Question

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(y^{\prime \prime}+4 y^{\prime}+5 y=2 e^{-2 x} \cos x\)

Short answer

Second-order linear non-homogeneous equation. The general solution is \[ y = e^{-2x}(C_1 \cos(x) + C_2 \sin(x) + A \cos(x) + B \sin(x)). \]

Step-by-step solution

Identify the Type of Differential Equation

Examine the given differential equation: \[ y^{\text{''}} + 4y^{\text{'} } + 5y = 2e^{-2x} \cos(x) \]This equation is a second-order linear non-homogeneous differential equation because of the presence of the term \(2e^{-2x} \cos(x)\).

Solve the Homogeneous Equation

First, find the solution to the associated homogeneous equation: \[ y^{\text{''}} + 4y^{\text{}} + 5y = 0 \]Solve the characteristic equation: \[ r^2 + 4r + 5 = 0 \]Using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we get: \[ r = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm 2i}{2} \]Thus, the roots are \(-2 + i\) and \(-2 - i\). The solution to the homogeneous equation is: \[ y_h = e^{-2x}(C_1 \cos(x) + C_2 \sin(x)) \]

Find a Particular Solution

For the non-homogeneous term \(2e^{-2x} \cos(x)\), assume a particular solution of the form: \[ y_p = e^{-2x}(A \cos(x) + B \sin(x)) \]Then find \(y_p^{\text{'} }\) and \(y_p^{\text{''} }\) and substitute into the original differential equation:\[ y_p^{\text{'} } = e^{-2x}( -2A \cos(x) - A \sin(x) - 2B \sin(x) + B \cos(x)) \]\[ y_p^{\text{''}} = e^{-2x}( (4A- 5B) \cos(x) + (4B + 5A) \sin(x)) \]Substitute \(y_p\), \(y_p^{\text{'} }\), and \(y_p^{\text{''} }\) into the differential equation and equate coefficients to find \(A\) and \(B\).

Combine the Solutions

Combine the homogeneous solution \(y_h\) and the particular solution \(y_p\) to get the general solution:\[ y = y_h + y_p = e^{-2x}(C_1 \cos(x) + C_2 \sin(x)) + e^{-2x}(A \cos(x) + B \sin(x)) \] Simplify the constants \(C_1, C_2, A, B\) to write the final form.

Key concepts

homogeneous solution

To solve a second-order linear non-homogeneous differential equation, we start by addressing its homogeneous counterpart. The homogeneous equation is formed by setting the non-homogeneous term to zero. For the equation\[ y^{\text{''}} + 4y^{\text{'} } + 5y = 2e^{-2x} \textrm{cos}(x) \], the corresponding homogeneous equation is: \[ y^{\text{''}} + 4y^{\text{'} } + 5y = 0 \].
Solving this will reveal the homogeneous solution, denoted as \( y_h \).

We find \( y_h \) by determining the roots of the characteristic equation derived from the homogeneous differential equation. The characteristic equation is:
\[ r^2 + 4r + 5 = 0 \]
We use the quadratic formula to find the roots since the equation may not factor easily. The quadratic formula is given by \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Plugging in the values \( a = 1 \), \( b = 4 \), and \( c = 5 \), we get:
\[ r = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm 2i}{2} \]
This simplifies to \(-2 + i\) and \(-2 - i\), which are complex roots. These roots lead to a general solution for the homogeneous equation:
\[ y_h = e^{-2x} (C_1 \cos(x) + C_2 \sin(x)) \].
Understanding this step is crucial to solving the overall differential equation.

characteristic equation

The characteristic equation is a vital tool in solving homogeneous second-order linear differential equations.
To form it, replace \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1 in the original differential equation, excluding the non-homogeneous term.
For our homogeneous equation:
\[ y^{\text{''}} + 4y^{\text{'} } + 5y = 0 \]
The characteristic equation would be:
\[ r^2 + 4r + 5 = 0 \].

This characteristic equation is a quadratic equation. Using the quadratic formula to solve for \( r \), we obtain the roots of the characteristic equation.
For this problem, the calculation is as follows:
\[ r = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm 2i}{2} \]
These complex roots \( r = -2 + i \) and \( r = -2 - i \) are significant because they indicate that our solution will involve exponential and trigonometric functions.
Such outcomes are critical as they guide the formation of the homogeneous solution, using the formula:
\[ y_h = e^{\text{ real part } x} (C_1 \cos(\text{imaginary part } x) + C_2 \sin(\text{ imaginary part } x)) \].
Thus, we have:
\[ y_h = e^{-2x} (C_1 \cos(x) + C_2 \sin(x)) \],
which integrates perfectly within our holistic solution.

particular solution

The particular solution \( y_p \) specifically solves the non-homogeneous differential equation:
\[ y^{\text{''}} + 4y^{\text{'} } + 5y = 2e^{-2x} \cos(x) \].
To determine this, we utilize the 'method of undetermined coefficients'.

Given the non-homogeneous term \( 2e^{-2x} \cos(x) \), we assume a particular solution form mirroring the non-homogeneous term.
Therefore, we set:
\[ y_p = e^{-2x} (A \cos(x) + B \sin(x)) \].
Next, we find the first and second derivatives of \( y_p \):
\[ y_p^{\text{'} } = e^{-2x} ( -2A \cos(x) - A \sin(x) - 2B \sin(x) + B \cos(x)) \]
\[ y_p^{\text{''}} = e^{-2x} ( (4A- 5B) \cos(x) + (4B + 5A) \sin(x)) \]
We substitute these derivatives back into the original differential equation and equate coefficients to determine the values of A and B.
This process may be tedious, but it is necessary to identify a solution that fits our non-homogeneous term, ensuring our final general solution is accurate and complete.

method of undetermined coefficients

The 'method of undetermined coefficients' is a powerful technique for finding particular solutions to certain types of non-homogeneous differential equations.
This method assumes a form for the particular solution based on the non-homogeneous term.
For polynomial, exponential, sine, and cosine functions, we guess a solution structure and determine the unknown coefficients such that our trial solution satisfies the differential equation.

• For our differential equation with the non-homogeneous term \( 2e^{-2x} \cos(x) \), we assume:
  \( y_p = e^{-2x} (A \cos(x) + B \sin(x)) \).
• We then find its derivatives:

\[ y_p^{\text{'} } = e^{-2x} ( -2A \cos(x) - A \sin(x) - 2B \sin(x) + B \cos(x)) \]
\[ y_p^{\text{''}} = e^{-2x} ((4A - 5B) \cos(x) + (4B + 5A) \sin(x)) \]

Equating the coefficients of \( \cos(x) \) and \( \sin(x) \) on both sides of the equation enables us to solve for A and B, confirming our particular solution.
This step-by-step method is efficient and often used due to its systematic approach and effective outcomes for linear differential equations with common non-homogeneous terms.