- #1
Lancelot59
- 646
- 1
Given [tex]y''-4y=2[/tex]
and that one solution is [tex]y_{1}=e^{-2x}[/tex]
I need to find a second solution of the homogeneous equation, and then a particular solution of the homogeneous equation.
Here's how I solved the homogeneous equation:
[tex]y=ue^{-2x}, y'=u'e^{-2x}-2ue^{-2x}, y''=u''e^{-2x}-2u'e^{-2x}+4ue^{-2x}-2u'e^{-2x}[/tex]
plugging into the equation:
[tex]u''e^{-2x}-2u'e^{-2x}+4ue^{-2x}-2u'e^{-2x}-4ue^{-2x}=0[/tex]
[tex]u''-4u=0[/tex]
Then using [tex]w=u', w'=u''[/tex]
[tex]w'=4w[/tex]
and I eventually got y to be [tex]y_{2}=\frac{e^{2x}/4}c[/tex]
The book says the answer is [tex]y=e^{2x}[/tex], so I'm not sure whether or not that factor of 1/4 should be there. Then for the second part of finding a particular solution, the answer is -1/2, but I'm not sure how that's arrived at either.
and that one solution is [tex]y_{1}=e^{-2x}[/tex]
I need to find a second solution of the homogeneous equation, and then a particular solution of the homogeneous equation.
Here's how I solved the homogeneous equation:
[tex]y=ue^{-2x}, y'=u'e^{-2x}-2ue^{-2x}, y''=u''e^{-2x}-2u'e^{-2x}+4ue^{-2x}-2u'e^{-2x}[/tex]
plugging into the equation:
[tex]u''e^{-2x}-2u'e^{-2x}+4ue^{-2x}-2u'e^{-2x}-4ue^{-2x}=0[/tex]
[tex]u''-4u=0[/tex]
Then using [tex]w=u', w'=u''[/tex]
[tex]w'=4w[/tex]
and I eventually got y to be [tex]y_{2}=\frac{e^{2x}/4}c[/tex]
The book says the answer is [tex]y=e^{2x}[/tex], so I'm not sure whether or not that factor of 1/4 should be there. Then for the second part of finding a particular solution, the answer is -1/2, but I'm not sure how that's arrived at either.