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Magazine Chapter 39: Problem 6
Solve the following: \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+4 \frac{\mathrm{d} y}{\mathrm{~d} x}+5 y=2 e^{-2 \mathrm{x}}\), given that \(x=0, y=1\) and \(\frac{\mathrm{d} y}{\mathrm{~d} x}=-2\).
Short Answer
Expert verified
The particular solution of the given differential equation is \(y(x) = e^{-2x}(-1\cos(x) - 4\sin(x)) + 2e^{-2x}\).
Step by step solution
01
Write the homogeneous part of the equation
The homogeneous part is given by:
\[\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}+4 \frac{\mathrm{d} y}{\mathrm{d} x}+5 y=0\]
02
Find the auxiliary equation
The auxiliary equation is a quadratic equation obtained by replacing the derivatives of the function \(y\) with powers of a variable \(m\). The auxiliary equation is:
\[m^2 + 4m + 5 = 0\]
03
Solve the auxiliary equation
To find the roots of the auxiliary equation, you can use the quadratic formula:
\[m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In this case, \(a = 1\), \(b = 4\), and \(c = 5\).
Plug in the values and solve for \(m\):
\[m = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)}\]
\[m = \frac{-4 \pm \sqrt{-4}}{2}\]
\[m = -2 \pm \sqrt{-1}\]
The auxiliary equation has complex roots. Therefore, the general solution to the homogeneous equation is given by:
\[y_h(x) = e^{-2x}(A\cos(x) + B\sin(x))\]
**Step 2: Find a particular solution for the inhomogeneous part**
04
Identify the form of the particular solution
Given that the inhomogeneous term is \(2e^{-2x}\), the particular solution has the form:
\[y_p(x) = Ce^{-2x}\]
05
Differentiate \(y_p(x)\) twice
To find a particular solution, we calculate the first and second derivatives of \(y_p(x)\) and substitute it into the original inhomogeneous equation.
First derivative:
\[y_p'(x) = -2Ce^{-2x}\]
Second derivative:
\[y_p''(x) = 4Ce^{-2x}\]
06
Substitute the derivatives into the inhomogeneous equation
Plug in the derivatives of the particular solution into the inhomogeneous equation:
\[(4Ce^{-2x}) + 4(-2Ce^{-2x}) + 5(Ce^{-2x}) = 2e^{-2x}\]
07
Solve for \(C\)
To solve for \(C\), we can cancel out the common term of \(e^{-2x}\):
\[4C - 8C + 5C = 2\]
Solve for \(C\):
\[C = 2\]
Thus, the particular solution is:
\[y_p(x) = 2e^{-2x}\]
**Step 3: Combine the general and particular solutions and apply the given initial conditions**
08
Write the general inhomogeneous solution
Add the homogeneous and particular solutions to create the general inhomogeneous solution:
\[y(x) = y_h(x) + y_p(x) = e^{-2x}(A\cos(x) + B\sin(x)) + 2e^{-2x}\]
09
Apply the initial conditions
We鈥檙e given that \(y(0) = 1\), so:
\[1 = e^{0}(A\cos(0) + B\sin(0)) + 2e^{0}\]
\[1 = A + 2\]
\[A = -1\]
Now we need to find the derivative of our general inhomogeneous solution \(y(x)\) and apply the second initial condition.
Differentiate \(y(x)\):
\[y'(x) = -2e^{-2x}(-1\cos(x) + B\cos(x) - 2\sin(x)) - 4e^{-2x}\]
We鈥檙e given that \(y'(0) = -2\), so:
\[-2 = -2e^{0}(-1\cos(0) + B\cos(0) - 2\sin(0)) - 4e^{0}\]
\[-2 = 2 + B\]
\[B = -4\]
10
Write the particular solution
Finally, substitute the values of \(A\) and \(B\) back into the general solution of the inhomogeneous equation to find the particular solution:
\[y(x) = e^{-2x}(-1\cos(x) - 4\sin(x)) + 2e^{-2x}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Equations
In differential equations, a homogeneous equation has the same structure on the left side and zero on the right side, resulting in a straightforward setup. For our problem, this translates into:
- Take the differential equation provided.
- Replace any non-zero function on the right-hand side with zero.
Particular Solution
Finding a particular solution involves identifying a specific solution that satisfies the entire non-homogeneous differential equation. The right-hand side of the differential equation isn't zero. Instead, it contains a function, which adds complexity. Here are the steps:
- Identify the form of the non-zero function on the right side. In our case, it's \(2e^{-2x}\).
- Based on this form, guess a solution that mirrors this structure. For our equation, we chose \(y_p(x) = Ce^{-2x}\).
- Substitute this guessed form into the differential equation to solve for constants like \(C\).
Complex Roots
Complex roots arise when solving the auxiliary equation for the homogeneous part of a differential equation. To find these, we use the quadratic formula method. Consider the auxiliary equation \(m^2 + 4m + 5 = 0\). Plugging in the coefficients into the quadratic formula:
- Calculate the discriminant \(b^2 - 4ac\). If it's negative, you will face complex solutions.
- For our equation: \(b = 4\), \(a = 1\), \(c = 5\).
- The discriminant is \(16 - 20 = -4\), yielding complex roots \(m = -2 \pm i\).
Initial Conditions
Initial conditions are specific values for a function and its derivatives at a particular point, used to find specific constants in the general solution of a differential equation. Imagine them as starting points that guide the differential solution where it needs to go. In our example, the given conditions were:
- \(y(0) = 1\)
- \(\frac{\mathrm{d}y}{\mathrm{d}x}(0) = -2\)
- A system of equations for \(A\) and \(B\) in the general solution \(y(x) = e^{-2x}(A\cos(x) + B\sin(x)) + 2e^{-2x}\).
