D.E. Reduction of Order: Can't integrate

[Jeff12341234]

see image

http://i.imgur.com/OCj7Q5f.jpg

 

[rock.freak667]

For your shortcut formula to work, what form must your DE be in i.e. how would you need to rewrite it as?

 

[Jeff12341234]

oh yea. I need to divide everything by 6x^2. Thanks.

 

[rock.freak667]

oh yea. I need to divide everything by 6. Thanks.

You need to not only divide by 6, but by x² as well

 

[Jeff12341234]

it still won't integrate..

 

[rock.freak667]

By virtue of one solution being y = x^(1/2) you can say that x > 0.

So that |x| = x

If you remember how |x| is defined:

|x| = -x for x<0
|x| = x for x > 0

 

[Jeff12341234]

So going on that concept, are these two answers correct?

 

[rock.freak667]

For 9: The actual integration is correct, however, your formula is wrong as you are multiplying by √2 instead of √x (y₁).

Number 10 should be correct. Not quite sure how you deduced the integral was equal to x²sin(lnx) but it is correct.

 

[Jeff12341234]

Thanks. I fixed #9. For #10 I just used the Ti-nSpire CAS.

 

[rock.freak667]

Thanks. I fixed #9. For #10 I just used the Ti-nSpire CAS.

AH okay then. But for other integrals just make sure you can do them by hand if you ever need to.

Also when you are putting together your general solution, you can combine the constants so if you have y=c₁x+25c₂x², you can rewrite it as y=c₁x+c₃x².

 

[Jeff12341234]

noted. thanks

 

[Jeff12341234]

What is the "largest interval of definition"? (0,inf)?

 

[rock.freak667]

What is the "largest interval of definition"? (0,inf)?

I believe that would be it as x > 0 would make the equation valid.

 

[Jeff12341234]

What are the steps you go through to answer that part of the question? You look in the denominator of every step to try to see if any x value would make the equation undefined? or do you just see if any x value would make the equation undefined for the final answer? or do you see if any x value would make the equation undefined for y1 and y2?

 

[rock.freak667]

What are the steps you go through to answer that part of the question? You look in the denominator of every step to try to see if any x value would make the equation undefined? or do you just see if any x value would make the equation undefined for the final answer? or do you see if any x value would make the equation undefined for y1 and y2?

Yes, you would check to see where the function exists. The function will not exist for x=0 as seen in the solution of y2 and not exist if x < 0 as seen in y1 and y2.

 

[Jeff12341234]

ok. So to be specific, you only look at the y1 and y2 when checking where the function exists, not the any of the work done to get the solutions.

 

[rock.freak667]

ok. So to be specific, you only look at the y1 and y2 when checking where the function exists, not the any of the work done to get the solutions.

Basically.