Finding a second solution of a Differential Equation, Given a first solution.?

Question

Allright, my question is simple (problem isn't though) - can anyone give me in good detail how to solve this equation:

(1 + X)*(d^2y/dx^2) +x*(dy/dx) -y = 0; y1=x

The idea is that, there is a solution of this equation already (y=x), but I am looking for a second solution. And the technique I am supposed to use, is to substitude in a "v" in front of the function, ie, "y2=xv", then "dy2/dx=xdv +v" and "dy2^2/dx^2=xd^2v + 2dv". Then you substitue this information in for the original equation, factor out DV and D^2V, then create another substitution for w=dV and dw=d^2v. Finally, you can set w and x on seperate sides, integrate, solve for w, then solve backwards back to v, and finally come up with the second solution y2 from the equation y=xv.

I've worked the problem, and I know how to solve these problems, it just seems I keep making the same mistakes and can't solve it. If anyone could actually go through the entire process and tell me how they did it, it would be greatly appreciated.

Answer 1

(1+x)y'' + xy' - y = 0

First, let us assume the solution will be of the form vy_1, which is vx. Finding the first two derivatives of vx:

(vx)' = v'x + v
(vx)'' = v''x + 2v'

Substituting into the original equation:

(1+x)(v''x + 2v') + x(v'x + v) - vx = 0

Simplifying:

v''x + 2v' + v''x² + 2v'x + v'x² + vx - vx = 0
(x²+x)v'' + (x²+2x+2)v' = 0

Making the substitution w=v':

(x²+x)w' + (x²+2x+2)w = 0

This is a linear equation in w. Moreover, it is a separable equation in w, so let us separate it:

(x²+x)w' = -(x²+2x+2)w
w'/w = -(x²+2x+2)/(x²+x)

Now, let us break up the right-hand side a bit:

w'/w = -(x²+2x+1)/(x²+x) - 1/(x²+x)
w'/w = -(x+1)²/(x²+x) - 1/(x(x+1))
w'/w = -(x+1)/x + 1/(x+1) - 1/x
w'/w = -(x+2)/x + 1/(x+1)
w'/w = -1 - 2/x + 1/(x+1)

Now, let us integrate:

ln |w| = -x - 2 ln |x| + ln |x+1| + C

Exponentiate:

w=Ke^(-x)(x+1)/x²

Now, since w=v':

v' = Ke^(-x)(x+1)/x²
v = ∫Ke^(-x)(x+1)/x² dx
v = K∫e^(-x)/x dx + K∫e^(-x)/x² dx

Integrating K∫e^(-x)/x² dx by parts:

v = K∫e^(-x)/x dx - Ke^(-x)/x - K∫e^(-x)/x dx

And now the more annoying integrals cancel (up to an additive constant):

v = - Ke^(-x)/x + C

Multiplying by x:

y=-Ke^(-x) + Cx

Let us check this solution:

-(1+x)Ke^(-x) + Kxe^(-x) + Cx + Ke^(-x) - Cx = 0

So we have found the correct solution.

Answer 2

purely a small addendum to hfshaw's remarkable answer. That sinusoidal section: 3sin(?*t) - ?*sin(3t) has a particularly interesting particular shape - an interference trend, which (regrettably) is honestly washed out by the solid damping (e^(-t)). 3sin(?*t) - ?*sin(3t) = (3+?) Sin[(?-3)/2*t] Cos[(?+3)/2*t] + (3-?) Sin[(?+3)/2*t] Cos[(?-3)/2*t] With the damping bumped off, we see 2 words (graph each and each physique by one), each and each representing an oscillation at known frequency, (0.40 9), between the organic (0.40 8) and forcing (0.50) frequencies. even nevertheless, the interference reasons "beats" because of the fact the sign fades out and in with a frequency, (0.01). equaling a million/2 the version between the organic and forcing frequencies. because of the fact the organic and compelled aspects are actually not equivalent, the 2nd (smaller, section shifted) interference cancels out a small part of the interference.