The main issue we encounter here is that equation solvig functionality doesn't work quite seamlessly for various types of transcendental functions. However looking at the result of DSolve
it appears convenient to introduce a new variable $z=\log(1+\frac{y(x)}{x})$.
z[x] == Log[1+ y[x]/x]
Now we can find the exact solution, setting up the initial condition in terms of z[x]
, i.e. z[3] == -Log[3]
:
z[x]/.First @ DSolve[{0==(x - y[x]) + (3x + y[x])y'[x]/.{
y[x] -> x (Exp[z[x]] - 1),
y'[x] -> D[x(Exp[z[x]] - 1), x]}//Simplify,
z[3] == -Log[3]},
z[x], x] // Quiet
-6 - Log[x] + ProductLog[2 E^6 x]
that is Log[1+ y[x]/x] == -6 - Log[x] + ProductLog[2 E^6 x]
i.e. the solution of our equation is y[x] == f[x]
such, that
f[x_] := E^(-6 + ProductLog[2 E^6 x]) - x
because
FullSimplify[x (Exp[-6 - Log[x] + ProductLog[2 E^6 x]] - 1), x > 0]
% // TraditionalForm
E^(-6 + ProductLog[2 E^6 x]) - x
where W
is the Lambert function. Now we can plot the graph of the solution. However since it looks like a straight line, we find an example of a linear function to compare the both graphs. We assume that the graphs intersect at x == 0
and x == 3
and so:
g[x_] = a x + b /. First @ Solve[{ f[0] == b, f[3] == 3 a + b}, {a, b}]
1/E^6 - ((1 + 2 E^6) x)/(3 E^6)
Plot[{ f[x], g[x]}, {x, 0, 4}, PlotStyle -> {Thick, Dashed},
Epilog -> {Red, PointSize[0.02], Point[{3, -2}]}]
The graph is very close to a linear function as could be seen from ContourPlot
. It might be convenient to plot the difference of the both functions:
Plot[ Re @ f[x] - g[x], {x, -3, 5}, PlotStyle -> {Thick, Magenta},
Epilog -> {Red, PointSize[0.02], Point[{3, 0}]}]
For completeness we plot the real and imaginary values of f[x]
since it becomes complex for negative numbers (this is independendent from the physical (?) model discribed by the original equation)
Plot[ Flatten @ {ReIm @ f[x], g[x]}, {x, -20, 20},
PlotStyle -> {Thick, Thick, Dashed}, Evaluated -> True,
Epilog -> {Red, PointSize[0.01], Point[{3, -2}]}]
ContourPlot[ Log[1 + y/x] - 2/(1 + y/x) + 6 + Log[x] == 0, {x, 0, 1}, {y, -2, 1}]
? $\endgroup$