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Magazine Chapter 7: Problem 21
Find the particular solution indicated. \(\left(D^{2}+4 D+4\right) y=0 ;\) when \(x=0, y=1, y^{\prime}=-1\).
Short Answer
Expert verified
The particular solution is \(y(x) = (1 + x)e^{-2x}\).
Step by step solution
01
Identify the differential equation
The given differential equation is \( (D^2 + 4D + 4) y = 0 \). This is a linear homogeneous second-order differential equation with constant coefficients.
02
Solve the characteristic equation
To solve the differential equation, we first find the characteristic equation associated with it: \( m^2 + 4m + 4 = 0 \). This factors into \((m + 2)^2 = 0\), giving a repeated root at \(m = -2\).
03
Write the general solution
For a second-order linear differential equation with constant coefficients and a repeated root \(m = -2\), the general solution is \( y(x) = (C_1 + C_2 x)e^{-2x} \).
04
Apply initial condition for y(0) = 1
Substitute \(x = 0\) and \(y = 1\) into the general solution: \( y(0) = (C_1 + C_2 \times 0)e^{0} = C_1 = 1 \). Thus, \(C_1 = 1\).
05
Find the derivative of the general solution
Differentiate \(y(x) = (1 + C_2 x)e^{-2x}\) to get \(y'(x) = -2(1 + C_2 x)e^{-2x} + C_2 e^{-2x}\). This simplifies to \(y'(x) = (-2 - 2C_2 x + C_2)e^{-2x}\).
06
Apply initial condition for y'(0) = -1
Substitute \(x = 0\) and \(y'(0) = -1\) into the derivative: \((-2 + C_2)e^{0} = -1\). This gives \(-2 + C_2 = -1\), leading to \(C_2 = 1\).
07
Write the particular solution
With \(C_1 = 1\) and \(C_2 = 1\), the particular solution is \(y(x) = (1 + x)e^{-2x}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
A key step in solving second-order differential equations is finding the **characteristic equation**. This is derived from the original differential equation, which in our example is \((D^2 + 4D + 4)y = 0\). To determine the characteristic equation, replace the derivatives with powers of a dummy variable, typically \(m\). Thus, our equation becomes \(m^2 + 4m + 4 = 0\).
- The characteristic equation helps identify the roots which dictate the form of the solution.
- For homogeneous equations with constant coefficients, like this one, factorizing is common and aims to find the roots.
Repeated Roots
In differential equations, a repeated root in the characteristic equation requires special handling to find the **general solution**. Simply put, repeated roots occur when a quadratic like \((m + 2)^2 = 0\) has only one unique solution, in this case \(m = -2\).
- When a repeated root is discovered, the general solution includes an additional polynomial factor.
- This accounts for the second solution due to the order of the differential equation.
Initial Value Problem
To find a **particular solution** that fits specific conditions, we take the general solution and apply initial values. This transforms a differential equation into what's known as an **initial value problem**.
Suppose we know that at \(x = 0\), \(y = 1\) and \(y' = -1\). We use these conditions to find the specific values of constants \(C_1\) and \(C_2\).
Suppose we know that at \(x = 0\), \(y = 1\) and \(y' = -1\). We use these conditions to find the specific values of constants \(C_1\) and \(C_2\).
- Start by substituting \(x = 0\) into the general solution: \(y(0) = (C_1 + C_2 \times 0)e^{0} = C_1 = 1\). Thus, \(C_1 = 1\).
- Next, differentiate the general solution to find \(y'(x)\).
- Apply the initial condition for the derivative: \(y'(0) = (-2 + C_2)e^{0} = -1\), leading to \(C_2 = 1\).
