Find the general solution of each of the following differential equations.

• [math]\deriv2y + \dydx - 2y = 5{e^{-x}}[/math]
• [math](D+2)(D-1)y=6e^{-2x}[/math]
• [math](D^2-3D+2)y = 4x+3[/math]
• [math]\deriv2y + y = e^x[/math]
• [math](D^2+1)y = x^2+1[/math].

Using equations and, prove that, if [math]L[/math] is a linear operator, then

Show that equations and can be replaced by a single equation. That is, prove that a function [math]L[/math] is a linear operator if and only if

Prove \ref{thm 11.3.2}; i.e., if [math]L_1[/math] and [math]L_2[/math] are linear operators, then the composition [math]L_1L_2[/math] is also a linear operator.

Prove the second equation in \ref{thm 11.3.3}, i.e., the distributive law [math](L_1+L_2)L_3 = L_1L_3+L_2L_3[/math].

It might at first seem more natural to define the product of two linear operators [math]L_1[/math] and [math]L_2[/math] by the equation

(This is the way the product of two real-valued functions is defined.) Using this definition, show that, if [math]D[/math] is the derivative, the [math]D^2[/math] is not a linear operator.

Let [math]f(x)[/math] be a given function and [math]L[/math] a linear operator. Define [math]f(x)L[/math] by the equation

Show that [math]f(x)L[/math] satisfies equations and and hence is a linear operator.

• Show that the operation of multiplication by a given function [math]f(x)[/math] is a linear operator. That is, prove that, if [math]M[/math] is defined by
  [[math]] My = f(x)y , [[/math]]
  then [math]M[/math] is the linear operator.
• Show that the composition of a linear operator [math]L[/math] followed by the operation of multiplication by [math]f(x)[/math] is just the operator [math]f(x)L[/math] defined in Problem Exercise.

(See Problems Exercise and Exercise.) If [math]f(x)[/math] is a differentiable function and if [math]D[/math] is the derivative, then both linear operators [math]f(x)D[/math] and [math]Df(x)[/math] are examples of linear differential operators more general than the type discussed in the text. Show that

by applying both sides to the function [math]y=x[/math]. Thus the commutative law of multiplication fails.

Let [math]f[/math] and [math]g[/math] be differentiable complex-valued functions of a real variable. Show that the ordinary product rule for differentiation is still valid; i.e., prove that

[Hint: Let [math]f(x) = f_1(x)+if_2(x)[/math] and [math]g(x) = g_1(x) + ig_2(x)[/math], and apply the definitions of the derivative and of multiplication of complex numbers.]