Theory of Higher-Order Linear Differential Equations

In Problems 38 and 39, use the elimination method of Section to find a general solution to the given system.

$$\begin{gathered} {\mathit{d}}^{\mathbf{2}}\mathit{x}\mathbf{/}\mathit{d}{\mathit{t}}^{\mathbf{2}}\mathbf{-}\mathit{x}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathit{x}\mathbf{+}{\mathit{d}}^{\mathbf{2}}\mathit{y}\mathbf{/}\mathit{d}{\mathit{t}}^{\mathbf{2}}\mathbf{-}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{3}\mathbf{t}} \end{gathered}$$

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.

 ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{x}$

 In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.

${\mathit{z}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{z}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}\mathit{z}\mathbf{=}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{x}}$

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{t}\mathit{a}\mathit{n}\mathit{x}$

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation. ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{s}\mathit{e}\mathit{c}\mathit{\theta }\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{<}\mathit{\theta }\mathbf{<}\mathit{\pi }\mathbf{/}\mathbf{2}$

Deflection of a Beam Under Axial Force. A uniform beam under a load and subject to a constant axial force is governed by the differential equation

 ${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{-}{\mathit{k}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{q}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}\mathbf{ }\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}\mathit{L}\mathbf{,}$

where   is the deflection of the beam, L  is the length of the beam,  k² is proportional to the axial force, and q(x)  is proportional to the load (see Figure 6.2).

(a) Show that a general solution can be written in the form

$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{C}}_{\mathbf{1}}\mathbf{+}{\mathit{C}}_{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{C}}_{\mathbf{3}}{\mathit{e}}^{\mathbf{k}\mathbf{x}}\mathbf{+}{\mathit{C}}_{\mathbf{4}}{\mathit{e}}^{\mathbf{-}\mathbf{k}\mathbf{x}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{k}}^{\mathbf{2}}}\mathbf{\int }\mathit{q}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{x}\mathit{d}\mathit{x}\mathbf{-}\frac{\mathbf{x}}{{\mathbf{k}}^{\mathbf{2}}}\mathbf{\int }\mathit{q}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}\mathbf{+}\frac{{\mathbf{e}}^{\mathbf{k}\mathbf{x}}}{\mathbf{2}{\mathbf{k}}^{\mathbf{3}}}\mathbf{\int }\mathit{q}\mathbf{\left(}\mathit{x}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{k}\mathbf{x}}\mathit{d}\mathit{x}\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{k}\mathbf{x}}}{\mathbf{2}{\mathbf{k}}^{\mathbf{3}}}\mathbf{\int }\mathit{q}\mathbf{\left(}\mathit{x}\mathbf{\right)}{\mathit{e}}^{\mathbf{k}\mathbf{x}}\mathit{d}\mathit{x}$

 (b) Show that the general solution in part (a) can be rewritten in the form 

 $\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{3}}{\mathit{e}}^{\mathbf{k}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{4}}{\mathit{e}}^{\mathbf{-}\mathbf{k}\mathbf{x}}\mathbf{+}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\mathit{q}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathit{G}\mathbf{(}\mathit{s}\mathbf{,}\mathit{x}\mathbf{)}\mathit{d}\mathit{s}\mathbf{,}$

where 

$\mathit{G}\mathbf{(}\mathit{s}\mathbf{,}\mathit{x}\mathbf{)}\mathbf{:}\mathbf{=}\frac{\mathbf{s}\mathbf{-}\mathbf{x}}{{\mathbf{k}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{h}\mathbf{\left[}\mathbf{k}\mathbf{\right(}\mathbf{s}\mathbf{-}\mathbf{x}\mathbf{\left)}\mathbf{\right]}}{{\mathbf{k}}^{\mathbf{3}}}\mathbf{.}$

(c) Let q(x)=1 First compute the general solution using the formula in part (a) and then using the formula in part (b). Compare these two general solutions with the general solution

$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{B}}_{\mathbf{1}}\mathbf{+}{\mathit{B}}_{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{B}}_{\mathbf{3}}{\mathit{e}}^{\mathbf{k}\mathbf{x}}\mathbf{+}{\mathit{B}}_{\mathbf{4}}{\mathit{e}}^{\mathbf{-}\mathbf{k}\mathbf{x}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}{\mathbf{k}}^{\mathbf{2}}}{\mathit{x}}^{\mathbf{2}}\mathbf{,}$

which one would obtain using the method of undetermined coefficients. 

Find a general solution to the Cauchy-Euler equation ${\mathit{x}}^{\mathbf{3}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{3}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{6}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{-}\mathbf{1}}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathit{x}\mathbf{>}\mathbf{0}\mathbf{,}$

given that $\left\{x,x^2,x^3\right\}$is a fundamental solution set for the corresponding homogeneous equation

Find a general solution to the Cauchy-Euler equation ${\mathit{x}}^{\mathbf{3}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathit{x}\mathbf{>}\mathbf{0}\mathbf{,}$

given that $\left\{x,\mathrm{xlnx},x^3\right\}$ is a fundamental solution set for the corresponding homogeneous equation

Given that$\left\{e^x,e^{-x},e^{2x}\right\}$ is a fundamental solution set for the homogeneous equation corresponding to the equation ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}$

determine a formula involving integrals for a particular solution.

Given that $\left\{x,x^{-1},x^4\right\}$  is a fundamental solution set for the homogeneous equation corresponding to the equation${\mathit{x}}^{\mathbf{3}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathit{x}\mathbf{>}\mathbf{0}\mathbf{,}$ determine a formula involving integrals for a particular solution.

Find a general solution to the Cauchy-Euler equation ${\mathit{x}}^{\mathbf{3}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{3}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{4}}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathit{x}\mathbf{>}\mathbf{0}$

Derive the system (7) in the special case when $\mathit{n}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{3}$. [Hint: To determine the last equation, require that $\mathit{I}\mathbf{.}\left[y_p\right]\mathbf{=}\mathit{g}$  and use the fact that ${\mathit{y}}_{\mathbf{1}}\mathbf{,}{\mathit{y}}_{\mathbf{2}}$, and   satisfy the corresponding homogeneous equation.]

Show that\({W_k}(x) = {( - 1)^{(n - k)}}W\left[ {{y_1}, \ldots ,{y_{k - 1}},{y_{k + 1}}, \ldots ,{y_n}} \right](x){\rm{. }}\)

Determine the intervals for which Theorem guarantees the existence of a solution in that

(a)  ${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{-}\mathbf{\left(}\mathit{l}\mathit{n}\mathit{x}\mathbf{\right)}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{x}$

(b) $\left(x^2-1\right){\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{\left(}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{\right)}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\sqrt{\mathbf{x}\mathbf{+}\mathbf{4}}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}{\mathit{e}}^{\mathbf{x}}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{ }$

Determine whether the given functions are linearly dependent or linearly independent on the interval $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$ .

(a)  $\left\{e^{2x},x^2{e}^{2x},e^{-x}\right\}$

(b)  $\left\{e^{x}sin2x,x{e}^{x}sin2x,e^x,x{e}^x\right\}$

(c)   $\left\{2e^{2x}-e^x,e^{2x}+1,e^{2x}-3,e^x+1\right\}$

Find a general solution for the homogeneous linear differential equation with constant coefficients whose auxiliary equation is

(a) \({(r + 5)^2}{(r - 2)^3}{\left( {{r^2} + 1} \right)^2} = 0\).

(b) \({r^4}{(r - 1)^2}{\left( {{r^2} + 2r + 4} \right)^2} = 0\).

Given that \({y_p} = \sin \left( {{x^2}} \right)\) is a particular solution to \({y^{(4)}} + y = \left( {16{x^4} - 11} \right)\sin \left( {{x^2}} \right) - 48{x^2}\cos \left( {{x^2}} \right)\)on\((0,\infty )\), find a general solution.

Find a differential operator that annihilates the given function.

(a)  x² - 2x + 5

(b)  e^3x + x - 1

(c)  x sin2x

(d)  x²e^-2x cos3x

(e)  x² - 2x + xe^-x + sin2x - cos3x

Use the annihilator method to determine the form of a particular solution for the given equation.

(a) ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{6}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$

(b) ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{19}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{20}\mathit{y}\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$

(c) ${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{6}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{x}$

(d) ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{x}\mathit{s}\mathit{i}\mathit{n}\mathit{x}$

Find a general solution to the Cauchy-Euler equation

\(\begin{array}{l}{x^3}{y^{\prime \prime \prime }} - 2{x^2}{y^{\prime \prime }} - 5x{y^\prime } + 5y = {x^{ - 2}},\\x > 0,\end{array}\)

given that \(\left\{ {x,{x^5},{x^{ - 1}}} \right\}\) is a fundamental solution set to the corresponding homogeneous equation.

Find a general solution to the given Cauchy-Euler equation.

(a) \(4{x^3}{y^{\prime \prime \prime }} + 8{x^2}{y^{\prime \prime }} - x{y^\prime } + y = 0,\quad x > 0\)

(b) \({x^3}{y^{\prime \prime \prime }} + 2{x^2}{y^{\prime \prime }} + 2x{y^\prime } + 4y = 0,\quad x > 0\)