Solutions to 18.03 Problem Set 4.5

Part II.

17. (F 12 Mar) (a) Remember, we know y_{1} and wish to find a second solution, which we’ll call y for the moment. So Abel’s equation is

y′ − \frac{y′_{1}}{y_{1}} y = \frac{e^{− ∫ p(x)dx}}{y_{1}} .  (1)

Multiply by an integrating factor (badly named w, since it’s not the Wronskian) and demand of it that

wy′ + w′y = (wy)′ = wy′ − \frac{wy′_{1}}{y_{1}} y

or w′ = −\frac{y′_{1}}{y_{1}} w. This is separable, and leads to

\frac{dw}{w} = −\frac{y′_{1}}{y_{1}} dx,

or

ln |w| = − ln |y_{1}| + c

so, since we can choose c as we like, we can take w to be 1/y_{1}. Inspecting (1) again we see that we could have seen this without doing the work. Anyway, this leads to

(\frac{y}{y_{1}} )′ = \frac{1}{y_{1}^{2}} e^{− ∫ p(x)dx}

or

y = y_{1} ∫ \frac{e^{− ∫ p(x)dx}}{y_{1}^{2}} dx,

which is EP 2.6 (8).

(b) You have to know how to interpret (xD)^{2}: it’s the operator you get by doing xD twice in succession: (xD)^{2}u = xD(xDu) = xD(xu′) = x(xu′′ + u′) = x^{2}u′′ + xu′ = (x^{2}D^{2} + xD)u.

(c) Let x = e^{t}, and write ẏ = dy/dt, etc. Then ẋ = x,

y′ = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{ẏ}{x},

y′′ = \frac{dy′}{dx} = \frac{1}{x} \frac{d}{dt} \frac{ẏ}{x} = \frac{1}{x} \frac{ÿ − ẏ}{x^{2}} \frac{dx}{dt} = \frac{ÿ − ẏ}{x^{2}} .

Thus x^{2}y′′ + pxy′ + qy = r(x) is, in terms of t, (ÿ − ẏ) + pẏ + qy = r(e^{t}), or

ÿ + (p − 1)ẏ + qy = r(e^{t}).

This substitution converts an Euler-Cauchy equation to a constant coefficient equation. For example, x(ln x)^{2} sin(ln x) = e^{t}t^{2} sin t and the given equation reads

ÿ − 2ẏ + y = t^{2}e^{t} sin t.

Solve this by replacing it with another ODE having a (complex) exponential on the RHS:

Lz = z̈ − 2ż + z = t^{2}e^{st},  s = 1 + i;  y = Im z.

This calls for ESL: f(r) = (r − 1)^{2}, so replacing r by (D + sI) we get (D + (s − 1)I)^{2} = D^{2} + 2iD − I (since s − 1 = i). Now write z = e^{st}u and apply ESL:

t^{2}e^{st} = f(D)(e^{st}u) = e^{st}f(D + sI)u = e^{st}(D^{2} + 2iD − I)u

so we wish to solve (D^{2}+2iD−I)u = t^{2}. This we can handle by undetermined coefficients: u = at^{2} + bt + c implies

−z = −at^{2} −bt −c
2iz′ =  4iat +2bi
z′′ =  2a

Equating coefficients, −a = 1 so a = −1; −b + 4ia = 0 so b = −4i, −c + 2bi + 2a = 0 so c = 6: u = −t^{2} − 4it + 6; so z_{p} = (−t^{2} − 4it + 6)e^{(1+i)t} = (−t^{2} − 4it + 6)e^{t}(cos t + i sin t); so y_{p} = Im z_{p} = e^{t}((−t^{2} +6) sin t−4t cos t) = x(−(ln x)^{2} +6) sin(ln x)−4(ln x) cos(ln x)).

This is a particular solution. The homogeneous equation x^{2}y′′ −xy′ +y = 0 is equivalent, under the same substition, to ÿ − 2ẏ + y = 0. The characteristic polynomial r^{2} − 2r + 1 has repeated root r = 1, so we have independent solutions e^{t}, te^{t}. In terms of x these are x, x ln x; so the general solution is

= x(−(ln x)^{2} + 6) sin(ln x) − 4(ln x) cos(ln x)) + ax + bx ln x.

(I hope I didn’t make an arithmetic mistake!)