Solving ordinary differential equations

Diogo Freitas

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Jul 5, 2021
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Recently I decided to embrace my self in the world of differential equations and I'm trying to follow a Book that recently bought. But this exercise has nothing to do with the exemples provided by the book and previous exemples (normally the exercise was resolver by puttin x' ans x isolated and then integrating, but I can't se how I could do that here)

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Recently I decided to embrace my self in the world of differential equations and I'm trying to follow a Book that recently bought. But this exercise has nothing to do with the exemples provided by the book and previous exemples (normally the exercise was resolver by puttin x' ans x isolated and then integrating, but I can't se how I could do that here)

View attachment 28103
Can you solve the following:

\(\displaystyle \frac{1}{1+t^2} * x'(t) - \frac{2*t}{(1+t^2)^2} * x(t) \ = \frac{t^2 - 1}{t^2+1} \)

continue.......
 
Hello, Diogo. Let me help you a little bit.

I guess you want to use the separation of variables method, but in this equation, that's not possible. What I would do is to write the equation in the following form:

[MATH]x'(t)+p(t)x(t)=q(t)[/MATH]
That is called a linear differential equation, and there are many known methods to arrive at their solution. Let me show you one of them:

First, we need to find an integrator factor [MATH]\mu (t)[/MATH] so we can integrate the left side of the equation. This factor is given by

[MATH]\mu (t)=e^{\int p(t)dt}[/MATH]
Then, you multiply every term of the equation by [MATH]\mu (t)[/MATH] and you will get to

[MATH]e^{\int p(t)dt}x'(t)+e^{\int p(t)dt}p(t)x(t)=e^{\int p(t)dt}q(t)[/MATH]
And this may look pretty scary, but believe me, it's not.
The following step is to write the left side of the equation in an integrated form:

[MATH]\frac{\mathrm{d} }{\mathrm{d} x}(e^{\int p(t)dt}x(t))=e^{\int p(t)dt}q(t)[/MATH]
[MATH]e^{\int p(t)dt}x(t)=\int e^{\int p(t)dt}q(t)[/MATH]
And the general solution is:

[MATH]x(t)=e^{-\int p(t)dt}\int e^{\int p(t)dt}q(t)dt[/MATH]
My advice: don't use the general formula for the solution. Follow every step. I'll give you a hint:

[MATH]\mu (t)=\frac{1}{1+t^2}[/MATH]
I hope it could help you :)
 
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