Solving A Non-Homogenous DE Using Reduction Order

[Lancelot59]

Given y''-4y=2
and that one solution is y_{1}=e^{-2x}
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:
y=ue^{-2x}, y'=u'e^{-2x}-2ue^{-2x}, y''=u''e^{-2x}-2u'e^{-2x}+4ue^{-2x}-2u'e^{-2x}
plugging into the equation:

u''e^{-2x}-2u'e^{-2x}+4ue^{-2x}-2u'e^{-2x}-4ue^{-2x}=0
u''-4u=0

Then using w=u', w'=u''
w'=4w
and I eventually got y to be y_{2}=\frac{e^{2x}/4}c

The book says the answer is y=e^{2x}, 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.

 

[HallsofIvy]

Given y''-4y=2
and that one solution is y_{1}=e^{-2x}
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:
y=ue^{-2x}, y'=u'e^{-2x}-2ue^{-2x}, y''=u''e^{-2x}-2u'e^{-2x}+4ue^{-2x}-2u'e^{-2x}
plugging into the equation:

u''e^{-2x}-2u'e^{-2x}+4ue^{-2x}-2u'e^{-2x}-4ue^{-2x}=0
u''-4u=0

You mean u''- 4u'= 0. Typo?

Then using w=u', w'=u''
w'=4w
and I eventually got y to be y_{2}=\frac{e^{2x}/4}c

Yes, from that equation, w= ce^{4x} so u= (c/4)e^{4x}
so y_2= ue^{-2x}= (c/4)e^{2x}

The book says the answer is y=e^{2x}, so I'm not sure whether or not that factor of 1/4 should be there.

You were asked for a particular solution. Choosing C= 4 in your solution gives the book's. Since the general solution will be formed as C_1y_1+ C_2y_2 constants don't matter.

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.

There are two ways, both of which your text probably discusses:

1) "Variation of parameters". Similar to what you did above. Look for a solution of the form y= u(x)y_1(x)+ v(x)y_2(x)= u(x)e^{-2x}+ v(x)e^{2x} for some functions u and v. y'= u'e^{-2x}- 2ue^{-2x}+ v'e^{2x}+ 2ve^{2x}.

Since there are, in fact, an infinite number of solutions of that form, we narrow the search by requireing that u'e^{-2x}+ v'e^{2x}= 0. That is, that all terms involving the derivatives of u and v sum to 0. Now, we have y'= -2ue^{-2x}+ 2ve^{2x}. Differentiating again, y''= -2u'e^{-2x}+ 4ue^{-2x}+ 2v'e^{2x}+ 4ve^{2x}.

With that value of y'', y''- 4y= 2 becomes -2u'e^{-2x}+ 2v'e^{2x}= 2. That, together with u'e^{-2x}+ v'e^{2x}= 0 gives two equations we can solve algebraically for u' and v', then integrate to get u and v.

2) "Undetermined Coefficients". As long as the right hand side is relatively simple (the same kinds of functions we expect as solutions to "linear homogenous equations with constant coefficients" we can guess the general form of the solution. Here, the right hand side is just the constant "2" so we try a constant for y_p. Take y_p= A, a constant. Then y'= 0, y''= 0 and you can put that into the differential equation to solve for A.

 

[Lancelot59]

I see, I need to go over the undetermined coefficients method a bit. I'll come back if I don't get it after reading some more. Thanks!