Download Problems for 01.03. Chapter 8 6.3 The differential equation is

Problems for 01.03. Chapter 8
6.3 The differential equation is linear inhomogenous 2nd order. The solution is
most easily found by solving the homogenous equation by substituting y0 =
Cekx to find 2 lin. independent solutions and a finding a particular solution yp
by an educated guess (an exponential such as in the inhomogenous term); the
general solution is given by y = y0 + yp . The method in the book is to write
d
d
− a)( dx
− b)y = Q(x) and solving 2 first order eqs.
the equation in the form ( dx
d
d
( dx − b)u = Q(x) and ( dx − a)y = u(x). Also, a more general method is find
the solution to the particular solution by variation of coefficients yp = u(x)y0 .
Answer: y(x) = C1 ex + C2 e−2x + 41 e2x .
6.11 The same comments as above: one can work with sines and cosines, but the solution can also be found by considering the complex equation with ei4x instead
of cos 4x, finally one is to take the real part of the solution.
Answer: e−x (A cos 3x + B sin 3x) − 6 cos 4x + 8 sin 4x.
6.23 The equation is as the ones before except that one is to try yp = Q(x)ex , where
Q(x) is a polynomial of the same order as on the inhomogenous term. In the
degenerate case one has to ”crank up the order” of Q(x) to get yp , as before.
Answer: A sin x + B cos x + (x − 1)ex .
7.17-7.22 These equation are linear 2nd order but with non-constant coefficients. The
cases here are of the form a2 x2 y 00 + a1 xy 00 + a0 y = f (x); 2nd order equations
of this form are called Euler-Cauchy equations. These are defined for x 6= 0,
since y 00 (0) in general diverges, as one can see implicitly from the equation.
The equation therefore in general has solutions in the two regimes x > 0 and
x < 0. Intuitively, we should expect solutions of the form y = xm , since
hitting this with n derivatives reduces the order by n, i.e. gives powers xm−n .
Therefore, one can solve the homogenous equation by substituting y = xm and
solving the roots of the resulting second order polynomial for m. In degenerate
cases, one has to add ln |x|, and cos(ln |x|), sin(ln |x|) if there are no real roots.
The particular solution is found by inspired guess for yp or varying coefficients
for the solution Cy0 (x).
There is a way to reduce the Euler-Cauchy equation to one with constant
coefficients, which are perhaps easier to solve. This is simply done by making
the transformation x = ez > 0 and solving the equation for y(z). In the end
one is to transform back with z = ln x. The solution for x < 0 is found by
x = −ez . For more on this method, see M.L. Boas.
Answers: 7.17: y(x) = c1 x4 +c2 x−4 +x4 ln |x| ; x > 0, x < 0. 7.18: y(x) = c1 x+
1
ln |x| ; x > 0, x < 0. 7.22: A cos(ln |x|) + B cos(ln |x|) +
c2 x−1 + 21 x ln |x| + 2x
x; x > 0, x < 0.
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