Changing variables
Sometimes it is convenient to interchange dependent and independent variables. There are cases when the given
differential equation \( \frac{{\text d}y}{{\text d}x} = f(x,y) \) is very hard or
impossible to solve while the inverse equation
\( \frac{{\text d}x}{{\text d}y} = \frac{1}{f(x,y)} \) can be relatively easily handled.
In this case, we consider x = x(y) as a function of y.
Example: Consider the initial value problem
\[
y' = \left( 2x-3y+5 \right)^{-1} , \qquad y(0) = 1.
\]
To find its solution, we use standard
Mathematica command
ans1 = DSolve[{y'[x] == 1/(2*x - 3*y[x] + 5), y[0] == 1}, y[x], x]
Out[1]= {{y[x] -> 1/6 (7 + 4 x - 3 ProductLog[1/3 E^(1/3 + (4x)/3)])}}
Plot[y[x] /. ans1, {x, -1, 1}, PlotRange -> {0.4, 1.5}, AspectRatio -> 1]
StreamPlot[{1, 1/(2*x - 3*y + 5)}, {x, -3, 3}, {y, -2, 3},
StreamPoints -> {{{{0, 1}, Red}, Automatic}}, StreamStyle -> Blue,
Background -> LightGray]
So
Mathematica provides a solution expressed through a special function. Therefore, we try another approach
based on the inverse problem when
x is considered to be a dependent variable:
\[
\frac{{\text d}x}{{\text d}y} = \left( 2x-3y+5 \right) , \qquad x(1) = 0.
\]
Since the equation is linear, we can find its solution without a problem.
ans2 = DSolve[{x'[y] == 2*x[y] - 3*y + 5, x[1] == 0}, x[y], y] // Simplify
Out[3]= {{x[y] -> 1/4 (-7 + E^(-2 + 2 y) + 6 y)}}
Plot[x[y] /. ans2, {y, 0.4, 1.5}, PlotRange -> {-1, 1}, AspectRatio -> 1]
Here is a way to flip the above graph:
ParametricPlot[Evaluate[{x[y], y} /. ans2], {y, 0, 2}, PlotRange -> {{-1, 1}, {0.4, 1.5}}, AspectRatio -> 1]
Linear Substitution
We present several classes of equations that can be reduced to a separable one. We start with the differential
equation of the form
\[
y' = F(ax +by +c), \qquad b\ne 0,
\]
where
F(v) is a given continuous function of a variable
v,
and
a,b,c are some constants. This equation is reduced to a
separable one by substitution
\( v=ax+by +c . \)
Example: Consider the differential equation
\[
y' = 2x+3y+5 ,
\]
Using substitution
v = 2x+3y+5, we find its derivative to be
v' = 2 +3 y' = 2 + 3 v, which is a separable one. Separating variables and integrating,
we obtain
\[
\frac{{\text d}v}{2+3v} = {\text d}x \qquad \Longrightarrow \qquad \ln \left( 2+ 3v \right) = 3x+C ,
\]
where
C is an arbitrary constant. Exponentiation yields the general solution in implicit form:
\[
2+3v = C\,e^{3x} \qquad \Longrightarrow \qquad 2+ 3\left( 2x+3y +5 \right) = C\,e^{3x} . \qquad ■
\]
Another class of equations is
\[
x\,y' = y\,F(xy),
\]
where F(v) is a function of the product v=xy.
This differential equation is reduced to a separable one by substitution v=xy.
Example:
First, we find solutions to the equation \( xy'=x^2 y^3 -y \) using the standard
Mathematica command:
DSolve[y'[x] == x*(y[x])^3 -y[x]/x,y[x],x]
and then plot corresponding family of solutions:
solution = DSolve[y'[x] == x*(y[x])^3 - y[x]/x, y[x], x]
g[x_] = y[x] /. solution[[1]]
t[x_] = Table[g[x] /. C[1] -> j, {j, 1, 6}]
Plot[t[x], {x, 0.2, 3}]
Next we find its solution manually; we set
v=xy, and get
\[
v' = y + x\,y' = y + x^2 y^3 -y = x^2 \left( \frac{v}{x} \right)^3 = \frac{v^3}{x} ,
\]
which is a separable one. So we separate variables and integrate
\[
\frac{{\text d}v}{v^3} = \frac{{\text d}x}{x} \qquad \Longrightarrow \qquad - \frac{1}{v^2} = 2\,\ln Cx = \ln C\,x^2 ,
\]
where
C is an arbitrary positive constant. Upon returning for original dependent
variable
v=xy, we get the general solution in implicit form:
\[
\frac{1}{y^2} = -x^2 \ln C\,x^2 \qquad \mbox{or} \qquad 1 + x^2 y^2 \,\ln C x^2 =0.
\]
Equations with Homogeneous Slope Function
Let r be a real number. A function of two variables g(x,y) is called
homogeneous of degree r if
\( g( \lambda x, \lambda y ) = \lambda^r g(x,y) \) for any nonzero constant λ.
Usually homogeneous functions of zero degree are referred to as homogeneous or homogeneous-polar (because the ratio
y/x in polar coordinates is the tangent of the angle). Obviously, a
function of the ratio y/x is a homogeneous-polar function.
Our next class includes differential equations with homogeneous coefficients when the slope function is a
function depending on the ratio y/x:
\[
y' = F(y/x) .
\]
This equation is reduced to a separable one by substitution
v(x) = y/x or
y=xv.
Indeed, applying the product rule, we find
\[
\frac{{\text d}y}{{\text d}x} = \dfrac{\text d}{{\text d}x} \left( x\,v \right) = x\,\frac{{\text d}v}{{\text d}x} + v
= x\, v + v .
\]
Upon substituting this expression into the equation
\( y' = F(y/x) = F(v) , \) we get
\[
x\,\frac{{\text d}v}{{\text d}x} + v
= F( v) \qquad \mbox{or} \qquad x\, v' = F(v) -v .
\]
The latter is a separable equation, which upon separation of variables and integration yields
\[
\int \frac{{\text d}v}{F(v) -v} = \int \frac{{\text d}x}{x} = \ln Cx ,
\qquad F(v) \ne v \qquad\mbox{and} \qquad v = y/x ,
\]
where
C is an arbitrary constant (positive when
x positive and negative when
x is negative).
One particular important class of differential equations constitute the equations with linear coefficients:
\[
\left( Ax + By \right) {\text d}y = \left(ax+by \right) {\text d}x \qquad \mbox{or} \qquad
\frac{{\text d}y}{{\text d}x} = \dfrac{ax+by}{Ax+By} = \dfrac{a+by/x}{A+By/x} ,
\]
where
a,
b,
A, and
B are known real constants. Upon substitution
v = y/x, the above
equation is reduced to a separable one:
\[
x\, \frac{{\text d}v}{{\text d}x} = \dfrac{a+bv - Av - B v^2}{A+Bv} \qquad \Longrightarrow \qquad
\int \frac{A + Bv}{a + b(b-A) - B v^2}\, {\text d}v = \int \frac{{\text d}x}{x} = \ln Cx ,
\]
with some arbitrary constant
C. The value of the integral over
v depends on whether the roots of
quadratic equation
\( Bv^2 + (A-b) \,v - a =0 \) are distinct real, complex conjugate, or
one double root.
Example:
We consider the differential equation with homogeneous slope function
\[
y' = \frac{y-x}{y+x} = \dfrac{y/x -1}{y/x +1} .
\]
Changing the dependent variable to
v=y/x or
\( y(x) = x\,v(x) , \)
we obtain a separable equation in variable
v:
\[
y' = v(x) + x\,v' (x) = \frac{v-1}{v+1} \qquad\Longrightarrow \qquad x\,v' = \frac{v-1}{v+1} -v = - \frac{v^2 +1}{v+1} .
\]
Separation of variables yields
\[
\frac{v+1}{v^2 +1}\,{\text d}v = - \frac{{\text d}x}{x} \qquad\Longrightarrow \qquad \arctan v + \frac{1}{2}\,\ln (v^2 +1 ) = \int \frac{v+1}{v^2 +1}\,{\text d}v = - \int \frac{{\text d}x}{x} = -\ln Cx .
\]
points := Tuples[{-2, -1, 0, 1, 2}, 2]
Table[StreamPlot[{1, {(y - x)/(x + y)}}, {x, -3, 3}, {y, -3, 3},
StreamPoints -> {points, Automatic, maxlen},
Epilog -> Point[points]],
{maxlen, {2.5, Scaled[.05]}}]
Example: Consider the differential equation
\[
\left( 9x -y \right) {\text d}x + \left( 7x+y \right) {\text d}y =0 \qquad 7x+y \ne 0.
\]
Substitution
v = y/x yields
\[
v + x\, v' = \frac{v-9}{v+7} \qquad\Longrightarrow \qquad \frac{\left( v+7 \right) {\text d}v}{v^2 + 6v +9} = -\frac{{\text d}x}{x} .
\]
Integrating both sides, we get the general solution in implicit form:
\[
\frac{4}{v+3} + \ln x \left( v+3 \right) = C \qquad\Longrightarrow \qquad \frac{4x}{3x+y} +
\ln \left\vert y + 3x \right\vert = C ,
\]
which is the general solution. Of course, we have to exclude the singular solution
y = -3x
from the implicit formula. Finally, we plot the direction fields using VectorPlot command:
VectorPlot[{1, (y - 9*x)/(7*x + y)}, {x, -1, 1}, {y, -1, 1},
VectorPoints -> Fine, StreamPoints -> Coarse, StreamStyle -> Red,
StreamScale -> Full]
points = Join[{#, -1.} & /@ Range[-1, 1, .2], {#, 1.} & /@
Range[-1, 1, .2]];
VectorPlot[{1, (y - 9*x)/(7*x + y)}, {x, -1, 1}, {y, -1, 1},
VectorPoints -> Fine, StreamPoints -> {points, Automatic, 2},
StreamStyle -> Orange, StreamScale -> Large]
Manipulate[points = Table[{Cos@t, Sin@t}*r, {t, 0., 2. \[Pi], d}];
VectorPlot[{1, (y - 9*x)/(7*x + y)}, {x, -1, 1}, {y, -1, 1},
VectorPoints -> Fine, StreamPoints -> {points, Automatic, l},
StreamStyle -> Blue, StreamScale -> Large,
Epilog -> {Green, PointSize@Large, Point@points}], {{d, \[Pi]/20.,
"point spacing"}, 0, 1, \[Pi]/40,
Appearance -> "Labeled"}, {{l, 2, "stream length"}, 0, 2,
Appearance -> "Labeled"}, {{r, 1, "radius"}, 0, 1,
Appearance -> "Labeled"}]
Example: Consider the differential equation
\[
\frac{{\text d}y}{{\text d}x} = \frac{2x+4y}{3x+y} \qquad 3x\ne y .
\]
Introducing a new dependent variable
v = y/x, we reduce our differential equation to a
separable one:
\[
v + x\, v' = \frac{2+4v}{v+3} \qquad\Longrightarrow \qquad \frac{\left( v+3 \right) {\text d}v}{v^2 - v -2} =
-\frac{{\text d}x}{x} , \quad v \ne 2, \quad v\ne -1.
\]
Since
\( \frac{\left( v+3 \right) {\text d}v}{v^2 - v -2} = \frac{5/3}{v-2} - \frac{2/3}{v+1} , \)
we integrate to obtain the general solution
\[
\frac{5}{3}\, \ln |v-2| - \ln \left\vert v+1 \right\vert = -C\, \ln |x| \qquad\Longrightarrow \qquad
\frac{|v-2|^5}{(v+1)^2} = C\, |x|^{-3} .
\]
Upon returning to the original variable
y = xv, we get the general solution in implicit form:
\[
|y-2x|^5 = C\,(y+x)^2 \qquad (C \mbox{ is a constant}).
\]
Initially, we excluded two solutions
y = 2x and
y = -x from our
derivation because they appeared under the logarithm sign. Now we see that they could be obtained from the general solution
and, therefore, are not singular solutions.
We plot direction fields with StreamPlot command. First, we just use the regular approach by identifying points and then
apply the main command:
points = With[{d = .1},
Join[#, -#] &[{#, -2 + d} & /@
DeleteCases[Range[-2, 2, 1/5], x_ /; Not[FreeQ[x, -2 | 0 | 2]]]]];
Semicolumn suppresses output to be visible on the screen. We use the set of points in the next command to plot solutions through these points.
StreamPlot[{1, (2*x + 4*y)/(3*x + y)}, {x, -2, 2}, {y, -2, 2},
StreamPoints -> {points, Automatic},
StreamScale -> {{0.4,
0.01},(*or use Full to completely prevent segmentation*)All, 0.02,
Automatic}]
However,
Mathematica allows us to identify a point on this plot and then draw a particular trajectory through
this point:
StreamPlot[{1, (2*x + 4*y)/(3*x + y)}, {x, -5, 5}, {y, -5, 5},
StreamPoints -> {{{{0.5, 2}, Red}, Automatic}}]
sp = StreamPlot[{1, (2*x + 4*y)/(3*x + y)}, {x, -5, 5}, {y, -5, 5},
StreamPoints -> {{{{0, 2}, White}, Automatic}},
Epilog -> {Red, PointSize[Large], Point[{0.5, 2}]}]
sp2 = StreamPlot[{1, (2*x + 4*y)/(3*x + y)}, {x, -5, 5}, {y, -5, 5},
StreamPoints -> {{{{0.5, 2}, Red}}}, RegionFunction -> (# > 0 &)]
Finally, we unite two previous pictures into one:
Show[sp, sp2]
Example: We demonstrate reduction of the equation with homogeneous slope function,
y' =-(x^2 + y^2)/(5xy), to a separable equation. Of course,
Mathematica is an appropriate tool for such reduction. We start with defining the differential equation with differentials.
ode[x_, y_] = -(x^2 + y^2)*dx == 5*x*y*dy
Out[8] = dx (-x^2 - y^2) == 5 dy x y
The first step is to replace the differential
dy with the sum
dy=xdv + vdx according to the product rule.
ode[x, v*x] /. {dy -> x*dv + v*dx}
Out[9] = dx (-x^2 - v^2 x^2) == 5 v x^2 (dx v + dv x)
Then we substitute instead of
y the product
y=xv:
Map[Cancel, Map[Function[q, q/x^2], %]]
Out[10] = -dx (1 + v^2) == 5 v (dx v + dv x)
Finally, we simplify the output by collecting similar terms.
Map[Function[u, Collect[u, {dx, dv}]], %]
Out[11] = -dx (1 + v^2) == 5 dx v^2 + 5 dv v x
Example: Consider the initial value problem
y' = 1/(x-y+2), y(0) =1 .
We find its solution and plot it using the following Mathematica commands:
ans1 = DSolve[{y'[x] == 1/(x - y[x] + 2), y[0] == 1}, y[x], x]
Plot[y[x] /. ans1, {x, -1, 1}, PlotRange -> {0, 2}, AspectRatio ->1]
ans2 = DSolve[{x'[y] == x[y] - y + 2, x[1] == 0}, x[y], y] //Simplify
ParametricPlot[Evaluate[{x[y], y} /. ans2], {y, 0, 2},
PlotRange -> {{-1, 1}, {0, 2}}, AspectRatio -> 1]
LineIntegralConvolutionPlot[{{1, 1/(x - y + 2)},
Image[Table[((-1)^i (-1)^j + 1)/2, {i, 45}, {j, 45}]]}, {x, -3,
3}, {y, -3, 3}, LineIntegralConvolutionScale -> 2,
ColorFunction -> GrayLevel, RasterSize -> 300]
StreamPlot[{1, 1/(x - y + 2)}, {x, -3, 3}, {y, -3, 3},
StreamStyle -> "Drop"]
