\magnification=\magstep1
\centerline{Mathematics 304 -- Ordinary Differential Equations}
\centerline{Selected Solutions -- Problem Set 2}
\centerline{September 25, 2000}
\bigskip
\noindent
\S 2.6/35. To solve:
$$ (3t - x - 6)dt + (t + x + 2)dx = 0.\leqno(1)$$
As discussed in the text, for 1st order equations $M dt + N dx = 0$
where $M$ and $N$ are both linear in $x,t$, if we first translate
the origin, we can remove the constant terms.  Here let 
$$\cases{t = u + h&\cr
         x = v + k&\cr}$$
where $h,k$ are constants.  We get from (1):
$$(3u - v + (3h - k - 6))du + (u + v + (h + k + 2))dv = 0\leqno(2)$$
We want to choose $h, k$ to satisfy 
$$\cases{3h - k - 6 = 0 &\cr
         h + k + 2 = 0,&\cr}$$
so solving simultaneously, $h = 1$ and $k = -3$.  From (2) we get
the following ODE, {\it thinking of $u$ as the independent variable}:
$${dv\over du} = {v - 3u\over v + u} = {{v\over u} - 3\over {v\over u} + 1}.\leqno(3)$$
This is a {\it homogeneous first order equation}, so we solve it by 
changing dependent variables, letting $z = {v\over u}$, or $v = uz$, 
so 
$${dv\over du} = z + u {dz\over u}.$$
Substituting in (3) and simplifying we get:
$$\eqalign{z + u {dz\over du} &= {z - 3\over z + 1}, {\rm\ so}\cr
               u {dz\over du} &= {z - 3\over z + 1} - z\cr
                              &= {-z^2 - z + z - 3\over z + 1}\cr
                              &= - {z^2 + 3\over z + 1}\cr}$$
This is a {\it separable} equation; after separating variables, we
get:
$$\int{z+1\over z^2 + 3} dz = -\int{1\over u} du\leqno(4)$$
The integral on the left can be separated into two terms:
$$\int{z\over z^2 + 3} dz + \int{1\over z^2 + 3} dz$$
In the first one, letting $s = z^2 + 3$, we see that the form is 
${1\over 2} \int {1\over s} ds = {1\over 2} \ln |s| = {1\over 2}\ln(z^2 + 3)$.
The second integral can be obtained from the table in the endpapers
of the text (it is an arctangent form):
$\int{1\over z^2 + 3} dz = {1\over \sqrt{3}} \arctan\left({z\over \sqrt{3}}\right)$.
The right side of (4) is the usual logarithm integral.  Hence from (4) we
get
$${1\over 2}\ln(z^2 + 3) + {1\over \sqrt{3}} \arctan\left({z\over \sqrt{3}}\right) = -\ln(u) + c$$
Now we must ``backsubstitute'' to write the solution in terms of the original
variables.  First $z = {v\over u}$, so we get
$${1\over 2}\ln((v/u)^2 + 3) + {1\over \sqrt{3}} \arctan\left({v\over u\sqrt{3}}\right) = -\ln(u) + c.$$
Then from the original ``homogenizing substitution'', $t = u+1$ and $x = v-3$,
so $u = t - 1$ and $v = x + 3$.  We get
$${1\over 2}\ln\left(\left({x+3\over t-1}\right)^2 + 3\right) + {1\over \sqrt{3}} \arctan\left({x+3\over (t-1)\sqrt{3}}\right) = -\ln(t-1) + c.$$
The answer can be left in this form, or we can combine the two logarithm
terms to obtain the answer in the text.  Note:
$${1\over 2}\ln\left(\left({x+3\over t-1}\right)^2 + 3\right) + \ln(t-1)$$
can be rewritten as
$${1\over 2}\ln\left[\left(\left({x+3\over t-1}\right)^2 + 3\right)(t - 1)^2\right]$$
or 
$${1\over 2}\ln((x+3)^2 + 3(t-1)^2)$$
since ${1\over 2}\ln(A) + \ln(B) = {1\over 2}\ln(AB^2)$.
\bigskip
\noindent
\S 2.6/46.  a)  We assume that $u$ is a particular solution of the 
Riccati equation 
$${dy\over dx} = P(x)y^2 + Q(x)y + R(x),\leqno(5)$$
So 
$${du\over dx} = P(x)u^2 + Q(x)u + R(x).\leqno(6)$$
Then to find other solutions, we let $y = u + {1\over v}$ as suggested 
in the problem and substitute into (5).  We obtain 
$$\eqalign{
{du\over dx} - {1\over v^2}{dv\over dx} &= P(x)\left(u + {1\over v}\right)^2 + Q(x)\left(u + {1\over v}\right) + R(x)\cr
&= P(x) u^2 + 2P(x){u\over v} + {P(x)\over v^2} + Q(x)u + {Q(x)\over v} + R(x)\cr}$$
Using (6), we can cancel terms from both sides to obtain:
$$-{1\over v^2}{dv\over dx} = 2P(x){u\over v} + {P(x)\over v^2} + {Q(x)\over v}$$
or after clearing denominators: ${dv\over dx} = -2P(x)uv - P(x) - Q(x)v$, so 
$${dv\over dx} + (2P(x)u + Q(x))v = -P(x)$$
which is first order linear.
\bigskip
b)  To solve:
$${dy\over dx} = x^3(y - x)^2 + {y \over x} = x^3 y^2 + \left(-2x^4+{1\over x}\right)y + x^5$$
This is of the Riccati form (5) with $P(x) = x^3, Q(x) = -2x^4+{1\over x}, R(x) = x^5$
By the general derivation from part a), using the particular solution 
$u = x$ given in the problem, we see that $y = x + {1\over v}$ solves the 
equation when $v$ solves
$${dv\over dx} + (2P(x)u + Q(x))v = -P(x)$$
or 
$${dv\over dx} + \left(2x^3\cdot x + -2x^4 + {1\over x}\right)v = -x^3$$
or
$${dv\over dx} + {1\over x} v = -x^3.$$
The integrating factor is $e^{\int {1\over x}dx} = |x|$, and
$$v = {c\over |x|} + {1\over |x|}\int |x|\cdot(-x^3)dx$$
Assuming we want a solution for $x > 0$, this can be written 
as
$$ = {c\over x} - {x^4\over 5}$$
And then 
$$\eqalign{
y &= u + {1\over v} = x + {1\over {c\over x} - {x^4\over 5}}\cr
  &= x + {5x \over c - x^5}.\cr}$$
\bye

     