Question

Let \(\alpha, \beta\) and \(\gamma\) be fixed real numbers such that

Hint:

If the RHS term \(e^{ax}\) is already present in the complementary solution, multiply the particular integral trial by \(x\).

Answer 1

Correct Answer: -1

Step 1: Identify the complementary solution.
The complementary solution is
\[ y_c=C_1e^{-x}+C_2e^{2x} \]

Step 2: Find roots of auxiliary equation.
The roots are
\[ m=-1,\quad m=2 \]

Step 3: Form the auxiliary equation.
\[ (m+1)(m-2)=0 \]
\[ m^2-m-2=0 \]

Step 4: Compare with standard auxiliary equation.
For
\[ y''+\beta y'+\gamma y=0 \] auxiliary equation is
\[ m^2+\beta m+\gamma=0 \]
So, comparing,
\[ \beta=-1,\qquad \gamma=-2 \]

Step 5: Find particular integral form.
Since \(e^{-x}\) is already part of complementary solution, particular solution is of the form
\[ y_p=\alpha xe^{-x} \]

Step 6: Apply operator to \(xe^{-x}\).
For \(P(m)=m^2-m-2\), we use
\[ P'(m)=2m-1 \]
At \(m=-1\),
\[ P'(-1)=2(-1)-1=-3 \]
Thus,
\[ L(xe^{-x})=-3e^{-x} \]

Step 7: Compare with right hand side.
\[ \alpha(-3e^{-x})=-e^{-x} \]
\[ -3\alpha=-1 \]
\[ \alpha=\frac{1}{3} \]
Now,
\[ \alpha(\beta+\gamma)=\frac{1}{3}(-1-2) \]
\[ =\frac{1}{3}(-3)=-1 \]
\[ \boxed{-1.0} \]