- #1
amb123
- 98
- 0
I am trying to solve a linear 1st order diff eq. I have found the integrating factor, and multiplied it through the equation, now expecting it to be exact. However, I can't get it into an easy to solve form.
After multiplying by the integrating factor I have :
(e^-3x) dy/dx - (3e^-3x) y = 6e^-3x I can't figure out how to get this into form. The schaum's book puts this in the form d/dx [ye^-3x)] = 6e^-3x. I see that if you use the product rule on ye^-3x you get the original, but I'm not sure how you figure out how to get the original to the second form. Is there a trick? Can my TI89 do it? Any help is greatly appreciated. I have a test tomorrow and I'm not doing terribly well. I understand the problem before and after this part, but not this transition.
thx.
-A
After multiplying by the integrating factor I have :
(e^-3x) dy/dx - (3e^-3x) y = 6e^-3x I can't figure out how to get this into form. The schaum's book puts this in the form d/dx [ye^-3x)] = 6e^-3x. I see that if you use the product rule on ye^-3x you get the original, but I'm not sure how you figure out how to get the original to the second form. Is there a trick? Can my TI89 do it? Any help is greatly appreciated. I have a test tomorrow and I'm not doing terribly well. I understand the problem before and after this part, but not this transition.
thx.
-A