Q Review Problems-4E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution for the given differential equation.

(a) ${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{0}$

(b) ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{5}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}$

(c) ${\mathit{y}}^{\mathbf{\left(}\mathbf{5}\mathbf{\right)}}\mathbf{-}{\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$

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

Step-by-Step Solution

Answer

The general solution for the given differential equation is:

 

1. $y\left(x\right)=c_1{e}^x+c_{2}x{e}^x+c_3{e}^{-x}+c_4{e}^{-3x}$
   
2. $y\left(x\right)=c_1{e}^x+c_2{e}^{(-2+\sqrt{5})x}+c_3{e}^{(-2-\sqrt{5})x}$
   
3. $y\left(x\right)=c_1{e}^x+c_2\cos x+c_3\sin x+c_{4}x\cos x+c_{5}x\sin x$
   
4. $y\left(x\right)=c_1{e}^{2x}+c_2{e}^{-x}+c_3{e}^x-\frac{1}{2}x{e}^x+\frac{1}{4}+\frac{1}{2}x$
   

Step 1: Determine the general solution for the given differential equation .

The auxiliary equation is;

$r^4+2r^3-4r^2-2r+3=0$

By inspection, we find that r = 1 is a root, and using polynomial division we get

$$\begin{gathered} r^4+2r^3-4r^2-2r+3=(r-1)\left(r^3+3r^2-r-3\right) \\ =(r-1)\left[r^2(r+3)-(r+3)\right] \\ =(r-1)(r+3)\left(r^2-1\right) \\ =(r-1{)}^2(r+1)(r+3)=0 \end{gathered}$$

Now we see that the roots of the auxiliary equation are $r_1=r_2=1,r_3=-1$and $r_4=-3$.

Therefore, a general solution to the given equation is;

$y\left(x\right)=c_1{e}^x+c_{2}x{e}^x+c_3{e}^{-x}+c_4{e}^{-3x}$

Step 2: Determine the general solution for the given differential equation

The auxiliary equation is;

$r^3+3r^2-5r+1=0$

By inspection, we find that r = 1 is a root, and using polynomial division we get

$r^3+3r^2-5r+1=(r-1)\left(r^2+4r-1\right)=0$

Now we see that the roots of the auxiliary equation are $r_1=1,r_2=-2+\sqrt{5}$and $r_3=-2-\sqrt{5}$

Therefore, a general solution to the given equation is:

$y\left(x\right)=c_1{e}^x+c_2{e}^{(-2+\sqrt{5})x}+c_3{e}^{(-2-\sqrt{5})x}$

Step 3: Determine the general solution for the given differential equation

The auxiliary equation is;

$r^5-r^4+2r^3-2r^2+r+1=0$

By inspection, we find that r = 1 is a root, and using polynomial division we get,

$$\begin{gathered} r^5-r^4+2r^3-2r^2+r+1=(r-1)\left(r^4+2r^2+1\right) \\ =(r-1){\left(r^2+1\right)}^2=0 \end{gathered}$$

Now we see that the roots of the auxiliary equation are $r_1=1,r_2=r_3=i$and $r_4=r_5=-i$. Therefore, a general solution to the given equation is;

$y\left(x\right)=c_1{e}^x+c_2\cos x+c_3\sin x+c_{4}x\cos x+c_{5}x\sin x$

Step 4: Determine the general solution for the given differential equation

The corresponding homogeneous equation takes the operator form,

$$\begin{gathered} \left(D^3-2D^2-D+2\right)\left[y\right]=0 \\ \left[D^2(D-2)-(D-2)\right]\left[y\right]=0 \\ (D-2)\left(D^2-1\right)\left[y\right]=0 \\ (D-2)(D+1)(D-1)\left[y\right]=0 \end{gathered}$$

Now we see that a solution to the homogeneous equation is;

$y_h\left(x\right)=c_1{e}^{2x}+c_2{e}^{-x}+c_3{e}^x.$

Observe that D - 1 annihilates e^x and D² annihilates x. So, the non-homogeneity e^x + x is annihilated by D²(D - 1). 

Therefore, every solution to the given nonhomogeneous equation also satisfies the following homogeneous equation;

$(D-2)(D+1)(D-1{)}^2{D}^2\left[y\right]=0.$

A general solution to this homogeneous equation is:

$y\left(x\right)=c_1{e}^{2x}+c_2{e}^{-x}+c_3{e}^x+c_{4}x{e}^x+c_5+c_{6}x$

Comparing this general solution to y_h(x) shows that the first three terms give a general solution to the associated homogeneous equation and the last three terms constitute a particular solution form with undetermined coefficients. 

A particular solution and its derivative functions are,

$$\begin{gathered} y_p\left(x\right)=c_{4}x{e}^x+c_5+c_{6}x \\ {y}_p^{\text{'}}\left(x\right)=c_4{e}^x+c_{4}x{e}^x+c_6 \\ {y}_p^{\text{'}\text{'}}\left(x\right)=2c_4{e}^x+c_{4}x{e}^x \\ {y}_p^{\text{'}\text{'}\text{'}}\left(x\right)=3c_4{e}^x+c_{4}x{e}^x \end{gathered}$$

Substituting all this into our given equation yields,

$$\begin{gathered} y_p^{\text{'}\text{'}\text{'}}-2y_p^{\text{'}\text{'}}-y_p^{\text{'}}+2y_p=3c_4{e}^x+c_{4}x{e}^x-2\left(2c_4{e}^x+c_{4}x{e}^x\right)-\left(c_4{e}^x+c_{4}x{e}^x+c_6\right)+2\left(c_{4}x{e}^x+c_5+c_{6}x\right) \\ =-2c_4{e}^x+2c_5-c_6+2c_{6}x \\ =e^x+x \end{gathered}$$

Now we have the following system,

$$\begin{gathered} -2c_4=1\Rightarrow c_4=-\frac{1}{2}2 \\ {c}_6=1\Rightarrow c_6=\frac{1}{2}2 \\ {c}_5-c_6=0\Rightarrow c_5=\frac{1}{4}. \end{gathered}$$

Therefore, a particular solution to the given equation is:

$y_p\left(x\right)=-\frac{1}{2}x{e}^x+\frac{1}{4}+\frac{1}{2}x$

Finally, a general solution to the given non-homogeneous equation is:

$y\left(x\right)=c_1{e}^{2x}+c_2{e}^{-x}+c_3{e}^x-\frac{1}{2}x{e}^x+\frac{1}{4}+\frac{1}{2}x$