4x achange ofthe (a) Show that Eq (i)can be rewritten as dr 1--(y(x) (b) harNae

Question

4x achange ofthe (a) Show that Eq (i)can be rewritten as dr 1--(y(x) (b) harNae amowdependent variable so that ay/a, or y/x. or y-wa), Expr thus E4 (i) is hom ogeneous xvCx), Expressdy/41 / dyad. Show that the resulting differential equation (c) Replace y and dy/dr in Eq, (ii) by the expressicas from ess from part (b) that invol t terms ei.sand dv/ds. (c) Find the solution of Eq(i) by replacing v by y/x in the solution (d) Solve Eq, (iii), obtaining v implicitly in terms of x. Observe that Eq, (iii) is separable. ofEq, (i) actually depends only on the ratio y/x, This means that inte-at the right Draw a direction Seld and some integral curves-or Eq, (i). Recall that integral the same slope at all points on any given straight line through the i ral ana- n, although the origin, slope changes from one line to another. T herefore, the direction fieldand the curves are symmetric with respect to the origin. Is this symmet rigin. Is this symmetry property evidentt your plot? The method outlined in Problem 30 can be used for any homogeneous equation us equation. That is, the substitution y=xva) transforms a homogeneous equation into a equation. The latter equation can be solved by direct integration, and then r byyx gives the solution to the orginal equation. In each of Problems 31 thro pacing (a) Show that the given equation is homogeneous. (b) Solve the differential equation. (c) Draw a direction field and some integral curves. Are they symmetric with respect to the origin? e: 32, dy=x?+3y? 2xy dx 2x-y dy 4x+3y 34 dr 2x + y The equationsconsidered here have nothing to do wi word "homogeneous" has different meanings in different mathematical contexts.The homogeneous and elsewhere. ith the homogeneous equations that will occur in Chapter

Explanation / Answer

Solution: (35)

Make a substitution to form a new and simpler differential equation:

dy / dx = (x + 3y) / (x - y)
dy / dx = (1 + 3y / x) / (1 - y / x)

Let u = y / x, => y = ux => dy / dx = x(du / dx) + u

x(du / dx) + u = (1 + 3u) / (1 - u)

Solve this differential equation by separating the variables then integrating:
x(du / dx) = {(1 + 3u) / (1 - u)} - u
x(du / dx) = (1 + 3u - u + u²) / (1 - u)
x(du / dx) = (u² + 2u + 1) / (1 - u)
(1 - u) / (u² + 2u + 1) du = (1 / x) dx

Integrate both sides;

(1 - u) / (u² + 2u + 1) du = (1 / x) dx
(u - 1) / (u + 1)² = - 1 / x dx

Integrate the expression on the left by substitution:

(u - 1) / (u + 1)² du
Let v = u + 1, => dv / du = 1 and u = v - 1

(u - 1) / (u + 1)² du = (v - 2) / v² dv
                             = (1/v - 2/v²) dv
(u - 1) / (u + 1)² du = ln|v| + 2/v + C

Since v = u + 1,

(u - 1) / (u + 1)² du = ln|u + 1| + 2/(u + 1)

Find the solution to the differential equation using this result:

(u - 1) / (u + 1)² = - 1 / x dx
ln|u + 1| + 2/(u + 1) = -ln|x| + C
ln|u + 1| + 2/(u + 1) = C - ln|x|

Since u = y / x,

ln|y/x + 1| + 2/(y/x + 1) = C - ln|x|
ln|(y + x)/x| + 2x/(y + x) = C - ln|x|
ln|y + x| - ln|x| + 2x/(y + x) = C - ln|x|
ln|x + y| + 2x / (y + x) = C
2x/(x + y) + ln|x + y| = C

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