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: Read the homogeneous part.
The complementary solution is $y_c=C_1e^{-x}+C_2e^{2x}$, so the roots are $m=-1$ and $m=2$.

Step 2: Build the auxiliary equation.
\[ (m+1)(m-2)=m^2-m-2=0 \]
Comparing with $m^2+\beta m+\gamma=0$ gives $\beta=-1$ and $\gamma=-2$.

Step 3: Choose the particular form.
The right side has $e^{-x}$, which already solves the homogeneous part, so try $y_p=\alpha x e^{-x}$.

Step 4: Apply the operator.
For $P(m)=m^2-m-2$ with a repeated style input at $m=-1$, the rule gives $L(x e^{-x})=P'(-1)e^{-x}$, and $P'(-1)=2(-1)-1=-3$.

Step 5: Match the right side.
So $\alpha(-3)e^{-x}=-e^{-x}$, which gives $\alpha=\frac13$.

Step 6: Final combination.
\[ \alpha(\beta+\gamma)=\tfrac13(-1-2)=-1 \]
\[ \boxed{-1.0} \]