I need to find a general solution of the following system of differential equations: \begin{align*} \begin{cases} x'(t) &= -y(t) \\ y'(t) &= (1.01) x(t) - y(t) \end{cases} \end{align*}

I'm not really sure what method to apply here. In matrix form this gives me: \begin{align*} \begin{pmatrix} x \\ y \end{pmatrix} ' = \begin{pmatrix} 0 & -1 \\ 1.01 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \end{align*} The eigenvalue method doesn't seem correct since the eigenvalues are difficult to handle because of the $1.01$ component.

EDIT: Following one of the suggestions below, differentiating the second equation gives $y'' = (1.01) x' - y'$. By using the first equation for $x'$, this becomes the homogeneous second order equation $$ y'' + y' + 1.01 y = 0. $$ The characteristic equation of this is $$ r^2 + r + 1.01 = 0 $$ so that $$ r_{1,2} = - \frac{1}{2} \pm i \frac{1}{2} \sqrt{3.04} $$ Hence a general solution is $$ y(t) = e^{-1/2 t} \big( c_1 \cos (\frac{1}{2} \sqrt{3.04} t) + c_2 \sin ( \frac{1}{2} \sqrt{3.04} t) \big) $$


Solve $$\begin{cases} \dfrac{\mathrm dx(t)}{\mathrm dt}&=-y(t),\\\\ \dfrac{\mathrm dy(t)}{\mathrm dt}&=\dfrac{101}{100}x(t)-y(t); \end{cases}$$

for $\{x(t),\,y(t)\}$.

Integrate $\frac{\mathrm dx(t)}{\mathrm dt}=-y(t)$ with respect to $t$: $$\int\frac{\mathrm dx(t)}{\mathrm dt}\,\mathrm dt=\int\Big(-y(t)\Big)\,\mathrm dt;$$

evaluate the integrals: $$x(t)=-\int y(t)\,\mathrm dt;$$

substitute $x(t)=-\int y(t)\,\mathrm dt$ in $\frac{\mathrm dy(t)}{\mathrm dt}=\frac{101}{100}x(t)-y(t)$: $$\frac{\mathrm dy(t)}{\mathrm dt}=-\frac{101}{100}\int y(t)\,\mathrm dt-y(t);$$

substitute $u(t)=\int y(t)\,\mathrm dt \implies y(t)=\frac{\mathrm du(t)}{\mathrm dt}$: $$\frac{\mathrm d}{\mathrm dt}\frac{\mathrm du(t)}{\mathrm dt}=-\frac{101}{100}\int \frac{\mathrm du(t)}{\mathrm dt}\,\mathrm dt-\frac{\mathrm du(t)}{\mathrm dt};$$

simplify: $$\frac{\mathrm d^2u(t)}{\mathrm dt^2}=-\frac{101}{100}u(t)-\frac{\mathrm du(t)}{\mathrm dt};$$

rearrange the equation: $$\frac{\mathrm d^2u(t)}{\mathrm dt^2}+\frac{\mathrm du(t)}{\mathrm dt}+\frac{101}{100}u(t)=0\tag{1}\label{ode1}$$

for the homogeneous linear second-order ordinary differential equation $\eqref{ode1}$, substitute $u(t)=e^{\lambda t}$: $$\frac{\mathrm d^2e^{\lambda t}}{\mathrm dt^2}+\frac{\mathrm de^{\lambda t}}{\mathrm dt}+\frac{101}{100}e^{\lambda t}=0;$$

evaluate the differentials: $$\lambda^2e^{\lambda t}+\lambda e^{\lambda t}+\frac{101}{100}e^{\lambda t}=0;$$

factor out $e^{\lambda t}$: $$e^{\lambda t}\left(\lambda^2+\lambda+\frac{101}{100}\right)=0;$$

since for any finite $\lambda$ and $t$: $e^{\lambda t}\neq0$, the roots must come from the polynomial $\lambda^2+\lambda+\frac{101}{100}$: $$\lambda^2+\lambda+\frac{101}{100}=0;$$

multiply both sides by $4$: $$(2\lambda)^2+4\lambda+\frac{404}{100}=0;$$

rewrite $\frac{404}{100}=1+\frac{304}{100}$: $$(2\lambda)^2+4\lambda+1+\frac{304}{100}=0;$$

complete the square: $$(2\lambda+1)^2+\frac{304}{100}=0;$$

isolate the term $(2\lambda+1)^2$: $$(2\lambda+1)^2=-\frac{304}{100};$$

take the square roots of both sides: $$2\lambda+1=\pm\frac{2i}{5}\sqrt{19};$$

isolate the term $2\lambda$: $$2\lambda=-1\pm\frac{2i}{5}\sqrt{19};$$

divide both sides by two: $$\lambda=\frac{-1\pm\frac{2i}{5}\sqrt{19}}{2};$$

then the solutions to $\eqref{ode1}$ are given by: $$u(t)=c_1e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}+c_2e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t},$$

where $c_1$ and $c_2$ are arbitrary constants and $i$ is the imaginary unit, such that $i^2=-1$; substitute back $u(t)=\int y(t)\,\mathrm dt$: $$\int y(t)\,\mathrm dt=c_1e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}+c_2e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t};$$

differentiate both sides with respect to $t$: $$\frac{\mathrm d}{\mathrm dt}\int y(t)\,\mathrm dt=\frac{\mathrm d}{\mathrm dt}\left(c_1e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}+c_2e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}\right);$$

evaluate the differentials: $$y(t)=\frac{c_1}{2}\left(2i\sqrt{19}/5-1\right)e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-\frac{c_2}{2}\left(1+2i\sqrt{19}/5\right)e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t};$$

substitute for $y(t)$ in $\frac{\mathrm dx(t)}{\mathrm dt}=-y(t)$: $$\frac{\mathrm dx(t)}{\mathrm dt}=-\left(\frac{c_1}{2}\left(2i\sqrt{19}/5-1\right)e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-\frac{c_2}{2}\left(1+2i\sqrt{19}/5\right)e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}\right);$$

integrate both sides with respect to $t$: $$\int\frac{\mathrm dx(t)}{\mathrm dt}\mathrm dt=-\int\left(\frac{c_1}{2}\left(2i\sqrt{19}/5-1\right)e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-\frac{c_2}{2}\left(1+2i\sqrt{19}/5\right)e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}\right)\mathrm dt;$$

evaluate the integrals: $$x(t)=-c_1e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-c_2e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}+c_3,$$

where $c_3$ is an arbitrary constant; collect the solutions: $$\therefore\begin{cases} x(t)=\boxed{c_3-c_1e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-c_2e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}}\phantom{.},\\\\ y(t)=\boxed{\frac{c_1}{2}\left(2i\sqrt{19}/5-1\right)e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-\frac{c_2}{2}\left(1+2i\sqrt{19}/5\right)e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}}\phantom{.}. \end{cases}$$


Hint If you don't want to diagonalize the matrix, you could readily transform this into a first-order system by differentiating the second equation to produce a second-order equation in $x', y', y''$ from this, and then use the first equation to eliminate $x'$, leaving just as single, second-order equation in $y$ (alone).

  • $\begingroup$ Thanks. I edited my original post and applied your method. Can you check if my work is ok? I only found $y(t)$. Should I now find $x(t)$ from equation 1, by integrating ? $\endgroup$ – Kamil Oct 10 '15 at 16:38
  • $\begingroup$ Your solution looks good to me, though I might write the expression without decimals (note that $\sqrt{3.04} = \frac{2}{5}\sqrt{19}$, which looks better to my eye, anyway). You could integrate, but you could also differentiate your expression for $y$ and then substitute your expressions for $y, y'$ in the second equation and solve for $x$. For one, this spares you from thinking about how to resolve the new constant that appears during integration. $\endgroup$ – Travis Willse Oct 10 '15 at 16:54

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