Nonhomogeneous ODEs: y'' + y' + y = 1 and y'' + y' + y = x

Worked Solution

How to Think About It: Because the operator $L[y] = y'' + y' + y$ is linear, the general solution is $y = y_h + y_p$: the full homogeneous family plus a single particular solution. With polynomial forcing and $0$ not a root of $r^2 + r + 1$, a polynomial of matching degree always works as $y_p$.

Quick Estimate: For large $x$ the homogeneous part $e^{-x/2}(\cdots)$ dies out, so the particular solution is the long-run behavior. For (a) the long-run level must satisfy $y = 1$ (all derivatives zero). For (b), if $y \approx x + b$ then $y' = 1$, $y'' = 0$, and $1 + x + b = x$ forces $b = -1$.

Formal Solution:

Part (a): $y'' + y' + y = 1$

*Step 1 -- Particular solution.* Try $y_p = A$ (constant). Then $y_p'' = y_p' = 0$ and the equation gives $A = 1$. So $y_p = 1$.

*Step 2 -- General solution.* $$y = e^{-x/2}\left(c_1\cos\tfrac{\sqrt3}{2}x + c_2\sin\tfrac{\sqrt3}{2}x\right) + 1.$$

Part (b): $y'' + y' + y = x$

*Step 1 -- Particular solution.* Try $y_p = Ax + B$. Then $y_p' = A$, $y_p'' = 0$, and $$0 + A + (Ax + B) = x \quad \Longrightarrow \quad Ax + (A + B) = x.$$ Matching coefficients: $A = 1$ and $A + B = 0$, so $B = -1$. Thus $y_p = x - 1$.

*Step 2 -- Check.* $y_p'' + y_p' + y_p = 0 + 1 + (x - 1) = x$. $\checkmark$

*Step 3 -- General solution.* $$y = e^{-x/2}\left(c_1\cos\tfrac{\sqrt3}{2}x + c_2\sin\tfrac{\sqrt3}{2}x\right) + x - 1.$$

*Remark.* Why does a polynomial ansatz work? $L$ maps polynomials of degree $n$ to polynomials of degree $n$ (the $y$ term preserves the degree because $0$ is not a characteristic root), so matching coefficients is a solvable triangular system.

Answer: (a) $y = e^{-x/2}\left(c_1\cos\frac{\sqrt3}{2}x + c_2\sin\frac{\sqrt3}{2}x\right) + 1$. (b) $y = e^{-x/2}\left(c_1\cos\frac{\sqrt3}{2}x + c_2\sin\frac{\sqrt3}{2}x\right) + x - 1$.