General and specific solutions

We posited in (lec:unique_solution_exists?) that the general solution to the ODE eq. ¿eq:ioode? is

\[ y_g(t) = y_h(t) + y_p(t). \qquad{(1)}\]

We have not and will not prove this, but simply propose it to be the case.
Working through a proof of this from, for instance, your differential equations textbook is of some value.

So, we already have $y_h$ and $y_p$, so finding $y_g$ is trivial.
What type of object is $y_g$?
The particular solution contributes only determined coefficients, but the homogeneous solution contributes $n$ “unknown” constants $C_i$.
This means $y_g$ inherits those constants and therefore is a family of solutions.

This leads us to our final step: applying the initial conditions to find the specific constants $C_i$ and thereby our specific solution $y$.

“Applying” the initial conditions is simply to subject $y_g$ to each of them.
For instance, if we have two initial conditions, such as

\[ y(0) = 2 \quad \text{and} \quad \left.\frac{d y}{d t}\right|_{t=0} = 0, \]

we construct two algebraic equations

\[ y_g(0) = 2 \quad \text{and} \quad \left.\frac{d y_g}{d t}\right|_{t=0} = 0, \]

which is a system of algebraic equations from which the two unknown constants $C_1$ and $C_2$ (from the homogeneous solution) can be solved.

Example 7.3

A general and a specific solution
Find the general solution for the equation

$$ \frac{d^2 y}{d t^2} + 5 \frac{d y}{d t} + 6 y = f(t), $$

with

f(t) = acos (ωt),

where a ∈ ℝ and ω = 5 rad/s.
Apply the following initial conditions to obtain a specific solution:

$$ y(0) = 3 \quad \text{and} \quad \left.\frac{d y}{d t}\right|_{t=0} = 0. $$

First, let’s find the homogeneous solution via the characteristic equation:

λ² + 5λ + 6 = 0

and its roots:

λ_1, 2 = − 2, − 3.

Both of these are real, so:

y_h(t) = C₁e^−2t + C₂e^−3t.

The particular solution should have the form:

y_p(t) = K₁cos (ωt) + K₂sin (ωt),

where K₁ and K₂ are our undetermined coefficients.

Substituting this into the ODE requires two derivatives:

$$ \begin{aligned} \dfrac{d y_p}{d t} &= - K_1 \omega \sin(\omega t) + K_2 \omega \cos(\omega t) \\ \dfrac{d^2 y_p}{d t^2} &= - K_1 \omega^2 \cos(\omega t) - K_2 \omega^2 \sin(\omega t) \end{aligned} $$

Substitution leads to the following algebraic system by equating cosine and sine terms:

$$ \begin{aligned} - K_1 \omega^2 + 5 K_2 \omega + 6 K_1 &= a \\ - K_2 \omega^2 - 5 K_1 \omega + 6 K_2 &= 0 \end{aligned} $$

Solving gives:

$$ \begin{aligned} K_1 &= -\frac{a(\omega^2 - 6)}{\omega^4 + 13 \omega^2 + 36} = -\frac{19a}{986} \\ K_2 &= \frac{5a\omega}{\omega^4 + 13 \omega^2 + 36} = \frac{25a}{986} \end{aligned} $$

So the particular solution is:

$$ y_p(t) = \frac{a}{986}(-19 \cos(5 t) + 25 \sin(5 t)). $$

The general solution is:

$$ \begin{aligned} y_g(t) &= y_h(t) + y_p(t) \\ &= C_1 e^{-2 t} + C_2 e^{-3 t} + \frac{a}{986}(-19 \cos(5 t) + 25 \sin(5 t)). \end{aligned} $$

Applying the initial conditions:

$$ \begin{aligned} C_1 + C_2 - \frac{19a}{986} &= 3 \\ -2 C_1 - 3 C_2 + \frac{125a}{986} &= 0 \end{aligned} $$

Solving:

$$ C_1 = \frac{-2a}{29} + 9, \qquad C_2 = \frac{3a}{34} - 6 $$

So the specific solution is:

$$ \begin{aligned} y(t) = \left(\frac{-2a}{29} + 9\right) e^{-2 t} + \left(\frac{3a}{34} - 6\right) e^{-3 t} + \frac{a}{986}(-19 \cos(5 t) + 25 \sin(5 t)). \end{aligned} $$

Online Resources for Section 7.5

No online resources.