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Chapter 8: Problem 15

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(y^{\prime \prime}-4 y^{\prime}+4 y=6 e^{2 x}\)

Short Answer

Expert verified
The differential equation is second-order, linear, non-homogeneous. General solution: \[y = (c_1 + c_2 x + \frac{3}{2} x) e^{2x}\]

Step by step solution

01

- Identify the Type of Differential Equation

This differential equation is a second-order, linear, non-homogeneous differential equation. It has the standard form: \[a(x) y^{\text{''}} + b(x) y^{\text{'}} + c(x) y = g(x)\]where \[a(x) = 1\], \[b(x) = -4\], \[c(x) = 4\], and \[g(x) = 6 e^{2x}\].
02

- Solve the Homogeneous Equation

First, solve the corresponding homogeneous equation: \[y^{\text{''}} - 4y^{\text{'}} + 4y = 0\]To solve this, find the characteristic equation: \[r^2 - 4r + 4 = 0\]Factoring, we get: \[(r - 2)^2 = 0\]This has a repeated root \[r = 2\]. Therefore, the general solution to the homogeneous equation is: \[y_h = (c_1 + c_2 x) e^{2x}\]
03

- Find a Particular Solution

Next, find a particular solution to the non-homogeneous equation. Since the right-hand side \[6 e^{2x}\] suggests a solution of the form \[y_p = A x e^{2x}\].We choose this form because it appears to fit the non-homogeneous part while accommodating the repeated root.Calculate the first and second derivatives:\[y_p = A x e^{2x}\]\[y_p^{'} = A e^{2x} (2x + 1)\]\[y_p^{\text{''}} = A e^{2x} (4x + 4)\]Substitute these into the left-hand side of the original differential equation: \[ (A e^{2x}(4x + 4)) - 4(A e^{2x}(2x + 1)) + 4A x e^{2x} = 6 e^{2x}\]Simplify the expression, \[ 4A e^{2x} = 6 e^{2x}\]\[A = 3/2\]Thus, the particular solution is: \[y_p = \frac{3}{2} x e^{2x}\]
04

- Write the General Solution

Combine the general solution of the homogeneous equation and the particular solution to get the general solution: \[y = y_h + y_p = (c_1 + c_2 x) e^{2x} + \frac{3}{2} x e^{2x}\]Simplify to: \[y = (c_1 + c_2 x + \frac{3}{2} x) e^{2x}\]
05

- Final Answer

The final general solution to the differential equation is: \[y = (c_1 + c_2 x + \frac{3}{2} x) e^{2x}\]where \[c_1\] and \[c_2\] are arbitrary constants.

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Key Concepts

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

second-order differential equations
A second-order differential equation involves the second derivative of a function. It has the general form: \(a(x)y^{\text{''}} + b(x)y^{\text{'}} + c(x)y = g(x)\). In our problem, the equation is \(y^{\text{''}} - 4y^{\text{'}} + 4y = 6e^{2x}\). Here, \(a(x) = 1\), \(b(x) = -4\), \(c(x) = 4\), and \(g(x) = 6e^{2x}\). The term \(y^{\text{''}}\) shows it's second-order. These equations describe systems where the acceleration or rate of change of the rate of change is significant. Common examples include motion and heat conduction.
linear differential equations
Linear differential equations have all derivatives of the unknown function and the function itself appearing linearly (i.e., not raised to any power other than one, and not inside other functions such as sin, cos, etc.). The general form is \(a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} +...+ a_{0}(x)y = g(x)\). Our given equation \(y^{\text{''}} - 4y^{\text{'}} + 4y = 6e^{2x}\) is linear because the terms \(y^{\text{''}}\), \(y^{\text{'}}\), and \(y\) are all first-degree terms.
non-homogeneous differential equations
A non-homogeneous differential equation includes a term \(g(x)\) that is not zero, called the 'forcing function'. This term represents an external influence on the system. Our equation \(y^{\text{''}} - 4y^{\text{'}} + 4y = 6e^{2x}\) is non-homogeneous because \(g(x) = 6e^{2x}\) is not zero. Solving such equations involves finding both the general solution to the related homogeneous equation (where \(g(x) = 0\)) and a particular solution that accounts for \(g(x)\).
characteristic equation
The characteristic equation is derived from a homogeneous linear differential equation and helps find the general solution. For the homogeneous part of our equation \(y^{\text{''}} - 4y^{\text{'}} + 4y = 0\), you assume solutions of the form \(y = e^{rx}\), leading to \(r^2 - 4r + 4 = 0\). Factoring it gives \((r - 2)^2 = 0\), which results in a repeated root \(r = 2\). This helps form the general solution to the homogeneous equation.
particular solution
A particular solution addresses the non-homogeneous part of the equation. We assume a form based on \(g(x)\), the right-hand side of the equation. Here, \(g(x) = 6e^{2x}\) suggests a solution of the form \(y_p = Ax e^{2x}\). Substituting \(y_p\) and its derivatives into the original equation, we solve for \(A\). In this exercise, we find \(A = \frac{3}{2}\), giving us the particular solution \(y_p = \frac{3}{2}xe^{2x}\).
general solution
The general solution of a non-homogeneous differential equation combines the general solution of the associated homogeneous equation and the particular solution. For our example, the homogeneous solution is \((c_1 + c_2 x)e^{2x}\), and the particular solution is \(\frac{3}{2} x e^{2x}\). Therefore, the general solution is \(y = (c_1 + c_2 x + \frac{3}{2} x)e^{2x}\), often simplified to \(y = (c_1 + (c_2 + \frac{3}{2}) x)e^{2x}\). This formula includes arbitrary constants \(c_1\) and \(c_2\) which can be determined by initial conditions.

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Most popular questions from this chapter

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(3 x^{3} y^{2} y^{\prime}-x^{2} y^{3}=1\)

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Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(x\left(y y^{\prime \prime}+y^{\prime 2}\right)=y y^{\prime}\) Hint: Iet \(u=1 x\).

By separation of variables, solve the differential equation \(d y / d x=\sqrt{1-y^{2}}\) to obtain solution containing one arbitrary constant. Although this solution may be referred to as the " general solution," show that \(y=1\) is a solution of the differential equation not obtainablc from the "general solution" by any choice of the arbitrary constant. The solution \(y=1\) is called a singular selution; \(y=-1\) is another singular solution. Sketch a number of graphs of the "general solution" for different values of the arbitrary constant and observe that \(y=1\) is tangent to all of them. This is characteristic of a singular solution -its graph is tangent at each point to one of the graphs of the "general solution." Note that the given differential equation is not linear; for linear equations, all solutions are contained in the general solution, but nonlinear equations may have singular solutions which cannot be obtained from the "general solution" by specializing the arbitrary constant (or constants). Thus a nonlinear first-order equation in \(x\) and \(y\) may have two (or more) solutions passing through a given point in the \((x, y)\) plane, whereas a linear first-order equation always has just one such solution. Show that any continuous curve made up of pieces of \(y=1, y=-1\), and the sinc curves of the "general solution," gives a solution of the above differential equation. Sketch such a solution curve on your graphs.
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