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Show that if \(y_{h}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=0\) and \(y_{p}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=f(t)\) then for any number \(C, y_{g}(t)=y_{p}(t)+C y_{h}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=f(t)\)

Short Answer

Expert verified
The function \( y_g(t) = y_p(t) + C y_h(t) \) solves the given non-homogeneous differential equation.

Step by step solution

01

Identify Given Functions

We are given two functions, \( y_h(t) \) and \( y_p(t) \). The function \( y_h(t) \) solves the homogeneous equation \( y''(t) + p y'(t) + q y(t) = 0 \). The function \( y_p(t) \) solves the non-homogeneous equation \( y''(t) + p y'(t) + q y(t) = f(t) \). We need to prove that \( y_g(t) = y_p(t) + C y_h(t) \) solves the non-homogeneous equation as well.
02

Calculate Derivatives of \( y_g(t) \)

First, calculate the first and second derivatives of \( y_g(t) = y_p(t) + C y_h(t) \). The first derivative is:\[ y_g'(t) = y_p'(t) + C y_h'(t) \]The second derivative is:\[ y_g''(t) = y_p''(t) + C y_h''(t) \]
03

Substitute Derivatives into the Equation

Substitute \( y_g(t), y_g'(t), \) and \( y_g''(t) \) into the differential equation \( y''(t) + p y'(t) + q y(t) \):\[y_g''(t) + p y_g'(t) + q y_g(t) = (y_p''(t) + C y_h''(t)) + p (y_p'(t) + C y_h'(t)) + q (y_p(t) + C y_h(t))\]
04

Simplify the Equation

Simplify the expression:\[ (y_p''(t) + p y_p'(t) + q y_p(t)) + C (y_h''(t) + p y_h'(t) + q y_h(t)) \]Using the fact that \( y_p(t) \) solves the non-homogeneous equation and \( y_h(t) \) solves the homogeneous equation, we get:\[ y_p''(t) + p y_p'(t) + q y_p(t) = f(t) \]\[ y_h''(t) + p y_h'(t) + q y_h(t) = 0 \]
05

Final Expression to Verify the Solution

Substitute back:\[ f(t) + C \cdot 0 = f(t) \]This confirms that \( y_g(t) = y_p(t) + C y_h(t) \) solves the differential equation \( y''(t) + p y'(t) + q y(t) = f(t) \).

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Key Concepts

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

Homogeneous Equations
A homogeneous differential equation is a vital concept in understanding differential equations. In these equations, all the terms depend on the function or its derivatives, and importantly, there is no independent term, which means it equals zero. So every solution curves pass through the origin. For instance, a second-order homogeneous differential equation can be written as:\[ y''(t) + p y'(t) + q y(t) = 0 \]Here, any solution of this equation is called the complementary function, often denoted as \( y_h(t) \). It mainly characterizes systems where forces act on one another without any external influence being present.
  • The general solution includes terms of possible sin, cos, and exponential functions, depending on the characteristic equation's roots.
  • The principle of superposition applies, meaning any linear combination of solutions is also a solution.
Understanding and solving homogeneous equations is crucial because they often form the foundation upon which we build solutions for more complex expressions in differential equations.
Non-Homogeneous Equations
Non-homogeneous differential equations are slightly more complex because they involve external forces or inputs. Simply put, they equal something other than zero. These equations model systems influenced by external forces, resulting in solutions that behave differently from homogeneous equations.The form of a non-homogeneous second-order differential equation can be expressed as:\[ y''(t) + p y'(t) + q y(t) = f(t) \]The function \( f(t) \) denotes the non-zero part of the equation, marking it as non-homogeneous.
  • Non-homogeneous equations require a particular solution, \( y_p(t) \), which accounts for the effect of the external term \( f(t) \).
  • The general solution combines both the particular solution \( y_p(t) \) and the complementary solution \( y_h(t) \): \( y(t) = y_p(t) + C y_h(t) \), where \( C \) is an arbitrary constant.
  • This formulation allows the general solution to incorporate the natural response of the system, as well as the forced response.
Understanding how to solve these equations helps in addressing real-world problems where systems are subjected to outside influences.
Solution Verification
Verification of solutions is a crucial step in dealing with differential equations. It ensures that the functions identified as solutions genuinely satisfy the equation in question.In the exercise above, we verify whether \( y_g(t) = y_p(t) + C y_h(t) \) qualifies as a solution for the differential equation \( y''(t) + p y'(t) + q y(t) = f(t) \).
  • First, identify derivatives of \( y_g(t) \) and substitute these into the original differential equation.
  • Upon substitution, each solution component corresponds to distinct terms in the equation: \( y_p(t) \) handles the non-zero part, \( f(t) \), while \( y_h(t) \) handles the zero homogeneous part.
  • The simplification demonstrates that \( f(t) + C \cdot 0 = f(t) \), confirming that the expression satisfies the equation.
Verification is essential as it ensures that theoretical solutions are actually applicable in modeling real systems, linking mathematical solutions with practical realities.

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Most popular questions from this chapter

Show that the for \(m, r,\) and \(k\) positive, the characteristic roots $$ r_{1}=\frac{-r+\sqrt{r^{2}-4 m k}}{2 m} \quad \text { and } \quad r_{2}=\frac{-r-\sqrt{r^{2}-4 m k}}{2 m} $$ for \(m y^{\prime \prime}(t)+r y^{\prime}(t)+k y(t)=0\) are both negative or have negative real parts.

Anderson and May \(^{4}\) give the following model of immune effector cells (helper and cytotoxic T-cells), \(E,\) that limit viral population, \(V\), growth in a human body. $$ \begin{array}{l} d E / d t=\Lambda-\mu E+\epsilon V E \\ d V / d t=r V-\sigma V E \end{array} $$ a. \(\Lambda\) is intrinsic production rate of effector cells from bone marrow. Give similar meaning to each of the other four terms on the RHS of Equations 18.64 . b. Find the equilibrium effector cell population, \(\hat{E},\) in the absence of virus. c. Suppose an inoculum \(V_{0}\) of virus is introduced into the body with \(E=\hat{E}\). Find conditions on \(r\), \(\sigma,\) and \(\hat{E}\) in order that the viral population will increase. d. The Jacobian matrix at any \((E, V)\) is $$ J(E, V)=\left[\begin{array}{rc} -\mu+\epsilon V & \epsilon E \\ -\sigma V & r-\sigma V \end{array}\right] $$ Show that $$ J(\hat{E}, 0)=\left[\begin{array}{rr} -\mu & \sigma \frac{\Lambda}{\mu} \\ 0 & r-\sigma \frac{\Lambda}{\mu} \end{array}\right] $$ e. The characteristic values of the upper diagonal matrix \(J(\hat{E}, 0)\) are the diagonal entries, \(-\mu\) and \(r-\sigma \frac{\Lambda}{\mu} .\) What happens to a small introduction of virus into a healthy individual if both characteristic values are negative? f. In order that the viral population to expand it is necessary that \(r-\sigma \frac{\Lambda}{\mu}>0 .\) What is the role of \(r\) in the model? g. If that condition is met and the viral population increases, show that there will be an equilibrium state, $$ E^{*}=r / \sigma, \quad V^{*}=\frac{\mu r-\Lambda \sigma}{\epsilon r} $$ It is clear that in order for \(V^{*}\) to be positive, we must have (again) $$ \mu r-\Lambda \sigma>0 \quad \text { so that } \quad r>\frac{\Lambda \sigma}{\mu} $$ h. Anderson and May report this system to be asymptotically stable, at \(\left(E^{*}, V^{*}\right),\) but only weakly so, meaning that it is subject to wide oscillations. Show that $$ J\left(E^{*}, V^{*}\right)=\left[\begin{array}{cc} -\frac{\Lambda \sigma}{r} & \epsilon \frac{r}{\sigma} \\ -\sigma \frac{\mu r-\Lambda \sigma}{\epsilon r} & 0 \end{array}\right] $$ i. The characteristic equation of a \(2 \times 2\) matrix \(M\) is is \(s^{2}-\operatorname{trace}(M) s+\operatorname{det}(M)=0\). Show that the characteristic equation of \(J\left(E^{*}, V^{*}\right)\) is $$ s^{2}+\frac{\Lambda \sigma}{r} s+\mu r-\Lambda \sigma=0 $$ j. The roots of the characteristic equation 18.65 are $$ \frac{-\frac{\Lambda \sigma}{r} \pm \sqrt{\left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)}}{2} $$ Argue that the real part is negative so that the system is stable. Note: Two cases: $$ \left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)>0 \quad \text { and } \quad\left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)<0 $$ k. Use \(\Lambda=1, \mu=0.5, \epsilon=0.02, r=0.25,\) and \(\sigma=0.01\) and compute \(E^{*}, V^{*},\) and the stability at \(\left(E^{*}, V^{*}\right)\) l. Let \(E_{0}=2, V_{0}=1, \Lambda=1, \mu=0.5, \epsilon=0.02, r=0.25,\) and \(\sigma=0.01 .\) Approximate the solutions to Equations 18.64 using the trapezoid rule. Observe that the rise in viral load precedes the increase in effector cells.

Exercise \(18.2 .3 \quad\) a. Show that for any number, \(x_{0},\) the solution to $$ \begin{array}{c} x(0)=x_{0} \quad x^{\prime}=-y \\ y(0)=0 \quad y^{\prime}=25 x \\ \text { is } x(t)=x_{0} \cos 5 t, \quad y(t)=5 x_{0} \sin 5 t \end{array} $$ b. Draw the graph in the \(x, y-\) plane of the solutions for \(x_{0}=1\) and \(x_{0}=0.5\). c. Show that the origin (0,0) is a stable equilibrium point of Equations 18.28 . d. Show that the origin (0,0) not an asymptotically stable equilibrium of Equations 18.28 .

Show that if \(y_{p, 1}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=f_{1}(t)\) and \(y_{p, 2}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=f_{2}(t),\) then for any numbers \(A\) and \(B\) $$ A y_{p, 1}(t)+B y_{p, 2} \quad \text { solves } \quad y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=A f_{1}(t)+B f_{2}(t) $$

Compute and graph the solutions to $$ y^{\prime \prime}+p y^{\prime}+y=0 \quad y(0)=1 \quad y^{\prime}(0)=0 $$ for \(p=4, p=2, p=1, p=0,\) and \(p=-2\)

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