Moved - ODE

[Ritch]

I would rewrite the ODE as:

\(\displaystyle \dfrac{1}{(y')^2+1}\,d(y')=dx\)

Integrating, we get:

\(\displaystyle \tan^{-1}(y')=x+c_1\)

or:

\(\displaystyle y'=\tan(x+c_1)\)

Integrating again, we find:

\(\displaystyle y=-\ln|\cos(x+c_1)|+c_2=-\ln|c_2\cos(x+c_1)|\)

That was very clever to see. Thanks for pointing that out! So short and simple if you see it.


And thanks for the alternative solution but I'm affraid it's a bit above my head to continue with the sec() formula's( we haven't seen those methods). There probably is a conventional solution though since my course exercices are all at the a certain level but still supposed to be solved('easily') with the substitution method i was trying to follow.

I've solved all my other 15 exercices given on 2nd order DE's(which is great news for me), only one left that I can't get the hang of it.


y'' + 2*x * y'² = 0
becomes after substitution y' = p , y'' = p' since x is in the DE
p' + 2*x*p² = 0

(1/p²)*dp = -2*x*dx

int => 1/p = x² - C₁

[1/ (x² - C₁)]*dx = dy

int(using the ln formula)=> gives a to complicated formula and made me think that i'm going the wrong again.




The problem is that I don't know the final answer for this DE(could have given me a hint what direction to work to).

They give however the particular solution(values of x and y to find C1 and C2) given y(1) = 1 and y'(1) = 1.

Filling in these values in the general solution would have to give y = - 1/x +2.


Thanks for your efforts so far.

 

[MarkFL]

For this latest ODE, I would write it in the form:

\(\displaystyle -(y')^{-2}\,d(y')=2x\,dx\)

Integrating, we find:

\(\displaystyle \dfrac{1}{y'}=x^2+c_1\)

Using the initial value, we find:

\(\displaystyle 1=1+c_1\,\therefore\,c_1=0\)

and then inverting, we get:

\(\displaystyle y'=x^{-2}\)

Integrating again, we obtain:

\(\displaystyle y=-\dfrac{1}{x}+c_2\)

Using the initial value, we find:

\(\displaystyle 1=-1+c_2\,\therefore\,c_2=2\)

And so the solution satisfying all of the given conditions is:

\(\displaystyle y=-\dfrac{1}{x}+2\)

 

[Deleted member 4993]

That was very clever to see. Thanks for pointing that out! So short and simple if you see it.


And thanks for the alternative solution but I'm affraid it's a bit above my head to continue with the sec() formula's( we haven't seen those methods). There probably is a conventional solution though since my course exercices are all at the a certain level but still supposed to be solved('easily') with the substitution method i was trying to follow.

I've solved all my other 15 exercices given on 2nd order DE's(which is great news for me), only one left that I can't get the hang of it.


y'' + 2*x * y'² = 0
becomes after substitution y' = p , y'' = p' since x is in the DE
p' + 2*x*p² = 0

(1/p²)*dp = -2*x*dx

int => 1/p = x² - C₁

[1/ (x² - C₁)]*dx = dy

int(using the ln formula)=> gives a to complicated formula and made me think that i'm going the wrong again.

Use partial fraction:

\(\displaystyle \displaystyle \frac{1}{x^2 - C_1} \ = \ \frac{1}{2√{C_1}} * \left [\frac{1}{x - √{C_1}} \ - \ \frac{1}{x + √{C_1}}\right ]\)

However, this method will not work when C_1 = 0. So before doing anything, you need to evaluate C_1 = ??


The problem is that I don't know the final answer for this DE(could have given me a hint what direction to work to).

They give however the particular solution(values of x and y to find C1 and C2) given y(1) = 1 and y'(1) = 1.

Filling in these values in the general solution would have to give y = - 1/x +2.


Thanks for your efforts so far.

.

 

[Ritch]

For this latest ODE, I would write it in the form:

\(\displaystyle -(y')^{-2}\,d(y')=2x\,dx\)

Integrating, we find:

\(\displaystyle \dfrac{1}{y'}=x^2+c_1\)

Using the initial value, we find:

\(\displaystyle 1=1+c_1\,\therefore\,c_1=0\)

and then inverting, we get:

\(\displaystyle y'=x^{-2}\)

Integrating again, we obtain:

\(\displaystyle y=-\dfrac{1}{x}+c_2\)

Using the initial value, we find:

\(\displaystyle 1=-1+c_2\,\therefore\,c_2=2\)

And so the solution satisfying all of the given conditions is:

\(\displaystyle y=-\dfrac{1}{x}+2\)

Sorry for this late reply first of all.

That trick of rewriting the ODE is something i will have to see for myself somehow. I understand what happens, I just (again) didn't see it for myself.


I have done my exam and there wasn't really much of ODE's in there strangely enough(the other parts where series and Laplace where most questions were asked from)

I can only hope for the best now.

Thanks to both for your answers, it was a great help!