Question

In Problems 39–44 proceed as in Examples 7 and 8 to find a solutionof the given boundary-value problem.

$\mathbf{43}\mathbf{.}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\left(}{\mathit{e}}^{\mathbf{-}\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution is $y_p\left(x\right)=\frac{1}{2}(\ln x{)}^2+\frac{1}{2}\ln x$.

Step-by-step solution

Step 1:Boundary value problem.

With boundary value problems we will have a differential equation and we will specify the function and/or derivatives at different points, which we’ll call boundary values. For second order differential equations, which will be looking at pretty much exclusively here, any of the following can, and will, be used for boundary conditions.

Find the roots.

We have the Cauchy Euler equation,

$x^{2}y\text{'}\text{'}+xy\text{'}=1$

Which is

$y\text{'}\text{'}+\frac{1}{x}y\text{'}=\frac{1}{x^2}$

As the form,

$g\left(x\right)y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=f\left(x\right)$,

With the initial conditions

$y\left(e^{-1}\right)=0$and $y\left(1\right)=0$,

And we have to get its solutions as the following technique:

First, we have to find the solutions of the corresponding homogeneous differential equation

$x^{2}y\text{'}\text{'}+xy\text{'}=0$...............(1)

By assuming that $y=x^m$is a solutions for homogeneous differential equation, then follow then differentiate this assumption with respect to x, then we have

$y\text{'}=m{x}^{(m-1)}$............(2)

$y\text{'}\text{'}=m(m-1)x^{(m-2)}$..........(3)

After that, substituting with equations (2) and (3) into the differential equation (1), then we obtain

$$\begin{gathered} x^{2}m(m-1)x^{(m-2)}+xm{x}^{(m-1)}=0 \\ m(m-1)x^m+m{x}^m=0 \\ m(m-1+1)x^m=0 \\ {m}^2{x}^m=0 \end{gathered}$$

Since $x^m$cannot be equal 0, then we have

$m^2=0$

Then the roots are

$m_{1,2}=0$

which are real and repeated.

Then we can obtain the solution of the corresponding homogeneous differential equation as

$y_c=k_1+k_2\ln x$................(4)

Apply boundary conditions.

Now we have to apply the initial boundary condition of the given differential equation as the following:

We have to apply the point $(x,y)=(e^{-1},0)$into the equation (4), then we have,

$$\begin{gathered} 0=k_1+k_2\ln e^{-1} \\ {k}_1-k_2=0 \end{gathered}$$

If we choose $k_2=1$then we have $k_1=1$, then by substituting with $k_1$and $k_2$into equation (4), then we have

$y_1=1+\ln x$

After that , we have to apply the point $(x,y)=(1,0)$into the above function of first derivative, then we have

$$\begin{gathered} 0=k_1+k_2\ln 1 \\ {k}_1=0 \end{gathered}$$

Let $k_2=1$then by substituting with $k_1$and $k_2$into equation (4), then we have

$$\begin{gathered} y_2=0+\ln x \\ {y}_2=\ln x \end{gathered}$$

Find the particular solution.

Second, now we have to find the particular solution $y_p\left(x\right)$of the given differential equation as the following technique:

Since we have $y_1=\ln x+1$and $y_2=\ln x$then we can obtain the wronskian $W(y_1,y_2)$as

$W\left(y_1,y_2\right)=\left|\begin{array}{cc}{y}_1& y_2\\ y{\text{'}}_{1}y\text{'}2& \end{array}\right|$

$W\left(y_y.y_2\right)=\left|\begin{array}{cc}lnx+1& \ln x\\ & \\ \frac{1}{x}& \frac{1}{x}\end{array}\right|$

$$\begin{gathered} =(\ln x+1)\left(\frac{1}{x}\right)-\left(\ln x\right)\left(\frac{1}{x}\right) \\ =\frac{\ln x}{x}+\frac{1}{x}-\frac{\ln x}{x} \\ =\frac{1}{x} \end{gathered}$$

After that, we have to find the Green’s function for the given differential equation as

$$\begin{gathered} G(x,t)=\left\{\frac{y_1\left(t\right)y_2\left(x\right)}{W\left(t\right)},e^{-1}⩽t⩽x \\ \frac{y_1\left(x\right)y_2\left(t\right)}{W\left(t\right)},x⩽t⩽1\right. \\ =\left\{\frac{(\ln t+1)\left(\ln x\right)}{t^{-1}},e^{-1}⩽t⩽x \\ \frac{(\ln x+1)\left(\ln t\right)}{t^{-1}},x⩽t⩽1\right. \end{gathered}$$

$$\begin{gathered} =\left\{\ln xt(\ln t+1),e^{-1}⩽t⩽x \\ (\ln x+1)t\ln t,x⩽t⩽1\right. \end{gathered}$$

Find the particular solution.

Now since we have $f\left(t\right)=\frac{1}{t^2}$, then we can obtain the particular solution $y_p\left(x\right)$as,

$$\begin{gathered} y_p\left(x\right)=\underset{e^{-1}}{\overset{1}{\int }}G(x,t)f\left(t\right)dt \\ =\underset{e^{-1}}{\overset{1}{\int }}G(x,t)\left(\frac{1}{t^2}\right)dt \end{gathered}$$

$$\begin{gathered} =\underset{e^{-1}}{\overset{x}{\int }}\ln xt(\ln t+1)\left(\frac{1}{t^2}\right)dt+\underset{x}{\overset{1}{\int }}(\ln x+1)t\ln t\left(\frac{1}{t^2}\right)dt \\ =\ln x\underset{e^{-1}}{\overset{x}{\int }}\left(\frac{\ln t}{t}+\frac{1}{t}\right)dt+(\ln x+1)\underset{x}{\overset{1}{\int }}\frac{\ln t}{t}dt \end{gathered}$$

The integration $\left(\int \frac{\ln t}{t}dt\right)$can be made using integration by substitution as the following:

If we set $u=\ln t$, we have $du=\frac{1}{t}dt$, then we can solve this integration as,

$$\begin{gathered} \int \frac{\ln t}{t}dt=\int udu \\ =\frac{1}{2}{u}^2 \\ =\frac{1}{2}(\ln t{)}^2 \end{gathered}$$

Then we obtain,

$\begin{array}{rcl}{y}_p\left(x\right)& =& \ln x{\left[\frac{1}{2}{\left(\ln t\right)}^2+\ln t\right]}_{e^{-1}}^x+(\ln x+1){\left[\frac{1}{2}{\left(\ln t\right)}^2\right]}_x^1\\ & =& \ln x\left[\frac{1}{2}{\left(\ln x\right)}^2+\ln x+\frac{1}{2}\right]+(\ln x+1)\left[0-\frac{1}{2}{\left(\ln x\right)}^2\right]\\ & =& \frac{1}{2}(\ln x{)}^3+(\ln x{)}^2+\frac{1}{2}\ln x-\frac{1}{2}(\ln x{)}^3-\frac{1}{2}(\ln x{)}^2\\ y_p\left(x\right)& =& \frac{1}{2}(\ln x{)}^2+\frac{1}{2}\ln x\end{array}$

Therefore,$y_p\left(x\right)=\frac{1}{2}(\ln x{)}^2+\frac{1}{2}\ln x$ $y_p\left(x\right)=\frac{1}{2}(\ln x{)}^2+\frac{1}{2}\ln x$.