Question

Solve the initial value problem and graph the solution. $$ y^{\prime \prime}+4 y=-e^{-2 x}[(4-7 x) \cos x+(2-4 x) \sin x], y(0)=3, \quad y^{\prime}(0)=1 $$

Short answer

Question: Find the solution to the given second-order nonhomogeneous differential equation with initial conditions \(y(0)=3\) and \(y^{\prime}(0)=1\): $$ y^{\prime \prime} + 4y = -e^{-2 x}[(4-7 x) \cos x +(2-4 x) \sin x] $$ Answer: The solution to the given second-order nonhomogeneous differential equation with initial conditions is: $$ y(x) = \frac{158}{45}\cos(2x)+\frac{67}{45}\sin(2x)+e^{-2x}\left(-\frac{5}{45}x\cos x+\frac{2}{45}x\sin x -\frac{23}{45}\cos x-\frac{22}{45}\sin x\right) $$

Step-by-step solution

Identify the complementary and particular solutions

For the given differential equation, the complementary equation is: $$ y^{\prime \prime} + 4y = 0 $$ This equation is a homogeneous second order linear differential equation with constant coefficients. We will solve it to find the complementary solution \(y_c(x)\). The particular solution \(y_p(x)\) will be found from the given nonhomogeneous equation, which is: $$ y^{\prime \prime} + 4y = -e^{-2 x}[(4-7 x) \cos x +(2-4 x) \sin x] $$ Ultimately, the general solution will be the sum of the complementary solution and particular solution \(y(x) = y_c(x) + y_p(x)\). We will then use the initial conditions to determine the coefficients in the general solution.

Solve the complementary equation

To solve the complementary equation \(y^{\prime \prime} + 4y = 0\), we assume a solution of the form \(y_c(x) = e^{rx}\). Differentiating twice, we get: $$ y^{\prime}(x) = re^{rx}, \quad y^{\prime \prime}(x) = r^2e^{rx} $$ Substituting this into the complementary equation, we obtain the characteristic equation: $$ r^2 + 4 = 0 $$ Solving for \(r\), we get \(r = \pm 2i\). The general solution to the complementary equation is: $$ y_c(x) = C_1 \cos(2x) + C_2 \sin(2x) $$

Find the particular solution

We use the method of undetermined coefficients to find the particular solution. Based on the given differential equation, we assume the particular solution will be in the form: $$ y_p(x) = e^{-2x}(A x \cos x + Bx \sin x + C \cos x + D \sin x ) $$ Now, we differentiate \(y_p(x)\) and substitute it into the given nonhomogeneous equation to find the coefficients A, B, C, and D.

Differentiate and substitute into the nonhomogeneous equation

Differentiating \(y_p(x)\) once: $$ y_p^\prime(x) = e^{-2x}(-A\sin(x)-B\cos(x)-(4-7A)x\sin(x)-(2-4A)\cos(x)\\ +(4-7B)x\cos(x)-(2-4B)\sin(x) ) $$ And differentiating \(y_p(x)\) again: $$ y_p^{\prime\prime}(x) = e^{-2x}[(-7A\sin(x)+4\cos(x)-B\sin(x))-(2\cos(x)+(4-7B)x)\sin(x)-(4A-2)\cos(x) \\ +(7A-4)x\cos(x)+(4B-2)\sin(x)] $$ Now, substituting \(y_p(x)\), \(y_p^\prime(x)\), and \(y_p^{\prime\prime}(x)\) into the nonhomogeneous equation and equating the coefficients of \(e^{-2x}\sin(x)\) and \(e^{-2x}\cos(x)\), we obtain a set of linear equations for A, B, C, and D.

Solve for coefficients A, B, C, and D

After substituting into the nonhomogeneous equation and equating the coefficients, the linear equations are: $$ \begin{cases} -7A+4 = -4 \\ -7B-2 = 2 \\ 2A-7B+4C = -7 \\ B+7A-4D = -4 \\ \end{cases} $$ Solving this linear system, we get: $$ A = -\frac{5}{45}, \quad B = \frac{2}{45}, \quad C = -\frac{23}{45}, \quad D = -\frac{22}{45} $$ Thus, the particular solution is: $$ y_p(x)=e^{-2 x}\left(-\frac{5}{45}x\cos x+\frac{2}{45}x\sin x -\frac{23}{45}\cos x-\frac{22}{45}\sin x\right) $$

Calculate the general solution and apply the initial conditions

The general solution is the sum of the complementary solution \(y_c(x)\) and the particular solution \(y_p(x)\): $$ y(x) = y_c(x) + y_p(x) = C_1 \cos(2x) + C_2 \sin(2x) + e^{-2 x}\left(-\frac{5}{45}x\cos x+\frac{2}{45}x\sin x -\frac{23}{45}\cos x-\frac{22}{45}\sin x\right) $$ We apply the initial conditions \(y(0)=3\) and \(y^{\prime}(0)=1\) to find the coefficients \(C_1\) and \(C_2\): $$ 3 = C_1 - \frac{23}{45}\implies C_1 = \frac{158}{45} $$ $$ 1 = 2C_2 - \frac{22}{45} \implies C_2 = \frac{67}{45} $$ The final solution is: $$ y(x) = \frac{158}{45}\cos(2x)+\frac{67}{45}\sin(2x)+e^{-2x}\left(-\frac{5}{45}x\cos x+\frac{2}{45}x\sin x -\frac{23}{45}\cos x-\frac{22}{45}\sin x\right) $$

Graph the solution

To graph the solution, plot the function \(y(x)\), with the domain restricted to \(x \ge 0\). The graph shows how the function behaves under the given initial conditions.

Key concepts

Second Order Linear Differential Equations

Second order linear differential equations form a fundamental part of calculus, dealing with equations that involve the second derivative of a function. In simple terms, these equations look like \( a y'' + b y' + c y = 0 \). The equation we examined is in this standard form. These equations are linear because they do not have the function or its derivatives multiplied by each other, and they are second order due to the highest derivative being the second one. These types of equations often model real-world situations like simple harmonic motion.
Understanding these equations requires finding solutions that satisfy both the equation and any boundary or initial conditions. One can often write the solution to such equations as a sum of two parts: the general solution of the complementary equation and a particular solution of the full equation.
The complementary equation, like the one in our problem, is solved by finding the roots of the so-called characteristic equation, derived by assuming a solution of an exponential form \( e^{rx} \).
When solving these, it's key to remember they play into many physical systems, helping us model and predict behavior like vibrations.

Nonhomogeneous Differential Equations

A nonhomogeneous differential equation includes an additional function on the right-hand side, distinct from zero. Our example problem can illustrate this effectively: \( y'' + 4y = -e^{-2x}[(4-7x)\cos x + (2-4x)\sin x] \). This additional function means solutions will largely consist of two parts: the complementary (or homogeneous) solution and the particular solution.
Solving nonhomogeneous differential equations involves more than just tackling a straightforward formula. We must decompose the problem into simpler parts, first solving the homogeneous part separately. Subsequently, we find the specific solution satisfying the nonhomogeneous part by methods like undetermined coefficients or variation of parameters.
In practical terms, nonhomogeneous differential equations may model systems where external factors influence behavior, such as forced mechanical systems. Each equation's unique right-hand function considerably shapes the particular solution, showcasing the inherent versatility and application of this concept in predicting behaviors of complex systems.

Initial Value Problems

Initial value problems present a specific challenge where you're asked not only to solve a differential equation but to find a solution meeting certain starting criteria. These problems appear in many forms in mathematics and physics, especially where predicting future behavior from a known state is necessary, like predicting population growth or cooling temperatures.
In the example we've covered, initial conditions are supplied as \( y(0)=3 \) and \( y'(0)=1 \). This means we find coefficients within the general solution that ensure these conditions hold. This process often involves substituting the initial conditions directly into the general solution and its derivative.
With these problems, having a solid understanding of both complementary and particular solutions ensures one can apply these initial conditions correctly to yield a final solution, giving us full predictive power over the system's behavior following its initial state.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique for finding the particular solution to certain nonhomogeneous differential equations, involving guessing a solution form based on the nonhomogeneous part of the equation. This method is especially useful when the right-side forcing function is a combination of polynomial, exponential, sine, and cosine functions. Our exercise showed it well with a nonhomogeneous part of \(-e^{-2x}[(4-7x)\cos x +(2-4x)\sin x]\).
The strategy begins by proposing a suitable form for \( y_p(x) \) that mimics the nonhomogeneous function’s structure. Next, derivatives are computed, substituted into the differential equation, and coefficients determined so that the assumed solution satisfies the equation.
This method stands out due to its reliance on an astute initial guess, allowing rapid construction of a workable particular solution. It streamlines many practical problems, simplifying otherwise complicated equations to a more manageable form.