Solve the system of differential equations: tdot x+6x-y=0…
Solve the system of differential equations: $$ t\dot x+6x-y=0,\qquad t\dot y+20x-6y=0 $$
Explanation of Method
As in Problem 61, the substitution $t=e^s$ removes the Euler-type $t\,d/dt$ operator, reducing this homogeneous system to a constant-coefficient eigenvalue problem whose solutions translate back into pure power-law functions of $t$.
- Substitute $t=e^s$ to convert to a constant-coefficient system in $s$.
- Find the eigenvalues and eigenvectors of the resulting matrix.
- Assemble the general solution in $s$.
- Translate back to $t$ using $e^{\pm4s}=t^{\pm4}$.
Step 1 · Substitution
With $t=e^s$, the system becomes the constant-coefficient homogeneous system $\mathbf u'=A\mathbf u$ with
$$ A=\begin{pmatrix}-6&1\\-20&6\end{pmatrix}. $$
Step 2 · Eigenvalues and eigenvectors
$\det(A-\lambda I)=\lambda^2-16=0$ gives $\lambda_1=4$ with eigenvector $(1,10)$ and $\lambda_2=-4$ with eigenvector $(1,2)$.
Step 3 · Translate back to $t$
Using $e^{4s}=t^4$ and $e^{-4s}=t^{-4}$,
$$ x(t)=C_1t^4+C_2t^{-4},\qquad y(t)=10C_1t^4+2C_2t^{-4}. $$
Step 4 · Verification
Substituting back into both original Euler-type equations confirms each holds identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues (in $s$) | $\lambda_1=4,\ \lambda_2=-4$ |
| Solution form (in $t$) | $t^4$ and $t^{-4}$ power-law modes |
| Wronskian check | $\det\begin{pmatrix}1&1\\10&2\end{pmatrix}=-8\neq0$ at $t=1$, confirming independence |
Additional Notes
- Being purely homogeneous, this Euler system needs no particular-solution step — only the eigenvalue/eigenvector analysis of the transformed constant-coefficient matrix.
- For $t<0$ the power functions $t^{\pm4}$ extend smoothly since the exponents are integers; for non-integer Euler exponents one would need $|t|^{\pm}$ instead.
- Verified by direct substitution into both original Euler-type equations.
Result
The general solution of the system is: