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Chapter 8: Problem 15

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

Short Answer

Expert verified
The general solution is y(x) = (C_1 + C_2x + 3x)e^{2x}

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is:y'' - 4y' + 4y = 6e^{2x}This equation is a second-order linear non-homogeneous differential equation.
02

Solve the Homogeneous Equation

First, solve the homogeneous part of the equation:y'' - 4y' + 4y = 0The characteristic equation for this is:r^2 - 4r + 4 = 0Factoring the characteristic equation gives:(r - 2)^2 = 0So, the repeated root is r = 2. The general solution to the homogeneous equation is:y_h(x) = (C_1 + C_2x)e^{2x}
03

Find a Particular Solution

To find the particular solution to the non-homogeneous equation, use the method of undetermined coefficients. The right-hand side of the equation is 6e^{2x}, so assume a particular solution of the form:y_p = A x e^{2x}Calculate y_p' and y_p'' and substitute them into the original equation:y_p' = A e^{2x} + 2Ax e^{2x}y_p'' = 2A e^{2x} + 2A e^{2x} + 4Ax e^{2x}Substituting these into the differential equation:( 2A e^{2x} + 2A e^{2x} + 4Axe^{2x} ) - 4 ( A e^{2x} + 2Axe^{2x} ) + 4 (Axe^{2x}) = 6e^{2x}Simplify and solve for A:2Ae^{2x} = 6e^{2x}A = 3, so the particular solution is y_p = 3xe^{2x}
04

Write the General Solution

Combine the homogeneous and particular solutions to find the general solution:y(x) = y_h(x) + y_p(x)y(x) = (C_1 + C_2x)e^{2x} + 3xe^{2x}

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Key Concepts

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

non-homogeneous differential equation
The equation given is a second-order linear non-homogeneous differential equation. Non-homogeneous means there's an extra term that doesn't depend on y or its derivatives. This term is referred to as the 'non-homogeneous' part. For example, in the given equation, the term 6e^{2x} makes it non-homogeneous.

In general, these types of equations can be written like this:
  • Homogeneous part: y'' - 4y' + 4y = 0
  • Non-homogeneous part: 6e^{2x}
The strategy to solve them involves dealing with each of these parts separately鈥攕olving the homogeneous equation first, then finding a particular solution that deals with the non-homogeneous part.

Combining these results gives the general solution.
method of undetermined coefficients
One popular technique to find the particular solution of a non-homogeneous differential equation is the method of undetermined coefficients. This method works well when the non-homogeneous term is a simple function like a polynomial, exponential, or sine/cosine.

Here's a step-by-step outline of how this method is used in the example:
  • Guess the form of the particular solution. Since we have 6e^{2x} as the non-homogeneous term, we'll guess a solution of the form Ax e^{2x}.
  • Differentiate the guessed form.
  • Substitute these derivatives back into the original equation.
  • Compare coefficients to solve for the unknowns鈥� in this case, we determined that A equals 3.
The key idea here is to make an educated guess about the form of the specific solution, then refine that guess using the original equation.
general solution
The general solution of a differential equation combines both the homogeneous solution and the particular solution.

In simpler terms:
  • The homogeneous solution, y_h(x), solves the equation without the non-homogeneous part.
  • The particular solution, y_p(x), solves the entire equation including the non-homogeneous part.
In this example, the homogeneous solution is \(y_h(x) = (C_1 + C_2x)e^{2x}\), and the particular solution found using the method of undetermined coefficients is \(y_p(x) = 3x e^{2x}\).

So, the overall general solution is: \[y(x) = (C_1 + C_2x)e^{2x} + 3x e^{2x}\]
characteristic equation
The characteristic equation helps solve the homogeneous part of the second-order linear differential equation. To form this equation, replace y, y', and y'' with their characteristic counterparts.

For the equation y'' - 4y' + 4y = 0, the characteristic equation is: \[r^2 - 4r + 4 = 0\] This is basically found by assuming solutions of the form e^{rt} and determining r.

In this example, this quadratic equation factors as: \(r - 2)^2 = 0\) yielding a repeated root r = 2. The solution to the homogeneous equation involving repeated roots takes the form: \[y_h(x) = (C_1 + C_2 x)e^{2x}\]
particular solution
The particular solution, y_p(x), addresses the non-homogeneous part of the differential equation. To find it, you use methods like the method of undetermined coefficients.

Given the term 6e^{2x} in the original equation, we hypothesize a solution of the form Ax e^{2x}. By differentiating this hypothesized solution and substituting it back into the original differential equation, we can solve for A.
  • First guess: y_p = Ax e^{2x}
  • First derivative: y_p' = A e^{2x} + 2Ax e^{2x}
  • Second derivative: y_p'' = 2A e^{2x} + 2A e^{2x} + 4Ax e^{2x}
These steps will yield the particular solution \(y_p(x) = 3xe^{2x}\).

Finally, by combining y_p(x) with the homogeneous solution y_h(x), we obtain the general solution to the differential equation.

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Most popular questions from this chapter

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$\left(x \cos y-e^{-\sin y}\right) d y+d x=0$$

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$d y-\left(2 y+y^{2} e^{3 x}\right) d x=0$$

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime}-y=2 e^{t}, \quad y_{0}=3$$

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

(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
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