# No result from DSolve

I don't get any answer when I evaluate the following expression:

DSolve[{x'[t] == -0.5*y[t] + 1 + 3*(e^(-2*t)),
y'[t] == 2*x[t] - y[t] - 4 - 4* (e^(-t)),
y == 0, x == 0}, {x[t], y[t]} , t]


Did I go wrong somewhere?

I tried solving this equation set with Laplace as well, but I didn't get any answer from that either.

• Is it e, or E ? – Dr. belisarius Oct 17 '12 at 17:28

Try replacing the .5by 1/2 and the e by E. Also, the asterisks aren't needed in Mathematica.

s = DSolve[
{x'[t] == -1/2 y[t] + 1 + 3 E^(-2 t),
y'[t] == 2 x[t] - y[t] - 4 - 4 E^(-t),
y  == 0,
x  == 0}, {x[t], y[t]}, t]

ParametricPlot[{x[t], y[t]} /. s, {t, 0, 10}, AxesOrigin -> {0, 0}] • and //FullSimplify ;) – Vitaliy Kaurov Oct 17 '12 at 17:32
• @VitaliyKaurov Tried that, but the result isn't compact enough to look nice here. So I kept it like that. – Dr. belisarius Oct 17 '12 at 17:34
• That was more a tip for @DSaad than an edit suggestion :) – Vitaliy Kaurov Oct 17 '12 at 17:36

Two problems here are e instead of E and 0.5 instead of 1/2. For a larger system it would be more reasonable to use Rationalize instead of substituting all numbers by hand. To get a wider class of solutions we could define a function being a solution of the system of differential equations, where the initial values are parametrized by x0:

ds[x0_] = Simplify @ DSolve[ Rationalize @ {x'[t] == -0.5*y[t] + 1 + 3*(E^(-2*t)),
y'[t] == 2*x[t] - y[t] - 4 - 4*(E^(-t)),
y == 0,
x == x0},     {x[t], y[t]}, t] originally x0 == 0, i.e.

ds To plot the wider class of solutions we could use ParametricPlot as in belisarius answer with Show emphasizing ds solution as a red thick curve :

Show[
ParametricPlot[{x[t], y[t]} /. Table[ds[x0], {x0, -2, 4, 0.2}],
{t, -1, 10}, Evaluated -> True],
ParametricPlot[{x[t], y[t]} /. ds, {t, -1, 10},
PlotStyle -> {Darker@Red, Thickness[0.007]}],
AxesOrigin -> {0, 0}, PlotRange -> {{-1, 6}, {-4, 5}}] 