Mathematics 304 – Ordinary Differential Equations Selected Solutions – Problem Set 2 September 25, 2000

§2.6/35. To solve:

(1)  (3t − x − 6)dt + (t + x + 2)dx = 0.

As discussed in the text, for 1st order equations Mdt + Ndx = 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

\{ \begin{matrix}t = u + h \\ x = v + k \end{matrix}

where h, k are constants. We get from (1):

(2)  (3u − v + (3h − k − 6))du + (u + v + (h + k + 2))dv = 0

We want to choose h, k to satisfy

\{ \begin{matrix}3h − k − 6 = 0 \\ h + k + 2 = 0, \end{matrix}

so solving simultaneously, h = 1 and k = −3. From (2) we get the following ODE, thinking of u as the independent variable:

(3)  \frac{dv}{du} = \frac{v − 3u}{v + u} = \frac{\frac{v}{u} − 3}{\frac{v}{u} + 1}.

This is a homogeneous first order equation, so we solve it by changing dependent variables, letting z = \frac{v}{u}, or v = uz, so

\frac{dv}{du} = z + u\frac{dz}{u} .

Substituting in (3) and simplifying we get:

z + u\frac{dz}{du} = \frac{z − 3}{z + 1}, so
u\frac{dz}{du} = \frac{z − 3}{z + 1} − z

= \frac{−z^{2} − z + z − 3}{z + 1}

= −\frac{z^{2} + 3}{z + 1}

This is a separable equation; after separating variables, we get:

(4)  ∫ \frac{z + 1}{z^{2} + 3}dz = − ∫ \frac{1}{u}du

The integral on the left can be separated into two terms:

∫ \frac{z}{z^{2} + 3}dz + ∫ \frac{1}{z^{2} + 3}dz

In the first one, letting s = z^{2} + 3, we see that the form is \frac{1}{2} ∫ \frac{1}{s}ds = \frac{1}{2} ln |s| = \frac{1}{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): ∫ \frac{1}{z^{2}+3}dz = \frac{1}{\sqrt{3}} arctan (\frac{z}{\sqrt{3}}). The right side of (4) is the usual logarithm integral. Hence from (4) we get

\frac{1}{2} ln(z^{2} + 3) + \frac{1}{\sqrt{3}} arctan (\frac{z}{\sqrt{3}}) = − ln(u) + c

Now we must “backsubstitute” to write the solution in terms of the original variables. First z = \frac{v}{u}, so we get

\frac{1}{2} ln((v/u)^{2} + 3) + \frac{1}{\sqrt{3}} arctan (\frac{v}{u\sqrt{3}}) = − 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

\frac{1}{2} ln ((\frac{x + 3}{t − 1})^{2} + 3) + \frac{1}{\sqrt{3}} arctan (\frac{x + 3}{(t − 1)\sqrt{3}}) = − 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:

\frac{1}{2} ln ((\frac{x + 3}{t − 1})^{2} + 3) + ln(t − 1)

can be rewritten as

\frac{1}{2} ln [((\frac{x + 3}{t − 1})^{2} + 3) (t − 1)^{2}]

or

\frac{1}{2} ln((x + 3)^{2} + 3(t − 1)^{2})

since \frac{1}{2} ln(A) + ln(B) = \frac{1}{2} ln(AB^{2}).

§2.6/46. a) We assume that u is a particular solution of the Riccati equation

(5)  \frac{dy}{dx} = P (x)y^{2} + Q(x)y + R(x),

So

(6)  \frac{du}{dx} = P (x)u^{2} + Q(x)u + R(x).

Then to find other solutions, we let y = u + \frac{1}{v} as suggested in the problem and substitute into (5). We obtain

\begin{aligned}\frac{du}{dx} − \frac{1}{v^{2}} \frac{dv}{dx} & = P (x) (u + \frac{1}{v} )^{2} + Q(x) (u + \frac{1}{v} ) + R(x) \\ & = P (x)u^{2} + 2P (x)\frac{u}{v} + \frac{P (x)}{v^{2}} + Q(x)u + \frac{Q(x)}{v} + R(x) \end{aligned}

Using (6), we can cancel terms from both sides to obtain:

−\frac{1}{v^{2}} \frac{dv}{dx} = 2P (x)\frac{u}{v} + \frac{P (x)}{v^{2}} + \frac{Q(x)}{v}

or after clearing denominators: \frac{dv}{dx} = −2P (x)uv − P (x) − Q(x)v, so

\frac{dv}{dx} + (2P (x)u + Q(x))v = −P (x)

which is first order linear.

b) To solve:

\frac{dy}{dx} = x^{3}(y − x)^{2} + \frac{y}{x} = x^{3}y^{2} + (−2x^{4} + \frac{1}{x}) y + x^{5}

This is of the Riccati form (5) with P (x) = x^{3}, Q(x) = −2x^{4} + \frac{1}{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 + \frac{1}{v} solves the equation when v solves

\frac{dv}{dx} + (2P (x)u + Q(x))v = −P (x)

or

\frac{dv}{dx} + (2x^{3} · x + −2x^{4} + \frac{1}{x}) v = −x^{3}

or

\frac{dv}{dx} + \frac{1}{x}v = −x^{3}.

The integrating factor is e^{∫ \frac{1}{x} dx} = |x|, and

v = \frac{c}{|x|} + \frac{1}{|x|} ∫ |x| · (−x^{3})dx

Assuming we want a solution for x > 0, this can be written as

= \frac{c}{x} − \frac{x^{4}}{5}

And then

\begin{aligned}y & = u + \frac{1}{v} = x + \frac{1}{\frac{c}{x} − \frac{x^{4}}{5}} \\ & = x + \frac{5x}{c − x^{5}} . \end{aligned}