Question

A solution of the differential equation $$ \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=4 e^{-x} $$ takes the value 1 when \(x=0\) and the value \(e^{-1}\) when \(x=1\). What is its value when \(x=2 ?\)

Short answer

The value when \(x=2\) is \(e^{-2}\).

Step-by-step solution

- Solve the Homogeneous Equation

First, solve the homogeneous equation: \[\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=0\]The characteristic equation is: \[r^2 + 2r + 1 = 0\]Factor to get: \[(r + 1)^2 = 0\]Thus, the solution to the homogeneous equation is: \[y_h = (C_1 + C_2x)e^{-x}\]

- Find a Particular Solution

To find a particular solution to the non-homogeneous equation \[\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=4 e^{-x}\]Assume a particular solution of the form: \[y_p = A e^{-x}\]Substitute this into the differential equation and solve for A: \[A(-e^{-x}) + 2A(-e^{-x}) + A e^{-x} = 4e^{-x}\]Simplify to get: \[-A = 4 \rightarrow A = -4\]Thus, the particular solution is: \[y_p = -4e^{-x}\]

- Combine General Solution

The general solution to the differential equation is: \[y = y_h + y_p\]Combine the homogeneous and particular solutions: \[y = (C_1 + C_2x)e^{-x} - 4e^{-x}\]Simplify to get: \[y = (C_1 - 4 + C_2x)e^{-x}\]

- Apply Initial Conditions

Use the given conditions to find the constants:1. When \(x=0\), \(y=1\):\[1 = (C_1 - 4)e^{0} \rightarrow C_1 - 4 = 1 \rightarrow C_1 = 5\]2. When \(x=1\), \(y=e^{-1}\):\[e^{-1} = (5 + C_2)e^{-1} - 4e^{-1}\]\[1 = 5 + C_2 - 4 \rightarrow C_2 = 0\]

- Solve for the Desired Value

The specific solution, incorporating the constants, is: \[y = (5 - 4)e^{-x} = e^{-x}\]Find the value at \(x=2\):\[y(2) = e^{-2}\]

Key concepts

Homogeneous Equation

To understand the solution to a differential equation, we first need to solve the homogeneous part of the equation. The homogeneous equation is:
\(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=0\).
We transform this into a characteristic equation by substituting \(\frac{dy}{dx} = r\).
This gives us \(r^2 + 2r + 1 = 0\).
By solving this characteristic equation, and factoring, we get \((r + 1)^2 = 0\).
This means we have a repeated root \((r = -1)\).
Therefore, the solution to the homogeneous equation is:
\(y_h = (C_1 + C_2x)e^{-x}\).
Here, \(C_1\) and \(C_2\) are constants that we'll determine using initial conditions later.

Solving the homogeneous equation is a crucial step because it helps us find the part of our solution that deals with the natural behavior of the system without external influences.

Particular Solution

Next, we need a particular solution to account for the non-homogeneous part of the differential equation: \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=4 e^{-x}\).
To find a particular solution, we assume a solution of the form \(y_p = A e^{-x}\), where \(A\) is a constant.
Substituting \(y_p\) into the differential equation, we get:
\(-Ae^{-x} + 2(-Ae^{-x}) + Ae^{-x} = 4e^{-x}\).
Simplifying this, we find \(-A = 4\), which leads us to \(A = -4\).
Thus, the particular solution is: \(y_p = -4e^{-x}\).

Finding the particular solution helps us understand how external forces or inputs (in this case, \(4 e^{-x}\)) affect the system.

Initial Conditions

Finally, we apply initial conditions to determine the constants \(C_1\) and \(C_2\) in our general solution \((y = y_h + y_p)\).
The general solution is:
\(y = (C_1 + C_2x)e^{-x} - 4e^{-x}\).
Simplifying, we get:
\(y = (C_1 - 4 + C_2x)e^{-x}\).

Using the initial condition \(y(0) = 1\): When \(x = 0\), \(y = 1\):
\(1 = (C_1 - 4)e^{0}\)
Solving this, we get: \(C_1 - 4 = 1 \rightarrow C_1 = 5\).

Using the second initial condition \(y(1) = e^{-1}\): When \(x = 1\), \(y = e^{-1}\):
\(e^{-1} = (5 + C_2)e^{-1} - 4e^{-1}\)
Simplifying, we find: \(1 = 5 + C_2 - 4 \rightarrow C_2 = 0\).

Thus, our specific solution is \(y = (5 - 4 + 0x)e^{-x} = e^{-x}\).
The final step is to find the value at \(x = 2\): \(y(2) = e^{-2}\).

Initial conditions are essential because they allow us to pinpoint the specific behavior of the system by determining the constants in the general solution.