Q21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A particular solution and a fundamental solution set are given for a nonhomogeneous equation and its corresponding homogeneous equation. 

(a) Find a general solution to the nonhomogeneous equation. 

(b) Find the solution that satisfies the specified initial condition.

$$\begin{gathered} {\mathbf{x}}^{\mathbf{3}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{-}\mathbf{lnx}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{>}\mathbf{0}\mathbf{;} \\ \mathbf{y}\left(1\right)\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(1\right)\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{;} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{=}\mathbf{lnx}\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\left\{x, \mathrm{xlnx}, x{\left(\mathrm{lnx}\right)}^2\right\} \end{gathered}$$

Step-by-Step Solution

Answer

(a) The value is $y=c_{1}x+c_{2}x\ln \left(x\right)+c_{3}x{\ln }^2\left(x\right)+\ln \left(x\right)$

(b) The value is $y=3x-x\ln \left(x\right)+x{\ln }^2\left(x\right)+\ln \left(x\right)$

(a) Step 1: Firstly solve for y n

The given equation is, $x^{3}y\text{'}\text{'}\text{'}+xy\text{'}-y=3-\ln x$

Solve for, $y_n$

$x^{3}y\text{'}\text{'}\text{'}+xy\text{'}-y=0$

Let, $y=x^r$

$x^3\left(x^r\right)\text{'}\text{'}\text{'}+x\left(x^r\right)\text{'}-\left(x^r\right)=0$

Solve the above equation,

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

Now one has,

$r = 1,1,1$

Then, $y_n=c_{1}x+c_{2}x\ln \left(x\right)+c_{3}x{\ln }^2\left(x\right)$

Step 2 :A general solution to the nonhomogeneous equation.

$$\begin{gathered} y=y_n+y_p \\ y=c_{1}x+c_{2}x\ln \left(x\right)+c_{3}x{\ln }^2\left(x\right)+\ln \left(x\right) \end{gathered}$$

(b) Step 3:Solve for given initial conditions.

Given initial conditions are, $y\left(1\right)=3, y\text{'}\left(1\right)=3, y\text{'}\text{'}\left(1\right)=0;$

Firstly, solve for, $y\left(1\right)=3$

One has, $y=c_{1}x+c_{2}x\ln \left(x\right)+c_{3}x{\ln }^2\left(x\right)+\ln \left(x\right)$

Substitute $y\left(1\right)=3$ in the above equation,

$$\begin{gathered} 3=c_1\left(1\right)+c_2\left(1\right)\ln \left(1\right)+c_3\left(1\right){\ln }^2\left(1\right)+\ln \left(1\right) \\ 3=c_1+c_2\left(0\right)+c_3\left(0\right)+0 \\ {c}_1=3 \end{gathered}$$

Step 4: Now, solve for y 1 (1)=3,

One has, $y\text{'}=c_1+c_2\left[1+\ln \left(x\right)\right]+c_3\left[\frac{x2\ln \left(x\right)}{x}+{\ln }^2\left(x\right)\right]+\frac{1}{x}$

Substitute $y\text{'}\left(1\right)=3$ in the above equation,

$$\begin{gathered} 3=c_1+c_2\left[1+\ln \left(1\right)\right]+c_3\left[\frac{\left(1\right)2\ln \left(1\right)}{1}+{\ln }^2\left(1\right)\right]+\frac{1}{1} \\ 3=c_1+c_2\left[1\right]+c_3\left[0+0\right]+1 \\ 3=c_1+c_2+1 \\ {c}_1+c_2=2 \end{gathered}$$

Substitute $c_1=3$ in the above equation,

$$\begin{gathered} 3+c_2=2 \\ {c}_2=-1 \end{gathered}$$

Step 5: Now, solve for y '' (1)=0,

One has, $y\text{'}\text{'}=c_2\left[\frac{1}{x}\right]+c_3\left[\frac{2}{x}+\frac{2\ln \left(x\right)}{x}\right]-\frac{1}{x^2}$

Substitute $y\text{'}\text{'}\left(1\right)=0$ in the above equation,

$$\begin{gathered} 0=c_2\left[\frac{1}{1}\right]+c_3\left[\frac{2}{1}+\frac{2\ln \left(1\right)}{1}\right]-\frac{1}{{\left(1\right)}^2} \\ 0=c_2+c_3\left[2\right]-1 \\ {c}_2+2c_3=1 \end{gathered}$$

Substitute $c_2=-1$ in the above equation,

$$\begin{gathered} -1+2c_3=1 \\ 2c_3=2 \\ {c}_3=1 \end{gathered}$$

Step 6: conclusion, the solution that satisfies the specified initial condition.

Substitute the value of $c_1, c_2$ and $c_3$ in the general solution.

$$\begin{gathered} y=\left(3\right)x+\left(-1\right)x\ln \left(x\right)+\left(1\right)x{\ln }^2\left(x\right)+\ln \left(x\right) \\ y=3x-x\ln \left(x\right)+x{\ln }^2\left(x\right)+\ln \left(x\right) \end{gathered}$$