Question

Consider the differential equation \(y^{\prime \prime}+\omega_{0}^{2} y=F \cos \omega t\). (a) Determine the complementary solution of this differential equation. (b) Use the method of undetermined coefficients to find a particular solution in each of the cases: (i) \(\omega=\omega_{1} \neq \omega_{0}\), (ii) \(\omega=\omega_{0}\).

Short answer

2. What is the particular solution for case (i) when \(\omega = \omega_1\)? 3. What is the particular solution for case (ii) when \(\omega = \omega_0\)? 4. What are the general solutions for each case?

Step-by-step solution

Determine the complementary solution

To find the complementary solution, we need to solve the homogeneous equation, which can be represented as: \[y^{\prime\prime} + \omega_0^2 y = 0\] To solve this equation, we'll try a solution of the form \[y = e^{r t}\] Taking derivatives: \[y' = r e^{r t}\] \[y'' = r^2 e^{r t}\] Now, we can substitute the function and its derivatives into the homogeneous equation: \[r^2 e^{rt} + \omega_0^2 e^{rt} = 0\] The exponential term is never 0, so we can divide both sides of the equation by it: \[r^2 + \omega_0^2 = 0\] Now, we can solve this equation for \(r\): \[r^2 = - \omega_0^2\] \[r = \pm \omega_0 i\] Putting this back into our original form for y, we get the complementary solution: \[y_c(t) = c_1 \cos(\omega_0 t) + c_2 \sin(\omega_0 t)\]

Find a particular solution when \(\omega=\omega_{1}\) (Case i)

For case (i), we use the method of undetermined coefficients. We know that the forcing function is of the form \(\cos\omega t\). Since the coefficients are constants, the complementary solution does not contain this form of function. We can then assume a particular solution of the form \[y_p(t) = A\cos(\omega_1 t) + B\sin(\omega_1 t)\] Now, let's determine the first and second derivatives of this function: \[y_p'(t) = -A\omega_1 \sin(\omega_1 t) + B\omega_1 \cos(\omega_1 t)\] \[y_p''(t) = -A\omega_1^2 \cos(\omega_1 t) - B\omega_1^2 \sin(\omega_1 t)\] Substitute \(y_p(t)\) and its derivatives into the given differential equation: \[ -A\omega_1^2 \cos(\omega_1 t) - B\omega_1^2 \sin(\omega_1 t) + \omega_0^2 (A\cos(\omega_1 t) + B\sin(\omega_1 t)) = F\cos(\omega_1 t) \] Now we can equate the like terms on both sides to determine the coefficients A and B: \[A(\omega_0^2 - \omega_1^2) = F\] \[B\omega_1^2 = 0\] \[A = \frac{F}{(\omega_0^2 - \omega_1^2)}\] \[B = 0\] So, we have a particular solution for this case: \[y_p(t) = \frac{F}{(\omega_0^2 - \omega_1^2)} \cos(\omega_1 t)\]

Find a particular solution when \(\omega=\omega_{0}\) (Case ii)

For case (ii), we cannot use the same method as in case (i), since the complementary solution already contains \(\cos(\omega_0 t)\) and \(\sin(\omega_0 t)\) terms. Instead, we try a particular solution of the form: \[y_p(t) = A t \cos(\omega_0 t) + B t \sin(\omega_0 t)\] Now, determine the first and second derivatives of this function: \[y_p'(t) = A (\cos(\omega_0 t) - \omega_0 t \sin(\omega_0 t)) + B (\sin(\omega_0 t) + \omega_0 t \cos(\omega_0 t))\] \[y_p''(t) = - 2 A \omega_0 \sin(\omega_0 t) - 2 B \omega_0 \cos(\omega_0 t) - A\omega_0^2 t \cos(\omega_0 t) - B\omega_0^2 t \sin(\omega_0 t)\] Substitute \(y_p(t)\) and its derivatives into the given differential equation: \[ - 2 A \omega_0 \sin(\omega_0 t) - 2 B \omega_0 \cos(\omega_0 t) - A\omega_0^2 t \cos(\omega_0 t) - B\omega_0^2 t \sin(\omega_0 t) + \omega_0^2 (A t \cos(\omega_0 t) + B t \sin(\omega_0 t)) = F\cos(\omega_0 t) \] Now we can equate the like terms on both sides to determine the coefficients A and B: \[- 2 B \omega_0 = F\] \[ - 2 A \omega_0 = 0\] \[A = 0\] \[B = -\frac{F}{2 \omega_0}\] So, we have a particular solution for this case: \[y_p(t) = -\frac{F}{2 \omega_0} t \sin(\omega_0 t)\] Now we can write the general solutions for each case: For case (i): \[y(t) = c_1 \cos(\omega_0 t) + c_2 \sin(\omega_0 t) + \frac{F}{(\omega_0^2 - \omega_1^2)} \cos(\omega_1 t)\] For case (ii): \[y(t) = c_1 \cos(\omega_0 t) + c_2 \sin(\omega_0 t) - \frac{F}{2 \omega_0} t \sin(\omega_0 t)\]

Key concepts

Complementary Solution

In the realm of differential equations, the complementary solution plays a pivotal role in solving homogeneous linear differential equations. To find the complementary solution of a differential equation like
\[ y'' + \omega_0^2 y = 0 \],
we first convert the differential equation into an algebraic equation by guessing a solution of the form \( y = e^{r t} \). This converts our equation into \( r^2 e^{rt} + \omega_0^2 e^{rt} = 0 \). Because the exponential term is never zero, we can divide it out to find \( r^2 + \omega_0^2 = 0 \).
From this, solving for \( r \) gives \( r = \pm \omega_0 i \), which leads us to the general complementary solution:
\[ y_c(t) = c_1 \cos(\omega_0 t) + c_2 \sin(\omega_0 t) \].
Here, \( c_1 \) and \( c_2 \) are constants determined by initial conditions. This solution represents the natural response of the system without external forces.

Method of Undetermined Coefficients

The method of undetermined coefficients is a clever technique used to find particular solutions to non-homogeneous linear differential equations like
\( y'' + \omega_0^2 y = F \cos\omega t \). It involves guessing a form for the particular solution and then determining the unknown coefficients by substitution.
For instance, consider the case \( \omega = \omega_1 eq \omega_0 \). A suitable guess would be \( y_p(t) = A\cos(\omega_1 t) + B\sin(\omega_1 t) \). By plugging this assumed form back into the differential equation and equating coefficients of like terms on both sides, we discover that:

• \( A = \frac{F}{(\omega_0^2 - \omega_1^2)} \)
• \( B = 0 \)

So, the particular solution becomes \( y_p(t) = \frac{F}{(\omega_0^2 - \omega_1^2)} \cos(\omega_1 t) \).
This method works beautifully for equations with constant coefficients and specific forms like sinusoids or exponentials.

Particular Solution

Finding a particular solution can be a bit trickier when your forcing function matches part of the complementary solution. This happens in our second case \( \omega = \omega_0 \). Since the standard forms of the complementary solution \( \cos(\omega_0 t) \) and \( \sin(\omega_0 t) \) match the given forcing term \( F \cos(\omega_0 t) \), we must modify our approach.
To resolve this, we propose using a solution with an added factor of \( t \), specifically
\( y_p(t) = A t \cos(\omega_0 t) + B t \sin(\omega_0 t) \).
Differentiating and substituting this into the differential equation, we equate coefficients to find:

• \( A = 0 \)
• \( B = -\frac{F}{2 \, \omega_0} \)

Thus, the particular solution becomes
\( y_p(t) = -\frac{F}{2 \, \omega_0} t \sin(\omega_0 t) \).
This solution accounts for the resonant frequency, highlighting how systems respond differently when driven at their natural frequency.