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Chapter 8: Problem 14

Identify each of the differential equations as 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

Expert verified
Non-homogeneous linear second-order; solution: \[ y = e^{-2x}((C_1 - 2) \cos x + (C_2 + 1) \sin x) \]

Step by step solution

01

Identify the type of differential equation

The given differential equation is \[ y'' + 4y' + 5y = 2e^{-2x} \, \cos x \] This is a non-homogeneous linear second-order differential equation because it contains both the second derivative and the first derivative of the unknown function y.
02

Solve the characteristic equation

First, solve the homogeneous part of the differential equation, which is \[ y'' + 4y' + 5y = 0 \] The characteristic equation is \[ r^2 + 4r + 5 = 0 \] Solving this quadratic equation using the quadratic formula \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 4 \), and \( c = 5 \), gives: \[ r = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = -2 \pm i \] So the complex roots are \[ r = -2 \pm i \]
03

Write the general solution of the homogeneous equation

Using the roots \( r = -2 \pm i \), the general solution for the homogeneous differential equation is: \[ y_h = e^{-2x}(C_1 \cos x + C_2 \sin x) \] where \(C_1\) and \(C_2\) are arbitrary constants.
04

Use the method of undetermined coefficients

Now find a particular solution to the non-homogeneous differential equation. For the non-homogeneous part, which is \[ 2e^{-2x} \cos x \], guess a solution of the form: \[ y_p = e^{-2x}(A \cos x + B \sin x) \]. Substitute \( y_p \) into the non-homogeneous differential equation and solve for \(A\) and \(B\).
05

Compute derivatives

Compute the first and second derivatives of \( y_p \): \[ y_p = e^{-2x}(A \cos x + B \sin x) \]\[ y_p' = e^{-2x}(-2A \cos x - 2B \sin x) + e^{-2x}(-A \sin x + B \cos x) \]\[ y_p'' = e^{-2x}(4A \cos x + 4B \sin x) + e^{-2x}(-2A \sin x + 2B \cos x) + e^{-2x}(-A \cos x - B \sin x) \]
06

Substitute back into the differential equation

Combine all terms and compare coefficients: \[ (4A - 4A) \cos x + (4B - 4B) \sin x + (-2A + B) \cos x + (-2B - A) \sin x = 2 \cos x \] This simplifies to: \[ 0 \cos x + 0 \sin x = 2 \cos x \] Therefore, substituting back, we find that \(A = -2\) and \(B = 1\).
07

Write the particular solution

The particular solution \( y_p \) is: \[ y_p = e^{-2x}(-2 \cos x + \sin x) \]
08

Write the general solution

Combine the homogeneous and particular solutions to get the general solution: \[ y = y_h + y_p \]\[ y = e^{-2x}(C_1 \cos x + C_2 \sin x) + e^{-2x}(-2 \cos x + \sin x) \] which simplifies to: \[ y = e^{-2x}((C_1 - 2) \cos x + (C_2 + 1) \sin x) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Non-Homogeneous Differential Equations
A non-homogeneous differential equation includes a term that does not depend on the solution function or its derivatives. For example, the given equation \[ y'' + 4y' + 5y = 2e^{-2x} \cos x \]; is non-homogeneous because of the term \; 2e^{-2x} \cos x \;.
This term makes the equation non-homogeneous because it does not involve \; y \; itself or any of its derivatives.
Understanding the nature of this term is crucial because it impacts the method for finding the solution.
Method of Undetermined Coefficients
This method is used to find a particular solution to non-homogeneous differential equations. The idea is to guess a form for the particular solution based on the non-homogeneous term.
In our case, the term is \[ 2e^{-2x} \ \cos x \ \];, so we guess the particular solution \[ y_p = e^{-2x}(A \ \cos x + B \ \sin x) \];.
We then differentiate \; y_p \; and substitute it into the original non-homogeneous equation. This allows us to solve for the constants \; A \; and \; B \;.
Finally, we combine this particular solution with the general solution of the homogeneous equation.
Characteristic Equation
To solve the homogeneous part of a differential equation, we use the characteristic equation. It is derived from the differential equation by assuming solutions of the form \; y = e^{rx} \;.
For \ y'' + 4y' + 5y = 0 \;, the characteristic equation is \[ r^2 + 4r + 5 = 0 \];.
Solving this quadratic equation with the quadratic formula \[ r = \frac{ -b \pm \sqrt{b^2-4ac}}{2a} \];, we find the roots are \; -2 \pm i \;.
These roots help define the general solution to the homogeneous equation.
General Solution of Differential Equations
The general solution of a non-homogeneous differential equation is the sum of the homogeneous solution and the particular solution.
For the homogeneous part \ y'' + 4y' + 5y = 0 \;, we determined the solution to be \[ y_h = e^{-2x}(C_1 \ \cos x + C_2 \ \sin x) \];.
For the non-homogeneous part \; y'' + 4y' + 5y = 2e^{-2x} \ \cos x \;, with the particular solution \[ y_p = e^{-2x}(-2 \ \cos x + \ \sin x) \];.
Combining these, the general solution is \[ y = e^{-2x}((C_1 - 2) \ \cos x + (C_2 + 1) \ \sin x) \];.
Homogeneous Solution
A homogeneous solution is obtained by setting the non-homogeneous term to zero. This gives us the homogeneous differential equation \ y'' + 4y' + 5y = 0 \;.
We solve this by finding the roots of the characteristic equation \; r \;, which we found to be \; r = -2 \pm i \;.
Thus, the homogeneous solution is \[ y_h = e^{-2x}(C_1 \ \cos x + C_2 \ \sin x) \];, where \; C_1 \; and \; C_2 \; are arbitrary constants.
This solution forms part of the overall general solution to the original non-homogeneous differential equation.

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Most popular questions from this chapter

Find the inverse Laplace transform of: $$\frac{p}{(p+a)^{3}}$$

(a) Show that $$\begin{aligned} D\left(e^{a x} y\right) &=e^{a x}(D+a) y \\ D^{2}\left(e^{a x} y\right) &=e^{a x}(D+a)^{2} y \end{aligned}$$ and so on; that is, for any positive integral \(n\) $$D^{n}\left(e^{a x} y\right)=e^{a x}(D+a)^{n} y$$ Thus show that if \(L(D)\) is any polynomial in the operator \(D,\) then $$L(D)\left(e^{a x} y\right)=e^{a x} L(D+a) y$$ This is called the exponential shift. (b) Use (a) to show that $$\begin{aligned} (D-1)^{3}\left(e^{x} y\right) &=e^{x} D^{3} y \\ \left(D^{2}+D-6\right)\left(e^{-3 x} y\right) &=e^{-3 x}\left(D^{2}-5 D\right) y \end{aligned}$$ (c) Replace \(D\) by \(D-a,\) to obtain $$e^{a x} P(D) y=P(D-a) e^{a x} y$$ This is called the inverse exponential shift. (d) Using (c), we can change a differential equation whose right-hand side is an exponential times a polynomial, to one whose right-hand side is just a polynomial. For example, consider \(\left(D^{2}-D-6\right) y=10 x e^{3 x} ;\) multiplying both sides by \(e^{-3 x}\) and using \((\mathrm{c}),\) we get $$\begin{aligned} e^{-3 x}\left(D^{2}-D-6\right) y &=\left[(D+3)^{2}-(D+3)-6\right] y e^{-3 x} \\\ &=\left(D^{2}+5 D\right) y e^{-3 x}=10 x \end{aligned}$$ Show that a solution of \(\left(D^{2}+5 D\right) u=10 x\) is \(u=x^{2}-\frac{2}{5} x ;\) then \(y e^{-3 x}=x^{2}-\frac{2}{5} x\) or \(y=e^{3 x}\left(x^{2}-\frac{2}{5} x\right) .\) Use this method to solve Problems 23 to 26

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}-2 y^{\prime}+5 y=5 x+4 e^{x}(1+\sin 2 x)$$

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$(y+2 x) d x-x d y=0$$

Compare the temperatures of your cup of coffee at time \(t\) (a) if you add cream and let the mixture cool; (b) if you let the coffee and cream sit on the table and mix them at time \(t\). Hints: Assume Newton's law of cooling (Problem 2.27) for both coffee and cream (where it is a law of heating). Combine \(n^{\prime}\) units of cream initially at temperature \(T_{0}^{\prime}\) with \(n\) units of coffee initially at temperature \(T_{0},\) and find the temperature at time \(t\) in (a) and in (b) assuming that the air temperature remains a constant \(T_{a},\) and that the proportionality constant in the law of cooling is the same for both coffee and cream.
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