Find the general solution of y'' - a^2y = e^(bx) , with a≠0…
Find the general solution of $y'' - a^2y = e^{bx}$, with $a\neq0$ fixed; $b$ an arbitrary real (or complex) parameter.
Explanation of Method
Whether r=b coincides with a root r=±a of the characteristic equation fixes the multiplicity s=0 or 1, splitting the particular solution into three cases.
Step 1: Homogeneous Solution
\[ r^2-a^2=0 \implies r_1=a,\ r_2=-a. \]
\[ Y_h = C_1e^{ax}+C_2e^{-ax}. \]
Step 2: Particular Solution — Three Cases
The right-hand side is $f(x)=e^{bx}$, with exponent $b$. Whether $r=b$ coincides with a root of the characteristic equation fixes the multiplicity $s$ (0 or 1) and hence how many factors of $x$ must be attached to the trial function $y^*=Ax^se^{bx}$. Three cases arise.
Case I · $b\neq a$ and $b\neq -a$.
Here $r=b$ is not a root of the characteristic equation, so $s=0$: try $y^*=Ae^{bx}$.
\[ y^{*\prime\prime}=Ab^2e^{bx} \implies Ab^2e^{bx}-a^2Ae^{bx}=e^{bx} \implies A(b^2-a^2)=1 \implies A=\frac{1}{b^2-a^2}. \]
\[ y^* = \frac{e^{bx}}{b^2-a^2}. \]
Case II · $b=a$.
Here $r=b=a$ is a simple root of the characteristic equation (multiplicity $s=1$), so the trial function must be multiplied by $x^s=x$ to avoid duplicating the homogeneous term $e^{ax}$: try $y^*=Axe^{ax}$, $y^{*\prime}=Ae^{ax}(1+ax), \qquad y^{*\prime\prime}=Ae^{ax}(2a+a^2x)$.
Substitute into $y''-a^2y=e^{ax}$: \[Ae^{ax}(2a+a^2x)-a^2Axe^{ax}=e^{ax} \implies 2aA=1 \implies A=\frac{1}{2a}.\]
\[ y^* = \frac{xe^{ax}}{2a}. \]
Case III · $b=-a$.
Here $r=b=-a$ is a simple root of the characteristic equation (multiplicity $s=1$), so again the trial function is multiplied by $x^s=x$, this time to avoid duplicating $e^{-ax}$: try $y^*=Axe^{-ax}$, $y^{*\prime}=Ae^{-ax}(1-ax), \qquad y^{*\prime\prime}=Ae^{-ax}(-2a+a^2x)$.
Substitute into $y''-a^2y=e^{-ax}$: \[Ae^{-ax}(-2a+a^2x)-a^2Axe^{-ax}=e^{-ax} \implies -2aA=1 \implies A=-\frac{1}{2a}.\]
\[ y^* = -\frac{xe^{-ax}}{2a}. \]
Step 3: Verification
Case I: $y^{*\prime\prime}-a^2y^*=\dfrac{b^2e^{bx}}{b^2-a^2}-\dfrac{a^2e^{bx}}{b^2-a^2} =\dfrac{(b^2-a^2)e^{bx}}{b^2-a^2}=e^{bx}. $\checkmark$$
Case II: $y^{*\prime\prime}-a^2y^*=Ae^{ax}(2a+a^2x)-a^2Axe^{ax}=2aAe^{ax} =2a\cdot\dfrac{1}{2a}\,e^{ax}=e^{ax}. $\checkmark$$
Case III: $y^{*\prime\prime}-a^2y^*=Ae^{-ax}(-2a+a^2x)-a^2Axe^{-ax}=-2aAe^{-ax} =-2a\left(-\dfrac{1}{2a}\right)e^{-ax}=e^{-ax}. $\checkmark$$
Step 4: Summary of the Three Cases
\[ \begin{array}{lccl} \textbf{Case} & \textbf{Condition} & \textbf{Particular solution } y^* & \textbf{General solution } y \\ \hline \text{I} & b\neq\pm a & \dfrac{e^{bx}}{b^2-a^2} & C_1e^{ax}+C_2e^{-ax}+\dfrac{e^{bx}}{b^2-a^2} \\[8pt] \text{II} & b=a & \dfrac{xe^{ax}}{2a} & C_1e^{ax}+C_2e^{-ax}+\dfrac{xe^{ax}}{2a} \\[8pt] \text{III} & b=-a & -\dfrac{xe^{-ax}}{2a} & C_1e^{ax}+C_2e^{-ax}-\dfrac{xe^{-ax}}{2a} \end{array} \]
Step 5: General Solution
Result
Final answer:
\[\boxed{\,y = C_1e^{ax}+C_2e^{-ax}+ \begin{cases} \dfrac{e^{bx}}{b^2-a^2}, & b\neq \pm a \\[8pt] \dfrac{xe^{ax}}{2a}, & b=a \\[8pt] -\dfrac{xe^{-ax}}{2a}, & b=-a \end{cases}\,}\]
Remark
Remark. Cases II and III are instances of resonance: the forcing frequency $b$ matches one of the system’s own characteristic exponents $\pm a$, so the naive trial $Ae^{bx}$ already solves the homogeneous equation and cannot be matched to a nonzero right-hand side — multiplying by $x^s$ is exactly what restores a valid particular solution.