Unit D4: Nonhomogeneous Second-Order ODEs

Given that $y_p = t - \tfrac12$ solves $y'' - y = -t + \tfrac12$, write the general solution.

The associated homogeneous equation is $y'' - y = 0$ with characteristic equation $r^2 - 1 = 0$, roots $r = \pm 1$. Hence $y_h = C_1 e^{t} + C_2 e^{-t}$.

Check: $y_p'' = 0$ and $-y_p = -t + \tfrac12$, so $y_p'' - y_p = -t + \tfrac12$, matching the right side.

Solve the initial value problem $y'' - y = -t + \tfrac12$ with $y(0) = 2$, $y'(0) = 0$, using the structure from Worked Example 1.1.

The general solution is $y(t) = C_1 e^{t} + C_2 e^{-t} + t - \tfrac12$. Differentiate: $y'(t) = C_1 e^{t} - C_2 e^{-t} + 1$. Now impose the data on the full solution, not on $y_h$.

Adding and subtracting gives $C_1 = \tfrac34$ and $C_2 = \tfrac74$. The solution of the initial value problem is

The particular part $t - \tfrac12$ contributes $-\tfrac12$ to $y(0)$ and $+1$ to $y'(0)$, which is exactly why the constants differ from what the bare homogeneous problem would give.

Suppose $y_p = \tfrac12 t^2$ is claimed to be a particular solution of $y'' + 4y = g(t)$ for some forcing $g$. Identify $g$ and write the general solution.

Apply the operator directly. Here $y_p' = t$ and $y_p'' = 1$, so

The homogeneous equation $y'' + 4y = 0$ has characteristic roots $r = \pm 2i$, so $y_h = C_1\cos 2t + C_2\sin 2t$. The general solution of $y'' + 4y = 1 + 2t^2$ is therefore

This reverse reading is a useful check: applying $L$ to any candidate $y_p$ must reproduce the forcing exactly, or the candidate is wrong.

Define $L[y] = a y'' + b y' + c y$. Linearity means $L[\alpha u + \beta v] = \alpha L[u] + \beta L[v]$ for constants $\alpha, \beta$, which follows because differentiation is linear and multiplication by the coefficients $a, b, c$ is linear.

Suppose $L[y_p] = g$. Let $y$ be any other solution, so $L[y] = g$. Then

so $w = y - y_p$ solves the homogeneous equation, meaning $w = y_h$ for suitable constants. Therefore $y = y_h + y_p$. Conversely, for any homogeneous $y_h$, $L[y_h + y_p] = 0 + g = g$, so every such sum is a solution. The solution set is the single particular solution shifted by the two-dimensional space of homogeneous solutions.

In linear-algebra terms, the solution set of $L[y] = g$ is an affine subspace: a fixed translate $y_p$ of $\ker L$. The kernel $\ker L = \{y : L[y] = 0\}$ is the two-dimensional space of homogeneous solutions (existence and uniqueness pins the dimension at exactly two). Choosing a different particular solution $\tilde y_p$ only re-labels constants, since $\tilde y_p - y_p \in \ker L$ is absorbed into $C_1 y_1 + C_2 y_2$. That is why we may take the simplest $y_p$ we can find.

Solve $y'' - 3y' + 2y = 4 e^{3t}$.

Homogeneous: $r^2 - 3r + 2 = (r-1)(r-2) = 0$, so $y_h = C_1 e^{t} + C_2 e^{2t}$. Since $3$ is not a root, no overlap, so try $y_p = A e^{3t}$.

so $2A = 4$ and $A = 2$. Hence

Find a particular solution of $y'' + y = 2t + \cos 2t$.

Split by superposition. For $g_1 = 2t$, try $y_1 = A t + B$. Then $y_1'' + y_1 = 0 + At + B = 2t$, forcing $A = 2$, $B = 0$, so $y_1 = 2t$.

For $g_2 = \cos 2t$, since $\pm 2i$ are not roots of $r^2 + 1 = 0$, try $y_2 = C\cos 2t + D\sin 2t$. Then $y_2'' = -4C\cos 2t - 4D\sin 2t$, so

giving $-3C = 1$, $D = 0$, so $C = -\tfrac13$. Therefore

Solve $y'' + 4y' + 5y = e^{-t}\cos 2t$.

Homogeneous: $r^2 + 4r + 5 = 0$ gives $r = -2 \pm i$, so $y_h = e^{-2t}(C_1\cos t + C_2\sin t)$. The forcing $e^{-t}\cos 2t$ does not appear in $y_h$, so there is no resonance. Use the product trial $y_p = e^{-t}(A\cos 2t + B\sin 2t)$. Write $u = A\cos 2t + B\sin 2t$ so $y_p = e^{-t}u$. Then

Substitute into $y'' + 4y' + 5y$:

With $u = A\cos 2t + B\sin 2t$: $u'' = -4u$ and $u' = -2A\sin 2t + 2B\cos 2t$. So $u'' + 2u' + 2u = -2u + 2u' = -2(A\cos 2t + B\sin 2t) + 2(-2A\sin 2t + 2B\cos 2t)$, which equals $(-2A + 4B)\cos 2t + (-2B - 4A)\sin 2t$. Setting this equal to $\cos 2t$:

The second gives $B = -2A$; substituting, $-2A - 8A = 1$, so $A = -\tfrac{1}{10}$ and $B = \tfrac15$. Hence

The key technique is factoring out $e^{-t}$ and shifting the operator onto $u$, which turns a messy product into a plain trigonometric match.

Solve $y'' - 3y' + 2y = e^{t}$.

Homogeneous roots are $r = 1, 2$, so $y_h = C_1 e^{t} + C_2 e^{2t}$. The forcing $e^{t}$ matches the root $r = 1$ (simple), so the plain trial $A e^{t}$ would give zero. Modify to $y_p = A t e^{t}$.

Compute $y_p' = A(1 + t)e^{t}$ and $y_p'' = A(2 + t)e^{t}$. Then

Setting $-A = 1$ gives $A = -1$, so

Solve $y'' + \omega^2 y = \cos\omega t$, the undamped oscillator driven at its natural frequency $\omega$.

Homogeneous: $r^2 + \omega^2 = 0$ gives $r = \pm i\omega$, so $y_h = C_1\cos\omega t + C_2\sin\omega t$. The forcing $\cos\omega t$ already appears in $y_h$, so modify: $y_p = t(A\cos\omega t + B\sin\omega t)$.

Differentiating twice and substituting (the cosine-only terms cancel by the homogeneous structure) yields

Matching, $2B\omega = 1$ and $-2A\omega = 0$, so $A = 0$ and $B = \dfrac{1}{2\omega}$. Hence

The amplitude $t/(2\omega)$ grows linearly in time: the hallmark of resonance.

Solve $y'' - 2y' + y = e^{t}$. The characteristic equation $r^2 - 2r + 1 = (r-1)^2 = 0$ has the double root $r = 1$, so $y_h = (C_1 + C_2 t)e^{t}$.

The forcing $e^{t}$ matches the double root. Both $e^{t}$ and $t e^{t}$ already live in $y_h$, so the smallest clean modification multiplies the base trial $A e^{t}$ by $t^2$: take $y_p = A t^2 e^{t}$.

Differentiate using the product rule. With $y_p = A t^2 e^{t}$,

Substitute:

The $t^2$ and $t$ terms cancel exactly (they must, since $(r-1)^2$ annihilates $e^{t}$ and $te^{t}$), leaving $2A e^{t} = e^{t}$, so $A = \tfrac12$. Hence

Write the operator in factored form using its characteristic roots: $L = a(D - r_1)(D - r_2)$, where $D = d/dt$. Suppose the forcing is $e^{\alpha t}$ and $\alpha = r_1$ is a simple root, while $r_2 \ne \alpha$.

The operator $(D - \alpha)$ has the shift property $(D - \alpha)\big(e^{\alpha t} h(t)\big) = e^{\alpha t} h'(t)$, which follows from the product rule since the $\alpha e^{\alpha t} h$ terms cancel. Apply $L$ to the candidate $y_p = t^s e^{\alpha t}$ (constant absorbed):

For this to produce a nonzero multiple of $e^{\alpha t}$ (matching the forcing $e^{\alpha t}$, which is $t^0 e^{\alpha t}$), we need $s\,t^{s-1}$ to contain a constant term, i.e. $s = 1$. With $s = 1$ the inner result is $e^{\alpha t}$, and applying $(D - r_2)$ gives $(\alpha - r_2)e^{\alpha t}$, a nonzero constant times $e^{\alpha t}$ because $\alpha \ne r_2$. So $s = 1$ works and $s = 0$ fails (it gets annihilated by $(D-\alpha)$).

If instead $\alpha$ is a double root, $L = a(D - \alpha)^2$, and $(D - \alpha)^2(t^s e^{\alpha t}) = e^{\alpha t}\,\dfrac{d^2}{dt^2}t^s = e^{\alpha t}\,s(s-1)t^{s-2}$. A constant term appears only when $s = 2$. This is precisely the modification rule: $s$ equals the multiplicity of the matched root, the smallest power that survives all the annihilating factors.

Start from $y_p = u_1 y_1 + u_2 y_2$. Differentiate:

We have two unknown functions but only one equation to satisfy, so we are free to impose one constraint. Choose

which kills the messy terms and leaves $y_p' = u_1 y_1' + u_2 y_2'$. Differentiate once more:

Substitute $y_p, y_p', y_p''$ into $y'' + p y' + q y = f$ and group by $u_1$ and $u_2$. Each group is $y_i'' + p y_i' + q y_i = 0$ because $y_1, y_2$ solve the homogeneous equation, so everything cancels except

Together with the constraint we have the linear system

By Cramer's rule, with $W = y_1 y_2' - y_2 y_1'$,

Integrate to recover $u_1, u_2$. Since $y_1, y_2$ are independent, $W \ne 0$, so the system is always solvable.

Solve $y'' + y = \sec t$ on an interval where $\sec t$ is continuous.

Homogeneous solutions: $y_1 = \cos t$, $y_2 = \sin t$. The equation is already in standard form, so $f = \sec t$. The Wronskian is

Then

Integrate: $u_1 = \ln|\cos t|$ and $u_2 = t$. Therefore

Solve $4y'' - 4y' + y = e^{t/2}\sqrt{1 - t^2}$ on $(-1, 1)$. The forcing has no finite trial form, so use variation of parameters, but first put the equation in standard form.

Divide by the leading coefficient $4$:

The characteristic equation $4r^2 - 4r + 1 = (2r - 1)^2 = 0$ has the double root $r = \tfrac12$, so $y_1 = e^{t/2}$ and $y_2 = t\,e^{t/2}$. Compute the Wronskian:

Then

Integrate: $u_1 = \tfrac{1}{12}(1 - t^2)^{3/2}$ (using $\int t\sqrt{1-t^2}\,dt = -\tfrac13(1-t^2)^{3/2}$), and $u_2 = \tfrac14\!\left(\tfrac{t}{2}\sqrt{1-t^2} + \tfrac12\arcsin t\right)$. Then $y_p = u_1 y_1 + u_2 y_2$. The point is structural: had we forgotten to divide by $4$ and used $f = e^{t/2}\sqrt{1-t^2}$, every coefficient would have been four times too large.

Solve $y'' - y = e^{2t}$ two ways. Homogeneous: $r^2 - 1 = 0$, so $y_1 = e^{t}$, $y_2 = e^{-t}$, and $y_h = C_1 e^{t} + C_2 e^{-t}$.

Undetermined coefficients. Since $2$ is not a root, try $y_p = A e^{2t}$. Then $y_p'' - y_p = (4A - A)e^{2t} = 3A e^{2t} = e^{2t}$, so $A = \tfrac13$ and $y_p = \tfrac13 e^{2t}$.

Variation of parameters. The Wronskian is $W = e^{t}(-e^{-t}) - e^{-t}(e^{t}) = -2$, and $f = e^{2t}$.

Integrate: $u_1 = \tfrac12 e^{t}$, $u_2 = -\tfrac16 e^{3t}$. Then

Both methods give the same $y_p = \tfrac13 e^{2t}$, as they must.

Solve $y'' - y = e^{t}$. Homogeneous: $r^2 - 1 = 0$, so $y_1 = e^{t}$, $y_2 = e^{-t}$, $y_h = C_1 e^{t} + C_2 e^{-t}$. The forcing $e^{t}$ matches the simple root $r = 1$: this is resonant.

Undetermined coefficients. The modification rule (Section 3) forces $y_p = A t e^{t}$. Then $y_p' = A(1+t)e^{t}$, $y_p'' = A(2+t)e^{t}$, so $y_p'' - y_p = A(2 + t - t)e^{t} = 2A e^{t} = e^{t}$, giving $A = \tfrac12$ and $y_p = \tfrac12 t e^{t}$.

Variation of parameters. No special case is needed. The Wronskian is $W = e^{t}(-e^{-t}) - e^{-t}(e^{t}) = -2$, and $f = e^{t}$.

Integrate: $u_1 = \tfrac12 t$ and $u_2 = -\tfrac14 e^{2t}$. Then

The extra $-\tfrac14 e^{t}$ is a homogeneous solution, so it merges into $C_1 e^{t}$. After absorbing it, the particular solution is again $\tfrac12 t e^{t}$. Notice the factor of $t$ appeared automatically from $\int u_1' = \int \tfrac12\,dt$, with no modification rule invoked.

Solve $y'' + 2y' + 2y = 5\cos t$ and identify the transient and steady-state parts.

Homogeneous: $r^2 + 2r + 2 = 0$ gives $r = -1 \pm i$, so $y_h = e^{-t}(C_1\cos t + C_2\sin t)$. This decays to zero as $t \to \infty$: it is the transient.

For the steady state, try $y_p = A\cos t + B\sin t$. Then $y_p'' = -A\cos t - B\sin t$ and $y_p' = -A\sin t + B\cos t$, so

Matching: $A + 2B = 5$ and $B - 2A = 0$, so $B = 2A$ and $A + 4A = 5$, giving $A = 1$, $B = 2$. Thus the steady state is $y_p = \cos t + 2\sin t$, and

A series circuit with $L = 1$, $R = 3$, $1/C = 2$ is driven by $E(t) = 10\sin t$. The charge obeys $q'' + 3q' + 2q = 10\sin t$. Find the steady-state charge.

Homogeneous roots: $r^2 + 3r + 2 = (r+1)(r+2) = 0$, both negative, so $q_h$ is transient. For steady state try $q_p = A\cos t + B\sin t$.

Match: $A + 3B = 0$ and $B - 3A = 10$. From the first, $A = -3B$; substituting, $B - 3(-3B) = 10B = 10$, so $B = 1$ and $A = -3$. The steady-state charge is

For the steady state $y_p = \cos t + 2\sin t$ found in Worked Example 6.1, express it as a single shifted cosine $R\cos(t - \phi)$.

The amplitude and phase come from $R = \sqrt{A^2 + B^2}$ and $\tan\phi = B/A$ for $y_p = A\cos t + B\sin t = R\cos(t - \phi)$:

So the steady-state oscillation has amplitude $\sqrt5 \approx 2.236$ and lags the drive by about $63.4^\circ$. Cross-check against the amplitude formula with $m = 1$, $c = 2$, $k = 2$, $F_0 = 5$, $\omega = 1$:

The amplitude formula and the direct coefficient computation agree, which is a reliable way to catch arithmetic slips.

Take $m y'' + c y' + k y = F_0\cos\omega t$ with steady-state amplitude

To find the resonant frequency, minimize the denominator $D(\omega) = (k - m\omega^2)^2 + c^2\omega^2$. Differentiate with respect to $\omega$ and set to zero:

Discarding $\omega = 0$, the bracket gives

When damping $c$ is small, $\omega_{\text{res}}^2 \approx k/m$, the square of the natural frequency, and the denominator nearly vanishes, so $A(\omega)$ has a tall sharp peak. As $c$ grows the peak shifts left and flattens, and once $c^2 \ge 2mk$ no interior peak survives: heavily damped systems do not resonate.

Construct $G(t, \tau)$ for $y'' + y = f(t)$.

Homogeneous solutions are $y_1 = \cos t$, $y_2 = \sin t$, with Wronskian $W = 1$. Substitute into the kernel:

Therefore the particular solution to $y'' + y = f$ is

a single clean integral that solves the equation for any continuous $f$.

Use the kernel $G(t,\tau) = \sin(t - \tau)$ from Worked Example 7.1 to solve $y'' + y = 1$ with $y(0) = 0$, $y'(0) = 0$, taking $t_0 = 0$.

The Green's-function form gives the solution that already satisfies the zero initial data:

Substitute $s = t - \tau$, so $ds = -d\tau$ and the limits run from $s = t$ down to $s = 0$:

Check directly: $y = 1 - \cos t$ gives $y'' = \cos t$, so $y'' + y = \cos t + 1 - \cos t = 1$, and $y(0) = 0$, $y'(0) = \sin 0 = 0$. The Green's function delivers the full solution of the initial value problem in one integral, with the homogeneous part already tuned to the zero initial conditions.

For $y'' + 0.1\,y' + y = \cos\omega t$ (so $m = 1$, $c = 0.1$, $k = 1$, $F_0 = 1$), estimate the peak steady-state amplitude and the frequency where it occurs.

From the resonance result, $\omega_{\text{res}}^2 = \dfrac{k}{m} - \dfrac{c^2}{2m^2} = 1 - \dfrac{0.01}{2} = 0.995$, so $\omega_{\text{res}} \approx 0.9975$. The peak amplitude is

With light damping the peak amplitude is roughly $F_0/(c\,\omega_{\text{res}}) \approx 1/(0.1\cdot 1) = 10$, ten times the static deflection $F_0/k = 1$. This factor, the quality factor $Q$, measures how sharply the system resonates: small $c$ produces a tall narrow peak.