Question

For the given differential equation, $$ y^{\prime \prime}+2 y^{\prime}+2 y=t^{3} $$

Short answer

Question: Find the general solution to the given inhomogeneous differential equation: $$ y^{\prime \prime} + 2y^{\prime} + 2y = t^3 $$ Answer: The general solution to the given inhomogeneous differential equation is: $$ y(t) = e^{-t}(C_1\cos(t) + C_2\sin(t)) + \frac{1}{2}t^3 - \frac{3}{4}t^2 + \frac{1}{2}t - \frac{1}{4} $$

Step-by-step solution

Find the complementary function for the homogeneous equation

To find the complementary function, we need to solve the homogeneous equation: $$ y^{\prime \prime} + 2y^{\prime} + 2y = 0 $$ First, assume a solution of the form \(y = e^{rt}\). Then, substitute this into the homogeneous equation: $$ r^2 e^{rt} + 2re^{rt} + 2e^{rt} = 0 $$ Factor out the \(e^{rt}\) term: $$ e^{rt}(r^2 + 2r + 2) = 0 $$ Since \(e^{rt}\) can't be zero, the characteristic equation is: $$ r^2 + 2r + 2 = 0 $$

Solve the characteristic equation

To find the roots of the characteristic equation, we can use the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Here, \(a = 1\), \(b = 2\), and \(c = 2\). Plugging in the values, we get: $$ r = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} = -1 \pm i $$ Since the roots are complex, the complementary function is of the form: $$ y_c(t) = e^{-t}(C_1\cos(t) + C_2\sin(t)) $$

Find a particular solution for the inhomogeneous equation

We will use the method of undetermined coefficients to find a particular solution. Since the inhomogeneous term is \(t^3\), we assume a particular solution of the form: $$ y_p(t) = At^3 + Bt^2 + Ct + D $$ Now, we need to find the first and second derivatives of \(y_p(t)\): $$ y_p^{\prime}(t) = 3At^2 + 2Bt + C \\ y_p^{\prime \prime}(t) = 6At + 2B $$

Substitute the assumed particular solution into the inhomogeneous equation

Plug \(y_p(t)\), \(y_p^{\prime}(t)\), and \(y_p^{\prime \prime}(t)\) into the given inhomogeneous equation: $$ (6At + 2B) + 2(3At^2 + 2Bt + C) + 2(At^3 + Bt^2 + Ct + D) = t^3 $$ Equating the coefficients of the same powers of \(t\), we have: $$ \begin{cases} 2A = 1 \\ 6A + 4B = 0 \\ 6At + 2B = 0 \\ 2C + 2D = 0 \end{cases} $$ Solving this system of equations, we get \(A = \frac{1}{2}\), \(B = -\frac{3}{4}\), \(C = \frac{1}{2}\), and \(D = -\frac{1}{4}\). So, the particular solution is: $$ y_p(t) = \frac{1}{2}t^3 - \frac{3}{4}t^2 + \frac{1}{2}t - \frac{1}{4} $$

Combine the complementary and particular solutions

To find the general solution, add the complementary function and the particular solution: $$ y(t) = y_c(t) + y_p(t) = e^{-t}(C_1\cos(t) + C_2\sin(t)) + \frac{1}{2}t^3 - \frac{3}{4}t^2 + \frac{1}{2}t - \frac{1}{4} $$ This is the general solution to the given inhomogeneous differential equation.

Key concepts

Homogeneous Equation

A homogeneous equation in differential equations is one where the right side of the equation is zero. This means that the equation only consists of expressions with the dependent variable and its derivatives. In our example, the homogeneous form of the equation is \(y'' + 2y' + 2y = 0\). Here, we focus on finding the complementary function, which involves solving this homogeneous equation first.
Understanding the homogeneous equation is important because it helps us grasp how the system behaves in the absence of external inputs. In this case, we seek solutions using the assumption \(y = e^{rt}\), a technique that leverages exponential functions due to their nice properties regarding differentiation."},{

Particular Solution

A particular solution to a differential equation addresses the non-homogeneous part, or the external inputs influencing the system. It doesn’t account for the general behavior over all conditions, but zeroes in on a specific case.
For the given equation, the non-homogeneous term is \(t^3\). To find the particular solution, one can assume a solution of a specific form, tailored to this external term. The assumed form helps in eliminating arbitrary constants, and focuses instead on fulfilling the condition set by the non-homogeneous part.
In the step-by-step solution provided, the particular solution is discovered by trial and error using assumed terms associated with the highest degree of the non-homogeneous term—a process crucial in accurately confronting actual external system forces."},{

Undetermined Coefficients

The method of undetermined coefficients is an effective and structured way to find a particular solution to non-homogeneous linear differential equations with constant coefficients. Here, the trick lies in guessing the form of the particular solution based on the form of the non-homogeneous term.
In our example, given the term \(t^3\), the guess for the particular solution is \(y_p(t) = At^3 + Bt^2 + Ct + D\). The coefficients \(A, B, C,\) and \(D\) are "undetermined" because they need to be calculated. These coefficients are found by ensuring all terms, when substituted back into the original equation, balance out to zero or fit the non-homogeneous term.
Using undetermined coefficients can be particularly straightforward for polynomial terms, exponential functions, or simple trigonometric terms which frequently appear in these problems."},{

Complex Roots

In the context of a differential equation's characteristic equation, complex roots are significant because they signal oscillatory solutions. Complex roots arise when the discriminant in the quadratic formula, \(b^2 - 4ac\), is negative, leading to solutions of the form \(r = -\alpha \pm i \beta\).
For the characteristic equation \(r^2 + 2r + 2 = 0\), this results in complex roots of \(-1 \pm i\).
The solution to such equations takes a form combining exponential decay/growth with oscillations, presented as \(y_c(t) = e^{-t}(C_1\cos(t) + C_2\sin(t))\). This expression showcases how complex parts translate into sine and cosine functions, depicting real-life phenomena like waves or circuits impacted by resistive, capacitive, or inductive elements."},{

Characteristic Equation

The characteristic equation is vital in evaluating the complementary function of a differential equation. It originates from substituting \(y = e^{rt}\) into the homogeneous equation, leading to a polynomial expression in \(r\). For our exercise, the characteristic equation was \(r^2 + 2r + 2 = 0\).

• This polynomial helps identify the roots, which dictate the form of the complementary solution.
• Real roots suggest simple exponential solutions like \(e^{rt}\), while complex roots indicate oscillatory components with sine and cosine.

The ability to derive and solve characteristic equations is crucial because it allows simplification of complex systems into manageable mathematical terms. These equations give insight into how different parts of the system interact under initial conditions, predicting behavior over time."}]} ố Mock-Up')} più una stringa del dispositivo creata da BABEL'].