Question

Find the general solution of the given differential equation. $$ u^{n}+\omega_{0}^{2} u=\cos \omega t, \quad \omega^{2} \neq \omega_{0}^{2} $$

Short answer

The general solution of the given linear ordinary differential equation is: $$ u(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t) + \frac{\cos(\omega t)}{\omega_0^2 - (-\omega^2)^{n-1}} $$ where A and B are arbitrary constants, and n is the order of the differential equation.

Step-by-step solution

Solve the homogeneous part

To find the complementary function, we need to solve the homogeneous equation: $$ u^{n}+\omega_{0}^{2} u=0 $$ This is a simple linear ODE with constant coefficients. We'll use the characteristic equation to solve for it: $$ r^n+\omega_{0}^{2} = 0 $$ Since we don't know the value of n, we can't explicitly solve the characteristic equation. However, we can write the complementary function in the general form: $$ u_c(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t) $$ where A and B are arbitrary constants.

Find the particular solution

Now, we need to find a particular solution for the nonhomogeneous part: $$ u^{n}+\omega_{0}^{2} u=\cos \omega t, \quad \omega^{2} \neq \omega_{0}^{2} $$ For this, we assume an ansatz (trial solution) of the form: $$ u_p(t) = C\cos(\omega t) + D\sin(\omega t) $$ where C and D are constants to be determined. Now, we'll find the derivatives of our ansatz and plug them into the given differential equation: $$ u_p'(t) = -\omega C\sin(\omega t) + \omega D\cos(\omega t) $$ $$ u_p''(t) = -\omega^2 C\cos(\omega t) - \omega^2 D\sin(\omega t) $$ and so on, until you find the nth derivative: $$ u_p^{(n)}(t) = (C\cos(\omega t) + D\sin(\omega t))\cdot(-\omega^2)^{n-1} $$ Now, we substitute the ansatz and its nth derivative into the given differential equation: $$ (C\cos(\omega t) + D\sin(\omega t))\cdot(-\omega^2)^{n-1}+\omega_{0}^{2}(C\cos(\omega t) + D\sin(\omega t)) = \cos \omega t $$ Next, we'll match the coefficients of the trigonometric terms on both sides of the equation to find the constants C and D.

Match the coefficients and solve for the constants C and D

From the equation above, we can match the coefficients of the cos and sin terms on both sides: $$ (-\omega^2)^{n-1} C + \omega_{0}^{2} C =1 $$ $$ (-\omega^2)^{n-1} D + \omega_{0}^{2} D =0 $$ Solving these equations, we get: $$ C = \frac{1}{\omega_0^2 - (-\omega^2)^{n-1}} $$ $$ D = 0 $$ So, our particular solution takes the form: $$ u_p(t) = \frac{\cos(\omega t)}{\omega_0^2 - (-\omega^2)^{n-1}} $$

Combine the complementary function and the particular solution to obtain the general solution

Finally, we combine the complementary function and the particular solution to find the general solution of the given differential equation: $$ u(t) = u_c(t) + u_p(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t) + \frac{\cos(\omega t)}{\omega_0^2 - (-\omega^2)^{n-1}} $$

Key concepts

Homogeneous Equation

In differential equations, a homogeneous equation is one where all terms involve the dependent variable or its derivatives, and it is set equal to zero. For our given problem, the homogeneous equation is:\[ u^n + \omega_0^2 u = 0 \]This expression means there are no external forces or inputs acting on the system. Solving this equation helps identify the natural responses of the system, forming the basis of the complementary function. By setting the right-hand side to zero, we find solutions that describe the behavior of the system without any external influence. The solutions to a homogeneous equation are essential as they reveal possible movements or states of the system that will persist unless influenced by a non-zero term.

Particular Solution

A particular solution is a solution to the non-homogeneous differential equation, incorporating the non-zero right side. This helps account for any external influences applied to the system. For our equation:\[ u^n + \omega_0^2 u = \cos \omega t \]The presence of \( \cos \omega t \) signifies an external input impacting the system. To find this particular solution, assume a trial solution that reflects the structure of the non-homogeneous term:\[ u_p(t) = C\cos(\omega t) + D\sin(\omega t) \]By substituting this trial solution and its derivatives into the differential equation, and solving, we deduce the values of constants \( C \) and \( D \). The particular solution represents how the system responds specifically to this external input.

Characteristic Equation

The characteristic equation is a crucial step in solving linear differential equations with constant coefficients. This equation emerges by replacing derivatives in the differential equation with powers of a variable, usually denoted by "r". For a differential equation like our homogeneous part: \[ u^n + \omega_0^2 u = 0 \]the characteristic equation would resemble:\[ r^n + \omega_0^2 = 0 \]Solving this helps find the roots of the equation—values of \( r \) that satisfy it. These roots dictate the form of the complementary function. Real roots sometimes lead to exponential solutions in time, while complex ones typically yield sinusoidal solutions, hence shaping the natural dynamic response of the system.

Complementary Function

The complementary function is derived from the homogeneous equation solutions. It represents the natural behavior of the system, assuming no external forces. This function uses the solutions from the characteristic equation.For the given differential equation, with the roots like `±iω₀`, the complementary function becomes:\[ u_c(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t) \]Here, \( A \) and \( B \) are arbitrary constants that depend on initial conditions. This function showcases a combination of cosine and sine terms, reflecting a typical oscillatory behavior found in systems resolved via these characteristic roots.