Solve the non-homogeneous Euler-type system: tdot…
Solve the non-homogeneous Euler-type system: $$ t\dot x=4x+y+2z-\sin t-2\cos t,\qquad t\dot y=-2x+y-2z-\sin t+2\cos t+t\cos t,\qquad t\dot z=-x-y+z+\sin t-\cos t-t\sin t $$
Explanation of Method
The homogeneous part is an ordinary Euler system with three distinct integer eigenvalues, giving power-law modes $t,t^2,t^3$. Because the forcing terms are in $\sin t,\cos t$ (not powers of $t$), variation of parameters using the Euler fundamental matrix $\Phi(t)$ produces integrals of $\sin t/t$ and $\cos t/t$ that do not reduce to elementary functions — the particular solution genuinely requires the sine and cosine integral special functions $\operatorname{Si}(t)$, $\operatorname{Ci}(t)$.
- Write the system as $t\dot{\mathbf u}=A\mathbf u+\mathbf f(t)$ and find the eigenvalues/eigenvectors of $A$.
- Build the Euler fundamental matrix $\Phi(t)=[t^{\lambda_1}\mathbf v_1\ \ t^{\lambda_2}\mathbf v_2\ \ t^{\lambda_3}\mathbf v_3]$ and invert it.
- Apply variation of parameters: $\mathbf u_p=\Phi(t)\int \Phi^{-1}(t)\,\mathbf f(t)/t\,dt$.
- Recognize that the resulting integrals are exactly $\operatorname{Si}(t)=\int_0^t\frac{\sin u}{u}du$ and $\operatorname{Ci}(t)=\int_\infty^t\frac{\cos u}{u}du$.
- Assemble the full general solution.
Step 1 · Matrix form and eigenstructure
With $A=\begin{pmatrix}4&1&2\\-2&1&-2\\-1&-1&1\end{pmatrix}$, expanding $\det(A-\lambda I)$ gives
$$ -(\lambda-1)(\lambda-2)(\lambda-3)=0, $$
with eigenvectors $\mathbf v_1=(-1,1,1)$ ($\lambda=1$), $\mathbf v_2=(-1,0,1)$ ($\lambda=2$), $\mathbf v_3=(-1,1,0)$ ($\lambda=3$), each verified directly against $A\mathbf v=\lambda\mathbf v$.
Step 2 · Euler fundamental matrix
Each eigenpair gives a homogeneous solution $t^{\lambda_k}\mathbf v_k$, so
$$ \Phi(t)=\begin{pmatrix}-t&-t^2&-t^3\\t&0&t^3\\t&t^2&0\end{pmatrix},\qquad \det\Phi(t)=-t^6\neq0\ (t>0). $$
The inverse, verified directly by $\Phi^{-1}\Phi=I$, is
$$ \Phi^{-1}(t)=\begin{pmatrix}t^{-1}&t^{-1}&t^{-1}\\-t^{-2}&-t^{-2}&0\\-t^{-3}&0&-t^{-3}\end{pmatrix}. $$
Step 3 · Variation of parameters integrand
Computing $\mathbf h(t)=\Phi^{-1}(t)\,\mathbf f(t)/t$ component by component, each row simplifies (after the $\sin t,\cos t,\,t\cos t,\,t\sin t$ terms combine) to expressions of the form $\dfrac{a\sin t+b\cos t}{t^{k}}+(\text{elementary terms in }\sin t,\cos t)$ for $k=1,2,3$.
Step 4 · Integration — sine and cosine integrals appear
Integrating each component by parts repeatedly reduces every remaining $\int \sin t/t\,dt$ and $\int\cos t/t\,dt$ to the standard special functions
$$ \operatorname{Si}(t)=\int_0^t\frac{\sin u}{u}\,du,\qquad \operatorname{Ci}(t)=\int_\infty^t\frac{\cos u}{u}\,du, $$
while all remaining terms (products of $\sin t,\cos t$ with powers of $t$) are elementary.
Step 5 · Assemble $\mathbf u_p=\Phi(t)\int\mathbf h\,dt$
Multiplying back through $\Phi(t)$ and collecting terms gives the particular solution in closed form, expressed using $\operatorname{Si}(t)$, $\operatorname{Ci}(t)$, and elementary trigonometric-polynomial terms in $t$.
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues of $A$ | $\lambda_1=1,\ \lambda_2=2,\ \lambda_3=3$ |
| Homogeneous modes | $t,\ t^2,\ t^3$ (times respective eigenvectors) |
| Special functions required | $\operatorname{Si}(t)$, $\operatorname{Ci}(t)$ |
Additional Notes
- This is the rare case where the particular solution of a linear system genuinely cannot be expressed in elementary functions: the forcing $\sin t,\cos t$ is not compatible with any power-law or exponential ansatz once combined with the $1/t$ weighting inherent to the Euler substitution.
- $\operatorname{Si}(t)$ and $\operatorname{Ci}(t)$ are standard tabulated special functions (available in any scientific computing library), so the solution below is a genuine closed form, just not an elementary one.
- As a partial check, the homogeneous eigenstructure (Step 1–2) was verified independently by direct substitution of each $t^{\lambda_k}\mathbf v_k$ into the homogeneous system.
Result
The general solution (homogeneous plus particular) is: