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Método directo con SAGE:

dxdt=−225x1+150x2dydt=225x1−225x2\begin{aligned} \frac{dx}{dt} & = -\frac{2}{25}x_{1} + \frac{1}{50}x_{2} \\ \frac{dy}{dt} & = \frac{2}{25}x_{1} - \frac{2}{25}x_{2} \end{aligned}

Solución General: 

# Sistema de ED's
t = var("t")
x1 = function("x1")(t)
x2 = function("x2")(t)
ec1 = diff(x1, t) == -2/25*x1 + 1/50*x2
ec2 = diff(x2, t) == 2/25*x1 - 2/25*x2
[solx1, solx2] = desolve_system([ec1, ec2], [x1, x2])
show(solx1)
show(solx2)
x1(t)=12 e(−125 t)x1(0)+12 e(−325 t)x1(0)\displaystyle x_{1}\left(t\right) = \frac{1}{2} \, e^{\left(-\frac{1}{25} \, t\right)} x_{1}\left(0\right) + \frac{1}{2} \, e^{\left(-\frac{3}{25} \, t\right)} x_{1}\left(0\right)
x2(t)=e(−125 t)x1(0)−e(−325 t)x1(0)\displaystyle x_{2}\left(t\right) = e^{\left(-\frac{1}{25} \, t\right)} x_{1}\left(0\right) - e^{\left(-\frac{3}{25} \, t\right)} x_{1}\left(0\right)

Solución Aplicando los valores iniciales:x1(0)=25x_{1}(0)=25, x2(0)=0x_{2}(0)=0

x1(t)=252 e(−125 t)+252 e(−325 t)\displaystyle x_{1}\left(t\right) = \frac{25}{2} \, e^{\left(-\frac{1}{25} \, t\right)} + \frac{25}{2} \, e^{\left(-\frac{3}{25} \, t\right)}
x2(t)=25 e(−125 t)−25 e(−325 t)\displaystyle x_{2}\left(t\right) = 25 \, e^{\left(-\frac{1}{25} \, t\right)} - 25 \, e^{\left(-\frac{3}{25} \, t\right)}

Grafica de la colución del sistema de ecuaciones anterior

Se utilizan las condiciones iniciales: x1(0)=25x_{1}(0)=25, x2(0)=0x_{2}(0)=0 

### Graficando los resultados:
solx1 = solx1.rhs()
solx2 = solx2.rhs()
fig1 = plot(solx1, 0, 50*pi, color="blue", legend_label="x1(t)")
fig2 = plot(solx2, 0, 50*pi, color="green", legend_label="x2(t)")
show(fig1 + fig2)

Ecuación auxiliar:

m2+425 m+3625\displaystyle m^{2} + \frac{4}{25} \, m + \frac{3}{625}
[m=(−325)\displaystyle m = \left(-\frac{3}{25}\right), m=(−125)\displaystyle m = \left(-\frac{1}{25}\right)]

Calculando los valores de: X1X_{1}, X1′X_{1}' y X2X_{2}

X1(t) =
C1e(−125 t)+C2e(−325 t)\displaystyle C_{1} e^{\left(-\frac{1}{25} \, t\right)} + C_{2} e^{\left(-\frac{3}{25} \, t\right)}
X1'(t) =
−125 C1e(−125 t)−325 C2e(−325 t)\displaystyle -\frac{1}{25} \, C_{1} e^{\left(-\frac{1}{25} \, t\right)} - \frac{3}{25} \, C_{2} e^{\left(-\frac{3}{25} \, t\right)}
X2(t) =
2 C1e(−125 t)−2 C2e(−325 t)\displaystyle 2 \, C_{1} e^{\left(-\frac{1}{25} \, t\right)} - 2 \, C_{2} e^{\left(-\frac{3}{25} \, t\right)}

Encontrando las constantes C1 y C2

Para x1(t)x_{1}(t) y x2(t)x_{2}(t)

X1(t) =
C1e(−125 t)+C2e(−325 t)\displaystyle C_{1} e^{\left(-\frac{1}{25} \, t\right)} + C_{2} e^{\left(-\frac{3}{25} \, t\right)}
X2(t) =
2 C1e(−125 t)−2 C2e(−325 t)\displaystyle 2 \, C_{1} e^{\left(-\frac{1}{25} \, t\right)} - 2 \, C_{2} e^{\left(-\frac{3}{25} \, t\right)}
C1=(252)\displaystyle C_{1} = \left(\frac{25}{2}\right)
C2=(252)\displaystyle C_{2} = \left(\frac{25}{2}\right)

Por tanto los resultados son:

x1(t)=252e−125t+252e−325tx_{1}(t)=\frac{25}{2}e^{-\frac{1}{25}t} + \frac{25}{2}e^{-\frac{3}{25}t} x2(t)=25e−125t−25e−325tx_{2}(t)=25e^{-\frac{1}{25}t} - 25e^{-\frac{3}{25}t}