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Chapter 8: Problem 16

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}-5 y^{\prime}+6 y=e^{2 x}$$

Short Answer

Expert verified
The general solution is \[y = C_1 e^{2x} + C_2 e^{3x} - x e^{2x}.\]

Step by step solution

01

Identify the type of differential equation

The given differential equation is \[y^{\text{''}} - 5y^{\text{'}} + 6y = e^{2x}.\] It is a non-homogeneous linear second order differential equation with constant coefficients.
02

Solve the corresponding homogeneous equation

First, solve the homogeneous part \[y^{\text{''}} - 5y^{\text{'}} + 6y = 0.\] The characteristic equation associated with this is \[r^2 - 5r + 6 = 0.\]This can be factored as \[(r - 2)(r - 3) = 0.\] Therefore, the solutions to the characteristic equation are \[r = 2\]and\[r = 3.\]The general solution to the homogeneous equation is \[y_h = C_1 e^{2x} + C_2 e^{3x},\] where \(C_1\) and \(C_2\) are arbitrary constants.
03

Find a particular solution

To find a particular solution \(y_p\) of the non-homogeneous equation \[y^{\text{''}} - 5y^{\text{'}} + 6y = e^{2x},\] we apply the method of undetermined coefficients. Assume a particular solution of the form \[y_p = A x e^{2x}.\]
04

Compute the derivatives of the particular solution

First compute \[y_p^{\text{'}} = A e^{2x} + 2Ax e^{2x},\] and then \[y_p^{\text{''}} = 2A e^{2x} + 2A e^{2x} + 4Ax e^{2x} = 4A e^{2x} + 4Ax e^{2x}.\]
05

Substitute the derivatives back into the original equation

We substitute \(y_p\), \(y_p^{\text{'}}\), and \(y_p^{\text{''}}\) back into the differential equation: \[(4A e^{2x} + 4Ax e^{2x}) - 5(A e^{2x} + 2A x e^{2x}) + 6(A x e^{2x}) = e^{2x}.\]Simplify the left side: \[4A e^{2x} + 4A x e^{2x} - 5A e^{2x} - 10Ax e^{2x} + 6Ax e^{2x} = e^{2x}.\] Combine like terms: \[(-A e^{2x}) + (0x e^{2x}) = e^{2x}.\]Thus, \[-A e^{2x} = e^{2x},\] giving \[A = -1.\]
06

Write the general solution

The general solution of the differential equation is given by the sum of the homogeneous solution and the particular solution: \[y = y_h + y_p = C_1 e^{2x} + C_2 e^{3x} + (-x e^{2x}).\]Therefore, \[y = C_1 e^{2x} + C_2 e^{3x} - x e^{2x}.\]

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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 differential equation is called non-homogeneous if it includes a term that is not dependent on the function or its derivatives. In the exercise provided, the presence of the term \(e^{2x}\) indicates the equation is non-homogeneous. For such an equation, the general solution is a combination of the homogeneous solution and a particular solution.
The given differential equation is:
\[y^{''} - 5y' + 6y = e^{2x}\].
Method of Undetermined Coefficients
This method is often used to find a particular solution for non-homogeneous linear differential equations with constant coefficients. The idea is to assume a specific form for the particular solution based on the non-homogeneous term. Let's break it down:
  • Identify the form of the non-homogeneous term, which in our case is \(e^{2x}\).
  • Based on this, we assume a particular solution of the form \(y_p = Ax e^{2x}\).
We then find the derivatives of this assumed solution and substitute them back into the differential equation to solve for the coefficient \(A\).
Characteristic Equation
To solve the homogeneous part of the differential equation, we first need to find the characteristic equation. For the equation:
\[y^{''} - 5y' + 6y = 0\],
the characteristic equation is obtained by replacing \(y\) with \(e^{rx}\) which gives us:
\[r^2 - 5r + 6 = 0\].
Solving this quadratic equation, we find the roots \(r = 2\) and \(r = 3\).
These roots help us form the general solution to the homogeneous part.
Homogeneous Solution
The homogeneous solution, \(y_h\), is derived from the roots of the characteristic equation. With roots \(r = 2\) and \(r = 3\), the solution is:
\[y_h = C_1 e^{2x} + C_2 e^{3x}\].
This represents the part of the solution that corresponds to the homogeneous differential equation.
Particular Solution
To find the particular solution \(y_p\), we assumed the form \(y_p = Ax e^{2x}\) since our non-homogeneous term is \(e^{2x}\). We then calculated the first and second derivatives:
\[y_p' = A e^{2x} + 2Ax e^{2x}\]
\[y_p'' = 2A e^{2x} + 2A e^{2x} + 4Ax e^{2x} = 4A e^{2x} + 4Ax e^{2x}\].
Substituting these derivatives back into the original differential equation and solving for \(A\), we found that \(A = -1\). Thus, the particular solution is:
\[y_p = -x e^{2x}\].
Combining y_h and y_p gives us the general solution:
\[y = C_1 e^{2x} + C_2 e^{3x} - x e^{2x}\].

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Most popular questions from this chapter

A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top and bottom of the block are parallel planes which remain horizontal during the oscillations and that the sides of the block are vertical. Show that the period of the motion (neglecting friction) is \(2 \pi \sqrt{h / g}\) where \(h\) is the vertical height of the part of the block under water when it is floating at rest. Hint: Recall that the buoyant force is equal to the weight of displaced water.

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

Solve the following sets of equations by the Laplace transform method. $$\begin{array}{ll} y^{\prime}-z^{\prime}-y=\cos t & y_{0}=-1 \\ y^{\prime}+y-2 z=0 & z_{0}=0 \end{array}$$

(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

Find a particular solution satisfying the given conditions. \(y y^{\prime \prime}+y^{\prime 2}+4=0 \quad y=3, y^{\prime}=0\) when \(x=1\)
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