Solve the system of differential equations: tdot x-x-3y=t…
Solve the system of differential equations: $$ t\dot x-x-3y=t,\qquad t\dot y-x+y=0 $$
Explanation of Method
The substitution $t=e^s$ converts the Euler-type operator $t\,d/dt$ into the constant-coefficient operator $d/ds$, turning the system into an ordinary constant-coefficient linear system that can be solved by the usual eigenvalue methods and then translated back to $t$.
- Substitute $t=e^s$ so that $t\dot x=x'$ (derivatives with respect to $s$).
- Solve the resulting constant-coefficient system for eigenvalues and eigenvectors.
- Find the particular solution for the $e^s$ forcing term.
- Translate the solution back to the original variable $t=e^s$.
Step 1 · Substitution
With $t=e^s$, $t\dot x=x'$ and $t\dot y=y'$ (primes denote $d/ds$), the system becomes
$$ x'-x-3y=e^s,\qquad y'-x+y=0. $$
Step 2 · Eigenvalues and eigenvectors
With $A=\begin{pmatrix}1&3\\1&-1\end{pmatrix}$, $\det(A-\lambda I)=\lambda^2-4=0$, giving $\lambda_1=2$ with eigenvector $(3,1)$ and $\lambda_2=-2$ with eigenvector $(-1,1)$.
Step 3 · Particular solution
Since $\mu=1$ is not an eigenvalue, trying $e^s(a,b)$ and solving $(A-I)(a,b)^\top=(-1,0)$ gives $a=-\tfrac23,\ b=-\tfrac13$.
Step 4 · Translate back to $t$
Using $e^s=t$, $e^{2s}=t^2$, $e^{-2s}=1/t^2$, the general solution in $s$ becomes
$$ x(t)=3C_1t^2-\frac{C_2}{t^2}-\frac{2}{3}t,\qquad y(t)=C_1t^2+\frac{C_2}{t^2}-\frac{1}{3}t. $$
Step 5 · Verification
Substituting back into both original Euler-type equations confirms each holds identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $t=e^s$, converting $t\,d/dt\to d/ds$ |
| Eigenvalues (in $s$) | $\lambda_1=2,\ \lambda_2=-2$ |
| Solution form (in $t$) | power functions $t^2,\ t^{-2}$, plus a linear term |
Additional Notes
- The Euler substitution $t=e^s$ is the standard technique for any linear system where every term has the form $t\,\dot{(\cdot)}$ or $(\cdot)$ — it always converts to constant coefficients in $s$.
- Because $\lambda=2,-2$ in $s$ correspond to $t^2$ and $t^{-2}$ in the original variable, Euler systems naturally produce power-law (not exponential) solutions in $t$.
- Verified by direct substitution into both original Euler-type equations.
Result
The general solution of the system is: