in standard linear form as:

$\displaystyle \frac{dr}{d\theta}+2\cot(\theta)r=-\csc(\theta)$

Next, we compute the integrating factor:

$\displaystyle \mu(\theta)=\exp\left(2\int \cot(\theta)\,d\theta\right)=\sin^2(\theta)$

And the ODE becomes:

$\displaystyle \sin^2(\theta)\frac{dr}{d\theta}+2\sin(\theta)\cos (\theta)r=-\sin(\theta)$

Or:

$\displaystyle \frac{d}{d\theta}\left(\sin^2(\theta)r\right)=-\sin(\theta)$

Can you proceed?

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