Q 6.4-14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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. 

Step-by-Step Solution

Answer

(a)   $$\begin{gathered} \left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}+\frac{1}{k^2}\int q\left(x\right)x \text{d}x-\frac{x}{k^2}\int q\left(x\right)\text{d}x \\ +\frac{e^{kx}}{2k^3}\int q\left(x\right)e^{-kx}dx-\frac{e^{-kx}}{2k^3}\int q\left(x\right)e^{kx}dx \end{gathered}$$

(b) The general solution in part (a) can be rewritten in the wanted form.

(c)  $$\begin{gathered} y_a\left(x\right)=c_1-\frac{1}{k^4}+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}-\frac{1}{2k^2}{x}^2 \\ {y}_b\left(x\right)=c_1-\frac{1}{k^4}+c_{2}x+\left(c_3+\frac{1}{2k^4}\right)e^{kx}+\left(c_4+\frac{1}{2k^4}\right)e^{-kx}-\frac{1}{2k^2}{x}^2 \end{gathered}$$

Step 1: Determine the corresponding homogeneous equation

Consider the given equation

 $y^{\left(4\right)}\left(x\right)-k^2{y}^{\text{'}\text{'}}\left(x\right)=q\left(x\right)$

First, we have to solve the corresponding homogeneous equation

 $y^{\left(4\right)}\left(x\right)-k^2{y}^{\text{'}\text{'}}\left(x\right)=0$

The auxiliary equation is

 $r^4-k^2{r}^2=r^2\left(r^2-k^2\right)=r^2(r-k)(r+k)=0$

We see that the solutions to the auxiliary equation are  $r_1=r_2=0,r_3=k$ and $r_4=-k$ . Therefore, a general solution to the corresponding homogeneous equation is

 $y_h\left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}$

and $\left\{1,x,e^{kx},e^{-kx}\right\}$  is a fundamental solution set to the corresponding homogeneous equation. Hence, we can obtain a particular solution of the form

 $y_p\left(x\right)=V_1\left(x\right)+V_2\left(x\right)x+V_3\left(x\right)e^{kx}+V_4\left(x\right)e^{-kx}.$

Let’s determine the functions V₁, V₂, V₃  and V₄ .

Step 2: Evaluate the five determinants

First, we must evaluate the five determinants

Now we can calculate the undetermined functions V₁, V₂, V₃  and V₄ 

$$\begin{gathered} V_1=\int \frac{q\left(x\right)·W_1}{W\left(x\right)}\text{d}x=\int \frac{q\left(x\right)·\left(-2k^{3}x\right)}{-2k^5} \text{d}x=\frac{1}{k^2}\int q\left(x\right)x \text{d}x \\ {V}_2=\int \frac{q\left(x\right)·W_2}{W\left(x\right)}\text{d}x=\int \frac{q\left(x\right)·\left(2k^3\right)}{-2k^5} \text{d}x=-\frac{1}{k^2}\int q\left(x\right)x \text{d}x \\ {V}_3=\int \frac{q\left(x\right)·W_3}{W\left(x\right)}\text{d}x=\int \frac{q\left(x\right)·\left(-k^2{e}^{-kx}\right)}{-2k^5} \text{d}x=\frac{1}{2k^3}\int q\left(x\right)x \text{d}x \\ {V}_4=\int \frac{q\left(x\right)·W_4}{W\left(x\right)}\text{d}x=\int \frac{q\left(x\right)·\left(k^2{e}^{kx}\right)}{-2k^5} \text{d}x=\frac{1}{2k^3}\int q\left(x\right)x \text{d}x \end{gathered}$$

Therefore, a particular solution to the given equation is

 $y_p\left(x\right)=\frac{1}{k^2}\int q\left(x\right)x \text{d}x-\frac{x}{k^2}\int q\left(x\right)\text{d}x+\frac{e^{kx}}{2k^3}\int q\left(x\right)e^{-kx} \text{d}x-\frac{e^{-kx}}{2k^3}\int q\left(x\right)e^{kx} \text{d}x$

Finally, a general solution to the given equation is

 $$\begin{gathered} y\left(x\right)=y_h\left(x\right)+y_p\left(x\right)y\left(x\right) \\ =c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}+\frac{1}{k^2}\int q\left(x\right)x \text{d}x-\frac{x}{k^2}\int q\left(x\right)\text{d}x+\frac{e^{kx}}{2k^3}\int q\left(x\right)e^{-kx} \text{d}x-\frac{e^{-kx}}{2k^3}\int q\left(x\right)e^{kx} \text{d}x \end{gathered}$$

We have shown that a general solution to the given equation can be written in the wanted form.

Step 3: Determine the general solution in the following form

Our task now is to show that this general solution can also be written in the following form

 $y\left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}+{\int }_0^{x}q\left(s\right)G(s,x)\text{d}s$

where

 $G(s,x)=\frac{s-x}{k^2}-\frac{\sinh \left[k\right(s-x\left)\right]}{k^3}.$

The first four terms are already the same, but let’s show that the rest is too. We start the transformation with substituting the function   and separating one integral into a sum of four integrals

$$\begin{gathered} {\int }_0^{x}q\left(s\right)G(s,x)\text{d}s={\int }_0^{x}q\left(s\right)\left[\frac{s-x}{k^2}-\frac{\sinh \left[k\right(s-x\left)\right]}{k^3}\right]\text{d}s \\ ={\int }_0^{x}q\left(s\right)\left[\frac{s-x}{k^2}\right]\text{d}s-{\int }_0^{x}q\left(s\right)\left[\frac{\sinh \left[k\right(s-x\left)\right]}{k^3}\right]\text{d}s \\ =\frac{1}{k^2}{\int }_0^{x}q\left(s\right)s \text{d}s-\frac{x}{k^2}{\int }_0^{x}q\left(s\right)\text{d}s-\frac{1}{k^3}{\int }_0^{x}q\left(s\right)\frac{e^{k(s-x)}-e^{-k(s-x)}}{2} \text{d}s \\ =\frac{1}{k^2}{\int }_0^{x}q\left(s\right)s \text{d}s-\frac{x}{k^2}{\int }_0^{x}q\left(s\right)\text{d}s-\frac{e^{-kx}}{k^3}{\int }_0^{x}q\left(s\right)e^{ks} \text{d}s+\frac{e^{kx}}{k^3}{\int }_0^{x}q\left(s\right)e^{-ks} \text{d}s \\ =\frac{1}{k^2}\int q\left(x\right)x \text{d}x-\frac{x}{k^2}\int q\left(x\right)\text{d}x-\frac{e^{-kx}}{k^3}\int q\left(x\right)e^{kx} \text{d}x+\frac{e^{kx}}{k^3}\int q\left(x\right)e^{-kx} \text{d}x \\ =\frac{1}{k^2}\int q\left(x\right)x \text{d}x-\frac{x}{k^2}\int q\left(x\right)\text{d}x+\frac{e^{kx}}{k^3}\int q\left(x\right)e^{-kx} \text{d}x-\frac{e^{-kx}}{k^3}\int q\left(x\right)e^{kx} \text{d}x \end{gathered}$$

Now we see that even the last four term are the same. Hence, we can conclude that the general solution in part (a) can be rewritten in the wanted form.

Let q(x)=1. First, we will compute the general solution using the formula in part (a):

$$\begin{gathered} y\left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}+\frac{1}{k^2}\int q\left(x\right)x \text{d}x-\frac{x}{k^2}\int q\left(x\right)\text{d}x+\frac{e^{kx}}{2k^3}\int q\left(x\right)e^{-kx} \text{d}x-\frac{e^{-kx}}{2k^3}\int q\left(x\right)e^{kx} \text{d}x \\ \backslash y\left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}+\frac{x^2}{2k^2}-\frac{x^2}{k^2}-\frac{e^{kx}{e}^{-kx}}{2k^4}-\frac{e^{-kx}{e}^{kx}}{2k^4} \\ {y}_a\left(x\right)=c_1-\frac{1}{k^4}+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}-\frac{1}{2k^2}{x}^2 \end{gathered}$$

Step 4: Determine the general solution using the formula

Now we compute the general solution using the formula in part (b)

$$\begin{gathered} y\left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}+{\int }_0^{x}q\left(s\right)\left[\frac{s-x}{k^2}-\frac{\sinh \left[k\right(s-x\left)\right]}{k^3}\right]\text{d}s \\ y\left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}+{\int }_0^x\left[\frac{s-x}{k^2}-\frac{\sinh \left[k\right(s-x\left)\right]}{k^3}\right]\text{d}s \\ y\left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}+\frac{x^2}{2k^2}-\frac{x^2}{k^2}-{\left.\frac{e^{-kx}}{k^3}\frac{e^{ks}}{k}\right|}_0+{\left.\frac{e^{kx}}{k^3}\frac{-e^{-ks}}{k}\right|}_0 \\ y\left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}-\frac{x^2}{2k^2}-\frac{e^{-kx}}{k^3}\left(\frac{e^{kx}}{k}-\frac{1}{k}\right)+\frac{e^{kx}}{k^3}\left(\frac{-e^{kx}}{k}+\frac{1}{k}\right)\backslash \\ y\left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}-\frac{1}{2k^2}{x}^2-\frac{1}{k^4}+\frac{1}{2k^4}{e}^{kx}+\frac{1}{2k^4}{e}^{-kx} \\ {y}_b\left(x\right)=c_1-\frac{1}{k^4}+c_{2}x+\left(c_3+\frac{1}{2k^4}\right)e^{kx}+\left(c_4+\frac{1}{2k^4}\right)e^{-kx}-\frac{1}{2k^2}{x}^2 \end{gathered}$$

Now we will compare these two solutions with the general solution one would obtain using the method of undetermined coefficients

 $y\left(x\right)=B_1+B_{2}x+B_3{e}^{kx}+B_4{e}^{-kx}-\frac{1}{2k^2}{x}^2.$

When we compare  y_a(x)  and y(x)    we see that these two solutions would be the same if

 $B_1=c_1-\frac{1}{k^4}, B_2=c_2, B_3=c_3 \text{ and } B_4=c_4.$

When we compare y_a(x)  and y(x)  we see that these two solutions would be the same if

 $B_1=c_1-\frac{1}{k^4}, B_2=c_2, B_3=c_3 \text{ and } B_4=c_4$

And when we compare y_a(x)  and y(x)  we see that these two solutions would be the same if

  $B_1=c_1-\frac{1}{k^4}, B_2=c_2, B_3=c_3+\frac{1}{2k^4} \text{ and } B_4=c_4+\frac{1}{2k^4}$

Hence, the solution 

(a) 

$$\begin{gathered} y\left(x\right)=c_1+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}+\frac{1}{k^2}\int q\left(x\right)x \text{d}x-\frac{x}{k^2}\int q\left(x\right)\text{d}x \\ +\frac{e^{kx}}{2k^3}\int q\left(x\right)e^{-kx}dx-\frac{e^{-kx}}{2k^3}\int q\left(x\right)e^{kx}dx \end{gathered}$$

(b) The general solution in part (a) can be rewritten in the wanted form.

(c)  $$\begin{gathered} y_a\left(x\right)=c_1-\frac{1}{k^4}+c_{2}x+c_3{e}^{kx}+c_4{e}^{-kx}-\frac{1}{2k^2}{x}^2 \\ {y}_b\left(x\right)=c_1-\frac{1}{k^4}+c_{2}x+\left(c_3+\frac{1}{2k^4}\right)e^{kx}+\left(c_4+\frac{1}{2k^4}\right)e^{-kx}-\frac{1}{2k^2}{x}^2 \end{gathered}$$