Problem 31
Question
$$ \text { In Problems 27-36, solve the given initial-value problem. } $$ $$ y^{\prime \prime}+4 y^{\prime}+5 y=35 e^{-4 x}, y(0)=-3, y^{\prime}(0)=1 $$
Step-by-Step Solution
Verified Answer
The solution is \( y = e^{-2x}(-10 \cos x + 9 \sin x) + 7e^{-4x} \).
1Step 1: Understand the Problem
We need to solve the second-order linear differential equation \( y'' + 4y' + 5y = 35e^{-4x} \) with initial conditions \( y(0) = -3 \) and \( y'(0) = 1 \). This is an initial-value problem (IVP) where we should find a particular solution to the non-homogeneous equation and solve the corresponding homogeneous equation.
2Step 2: Solve the Homogeneous Equation
The first step is to solve the homogeneous equation, \( y'' + 4y' + 5y = 0 \). We find its characteristic equation, \( r^2 + 4r + 5 = 0 \). Solving this quadratic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=1, b=4, c=5 \), gives the roots \( r = -2 \pm i \). So, the complementary (homogeneous) solution \( y_h \) is \( y_h = e^{-2x}(C_1 \cos x + C_2 \sin x) \).
3Step 3: Solve for the Particular Solution
To find a particular solution \( y_p \), use the method of undetermined coefficients. Assume a solution of the form \( y_p = Ae^{-4x} \). Substitute \( y_p \), \( y_p' = -4Ae^{-4x} \), and \( y_p'' = 16Ae^{-4x} \) into the differential equation and match coefficients:\[ 16A e^{-4x} - 4(4A)e^{-4x} + 5Ae^{-4x} = 35e^{-4x} \]\[ 16A - 16A + 5A = 35 \]\( 5A = 35 \) so \( A = 7 \). The particular solution is \( y_p = 7e^{-4x} \).
4Step 4: Find the General Solution
The general solution to the differential equation is the sum of the homogeneous and particular solutions: \[ y = y_h + y_p = e^{-2x}(C_1 \cos x + C_2 \sin x) + 7e^{-4x} \]
5Step 5: Apply Initial Conditions
Use the initial conditions to solve for \( C_1 \) and \( C_2 \). First, apply \( y(0) = -3 \):\[ e^0(C_1 \cdot 1 + C_2 \cdot 0) + 7e^0 = -3 \] \[ C_1 + 7 = -3 \] \[ C_1 = -10 \]Next, apply \( y'(0) = 1 \). Compute \( y' \):\[ y' = \, (-2e^{-2x}(C_1 \cos x + C_2 \sin x) + e^{-2x}(-C_1 \sin x + C_2 \cos x)) - 28e^{-4x} \]At \( x = 0 \):\[ y'(0) = -2C_1 + C_2 - 28 = 1 \]Substitute \( C_1 = -10 \):\[ y'(0) = 20 + C_2 - 28 = 1 \]\[ C_2 - 8 = 1 \] \\[ C_2 = 9 \].
6Step 6: Construct the Final Solution
Substitute \( C_1 = -10 \) and \( C_2 = 9 \) into the general solution:\[ y = e^{-2x}(-10 \cos x + 9 \sin x) + 7e^{-4x} \]
Key Concepts
Second-Order Linear Differential EquationCharacteristic EquationMethod of Undetermined CoefficientsGeneral Solution
Second-Order Linear Differential Equation
To begin with, let's clarify what a second-order linear differential equation is. This type of equation involves the second derivative of a function, which in this context is represented by \(y''\). The equation also includes terms with the first derivative \(y'\) and the function \(y\) itself.
An important characteristic of linear differential equations is that they can be expressed in the form \(a(x)y'' + b(x)y' + c(x)y = g(x)\), where \(a(x)\), \(b(x)\), and \(c(x)\) are functions of \(x\), and \(g(x)\) is any given function, often referred to as the non-homogeneous part.
In our example, the equation is \( y'' + 4y' + 5y = 35e^{-4x} \). Here, our coefficients are constants:
An important characteristic of linear differential equations is that they can be expressed in the form \(a(x)y'' + b(x)y' + c(x)y = g(x)\), where \(a(x)\), \(b(x)\), and \(c(x)\) are functions of \(x\), and \(g(x)\) is any given function, often referred to as the non-homogeneous part.
In our example, the equation is \( y'' + 4y' + 5y = 35e^{-4x} \). Here, our coefficients are constants:
- \(a = 1\)
- \(b = 4\)
- \(c = 5\)
- and \(g(x) = 35e^{-4x}\)
Characteristic Equation
The characteristic equation is a vital tool when solving homogeneous linear differential equations. It helps determine the form of the complementary (or homogeneous) solution.
In our case, the homogeneous equation derived from \( y'' + 4y' + 5y = 0 \) is transformed into a quadratic equation, known as the characteristic equation. This is done by substituting the derivatives with powers of \(r\), resulting in the equation \( r^2 + 4r + 5 = 0 \).
Quadratic equations are solved using the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For \(a = 1\), \(b = 4\), and \(c = 5\), the discriminant \(b^2 - 4ac\) is negative, leading to complex roots \(-2 \pm i\).
These roots indicate the form of the complementary solution:
In our case, the homogeneous equation derived from \( y'' + 4y' + 5y = 0 \) is transformed into a quadratic equation, known as the characteristic equation. This is done by substituting the derivatives with powers of \(r\), resulting in the equation \( r^2 + 4r + 5 = 0 \).
Quadratic equations are solved using the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For \(a = 1\), \(b = 4\), and \(c = 5\), the discriminant \(b^2 - 4ac\) is negative, leading to complex roots \(-2 \pm i\).
These roots indicate the form of the complementary solution:
- For real and distinct roots, use exponentials.
- For complex roots, use exponentials with sines and cosines.
Method of Undetermined Coefficients
The method of undetermined coefficients is a strategy to find a particular solution to a non-homogeneous differential equation. This method involves assuming a form for the particular solution and then determining the coefficients that satisfy the equation.
Given the non-homogeneous term in our problem, \(35e^{-4x}\), we assume that the particular solution \(y_p\) takes a similar form, say \(Ae^{-4x}\).
We derive the first and second derivatives of this assumed form:
After simplification, we find that \(A = 7\), so the particular solution is \( y_p = 7e^{-4x} \).
Given the non-homogeneous term in our problem, \(35e^{-4x}\), we assume that the particular solution \(y_p\) takes a similar form, say \(Ae^{-4x}\).
We derive the first and second derivatives of this assumed form:
- \(y_p' = -4Ae^{-4x}\)
- \(y_p'' = 16Ae^{-4x}\)
After simplification, we find that \(A = 7\), so the particular solution is \( y_p = 7e^{-4x} \).
General Solution
The general solution to a second-order linear differential equation is the sum of its complementary (homogeneous) solution and a particular solution.
In our exercise, we've determined the complementary solution as \( y_h = e^{-2x}(C_1 \cos x + C_2 \sin x) \) and the particular solution as \( y_p = 7e^{-4x} \).
Thus, the general solution is: \[ y = e^{-2x}(C_1 \cos x + C_2 \sin x) + 7e^{-4x} \] To find the specific solution that satisfies the initial conditions \(y(0) = -3\) and \(y'(0) = 1\), we substitute these conditions into the general solution.
This allows us to solve for the constants \(C_1\) and \(C_2\), leading to the final tailored solution. In the given problem, the constants are \(C_1 = -10\) and \(C_2 = 9\).
The tailored solution becomes: \[ y = e^{-2x}(-10 \cos x + 9 \sin x) + 7e^{-4x} \] This solution satisfies both the differential equation and the initial conditions. It is crucial that both aspects are met to solve the initial-value problem correctly.
In our exercise, we've determined the complementary solution as \( y_h = e^{-2x}(C_1 \cos x + C_2 \sin x) \) and the particular solution as \( y_p = 7e^{-4x} \).
Thus, the general solution is: \[ y = e^{-2x}(C_1 \cos x + C_2 \sin x) + 7e^{-4x} \] To find the specific solution that satisfies the initial conditions \(y(0) = -3\) and \(y'(0) = 1\), we substitute these conditions into the general solution.
This allows us to solve for the constants \(C_1\) and \(C_2\), leading to the final tailored solution. In the given problem, the constants are \(C_1 = -10\) and \(C_2 = 9\).
The tailored solution becomes: \[ y = e^{-2x}(-10 \cos x + 9 \sin x) + 7e^{-4x} \] This solution satisfies both the differential equation and the initial conditions. It is crucial that both aspects are met to solve the initial-value problem correctly.
Other exercises in this chapter
Problem 31
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