Help with first order differential equation

[bumpjump]

Help with the integration steps of a first order differential equation

I'll try to be as thorough as possible

The problem:

(2x^4 + x^3 - 2x^2 + 2x - 12)dy/dx = (2x^5 - 3x^4 + 3x^3 + 10x^2 - 17x - 22)y*Sqrt[ln^2(y) + ln(y) + 1]

I separated and got all the x's and y's on one side:

(1/y*Sqrt[ln^2(y) + ln(y) + 1])dy = ((2x^5 - 3x^4 + 3x^3 + 10x^2 - 17x - 22)/(2x^4 + x^3 - 2x^2 + 2x - 12))dx

Then I tried to integrate both sides:

Int[(1/y)(1/Sqrt[ln^2(y) + ln(y) + 1])dy] = Int[((2x^5 - 3x^4 + 3x^3 + 10x^2 - 17x - 22)/(2x^4 + x^3 - 2x^2 + 2x - 12))dx]

I made a u-substitution for the left side and got:

u = ln(y)
du = (1/y)dy

------> Int[(1/Sqrt[u^2 + u + 1])du

On the right side, I used polynomial division to get:

((2x^5 - 3x^4 + 3x^3 + 10x^2 - 17x - 22)/(2x^4 + x^3 - 2x^2 + 2x - 12)) = (x-2) ((3x^3 - 4x^2 - x + 2)/(2x^4 + x^3 - 2x^2 + 2x - 12))

I then factored the denominator and got:

((2x^5 - 3x^4 + 3x^3 + 10x^2 - 17x - 22)/(2x^4 + x^3 - 2x^2 + 2x - 12)) = (x-2) ((3x^3 - 4x^2 - x + 2)/((x^3(2x + 1)) - (2(x + 3)(x - 2)))

Where do I go from here? I'm not sure how to integrate the left side of the equation -- I looked for trig substitutions but couldn't find anything that fit the form

dx/(x^2 + x + 1)

and I don't know how to break down the right side of the equation now into partial fractions.

Thanks for your help.

 

[daon2]

The rational function you are trying to integrate is not pretty. Are you sure it has been copied correctly?

As for the y-integral:

\(\displaystyle \displaystyle \int \dfrac{1}{\sqrt{u^2+u+1}} du =\int \dfrac{1}{\sqrt{\left(u+\frac{1}{2}\right)^2+\frac{3}{4}}} du = \int \dfrac{1}{\sqrt{v^2+a^2}} dv \)

where

\(\displaystyle v= u+\frac{1}{2}, a=\frac{\sqrt{3}}{2}\)

 

[bumpjump]

I forgot to add a +2x term in the denominator of the x-integral but I've added that now. My DE professor is notorious for writing complex problems

I used your method for the y-integral and it ended up substituting neatly into sinh^-1 (v/a). Any ideas on the x-integral?

Thanks so much for your help.

 

[daon2]

I forgot to add a +2x term in the denominator of the x-integral but I've added that now. My DE professor is notorious for writing complex problems

I used your method for the y-integral and it ended up substituting neatly into sinh^-1 (v/a). Any ideas on the x-integral?

Thanks so much for your help.

Edit: I apologize, I did not see the h after the sin! You are correct about that.

The other integral is now relatively straight-forward. The denominator factors now: \(\displaystyle (2+x) (-3+2 x) (2+x^2)\)

After long-division, you can use partial fractions to obtain (I used a CAS):

\(\displaystyle -2+x+\dfrac{1}{3-2 x}+\dfrac{2}{2+x}+\dfrac{2 x}{2+x^2} + \dfrac{5}{2+x^2}\)

 

[bumpjump]

So adding that last step, I now have this equation:

sinh^-1((2ln(y) + 1)/Sqrt[3]) = ((x² - 4x)/2) + ((1/2)ln(3 - 2x) + (2ln(x + 2)) + ln(x² + 2) + ((5tan^-1(x/Sqrt[2]))/Sqrt[2] + K

where K =C₁ + C₂ + C₃ + C₄ + C₅

How do I isolate y now? Should I take the sinh of both sides to get rid of sinh^-1 on the left? Also how do I make posts with proper math notation?

Thanks for your help!

 

[daon2]

Yes about isolating y. The formatting used on this (and many other) sites as well as most professional mathematics papers is a language called LaTeX. It takes a bit of time to learn, I suggest quoting posts on this site that have used it to learn some. You may practice yourself using the "preview" function when making a post.

For example, typing

[tex]\frac{a^2+b^2}{2} \ge \sqrt{a^2b^2}=|ab| \iff 0 \le a^2+b^2-2|ab|[/tex]

[tex]\frac{d}{dx}\int_{a}^{x}f(t) dt = f(x)[/tex]

Results in

\(\displaystyle \frac{a^2+b^2}{2} \ge \sqrt{a^2b^2}=|ab| \iff 0 \le a^2+b^2-2|ab|\)

\(\displaystyle \frac{d}{dx}\int_{a}^{x}f(t) dt = f(x)\)