Question

In Problems 19–30 proceed as in Example 5 to ­find a solution of the given initial-value problem.

$\mathbf{29}\mathbf{.}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{=}\mathit{l}\mathit{n}\mathit{x}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{3}$

Short answer

$y=\frac{46}{45}{x}^3-\frac{1}{20}{x}^{-2}+\frac{1}{36}-\frac{1}{6}\ln x$

Step-by-step solution

The second order differential equation

We have the second order differential equation

$x^2{y}^{\text{'}\text{'}}-6y=\ln x$

which is

$y^{\text{'}\text{'}}-\frac{6}{x^2}y=\frac{\ln x}{x^2}$

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

$y\left(1\right)=1\text{and}{y}^{\text{'}}\left(1\right)=3$

and we have to get its solution as the following technique :

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

$x^2{y}^{\text{'}\text{'}}-6y=0,......\left(1\right)$

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

$$\begin{gathered} y^{\text{'}}=m{x}^{(m-1)} \\ {y}^{\text{'}\text{'}}=m(m-1)x^{(m-2)}......\left(2\right) \end{gathered}$$

After that, substitute with $y=e^{mx}$and equation (2) into the differential equation (1), then we obtain

$$\begin{gathered} x^2\left(m(m-1)x^{(m-2)}\right)-6x^m=0 \\ m(m-1)x^{(m-2)}\times x^2-6x^m=0 \\ m(m-1)x^m-6x^m=0 \\ \left(m^2-m-6\right)x^m=0 \end{gathered}$$

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

$$\begin{gathered} m^2-m-6=0 \\ (m-3)(m+2)=0 \end{gathered}$$

Then the roots are

$m_1=3\text{and}{m}_2=-2$

which are real and distinct.

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

$y_c=k_1{x}^3+k_2{x}^{-2}...........\left(a\right)$

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 $y_1=x^{3}and{y}_2=x^{-2}$, then we can obtain the wronskian$W\left(y_1,y_2\right)$ as

$$\begin{gathered} W\left(x^3,x^{-2}\right)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right| \\ =\left|\begin{array}{cc}{x}^3& x^{-2}\\ 3x^2& -2x^{-3}\end{array}\right| \\ =\left(x^3\right)\times \left(-2x^{-3}\right)-\left(x^{-2}\right)\times \left(3x^2\right) \\ =-2x^0-3x^0 \\ =-5 \end{gathered}$$

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

$$\begin{gathered} G(x,t)=\frac{y_1\left(t\right)y_2\left(x\right)-y_1\left(x\right)y_2\left(t\right)}{W\left(t\right)} \\ =\frac{t^3\times x^{-2}-x^3\times t^{-2}}{-5} \\ =\frac{x^3\times t^{-2}-x^{-2}\times t^3}{5} \\ =\frac{1}{5}{x}^3{t}^{-2}-\frac{1}{5}{x}^{-2}{t}^3 \end{gathered}$$

Using the Green’s function

Then, since$f\left(t\right)=\frac{\ln t}{t^2}$and using the Green's function we have to find the particular solution$y_p\left(x\right)$as

$$\begin{gathered} y_p\left(x\right)={\int }_{x_0}^{x}G(x,t)f\left(t\right)dt \\ ={\int }_{x_0}^x\left(\frac{1}{5}{x}^3{t}^{-2}-\frac{1}{5}{x}^{-2}{t}^3\right)\frac{\ln t}{t^2}dt \\ =\frac{1}{5}{x}^3{\int }_1^x\frac{\ln t}{t^4}dt-\frac{1}{5}{x}^{-2}{\int }_1^{x}t\ln tdt \end{gathered}$$

For the integration $\int \frac{\ln t}{t^4}dt$, we can made it using integration by parts as the following:

If we set $s=\ln t$, then $ds=\frac{1}{t}dt$, and if we set $dv=t^{-4}dt$, then $v=-\frac{1}{3t^3}$, then we obtain

$$\begin{gathered} \int \frac{\ln t}{t^4}dt=s\times v-\int vds \\ =-\frac{\ln t}{3t^3}-\int -\frac{1}{3t^4}dt \\ =-\frac{\ln t}{3t^3}-\frac{1}{9t^3} \end{gathered}$$

and for the integration , we can made it using integration by parts as the following :

If we set $s=\ln t$, then $ds=\frac{1}{t}dt$, and if we set $dv=tdt$, then $v=\frac{1}{2}{t}^2$, then we obtain

$$\begin{gathered} \int t\ln tdt=s\times v-\int vds \\ =\frac{1}{2}{t}^2\ln t-\int \frac{1}{2}tdt \\ =\frac{1}{2}{t}^2\ln t-\frac{1}{4}{t}^2 \end{gathered}$$

Then we can obtain the particular solution of the given differential equation as

$$\begin{gathered} y_p\left(x\right)=\frac{1}{5}{x}^3{\left[-\frac{\ln t}{3t^3}-\frac{1}{9t^3}\right]}_1^x-\frac{1}{5}{x}^{-2}{\left[\frac{1}{2}{t}^2\ln t-\frac{1}{4}{t}^2\right]}_1^x⩽ \\ =\frac{1}{5}{x}^3\left[\left(-\frac{\ln x}{3x^3}-\frac{1}{9x^3}\right)-\left(0-\frac{1}{9}\right)\right]-\frac{1}{5}{x}^{-2}\left[\left(\frac{1}{2}{x}^2\ln x-\frac{1}{4}{x}^2\right)-\left(0-\frac{1}{4}\right)\right] \\ =-\frac{1}{15}\ln x-\frac{1}{45}+\frac{1}{45}{x}^3-\frac{1}{10}\ln x+\frac{1}{20}-\frac{1}{20}{x}^{-2} \\ =-\frac{1}{45}{x}^3-\frac{1}{6}\ln x+\frac{1}{36}-\frac{1}{20}{x}^{-2}.......\left(b\right) \end{gathered}$$

Then from (a) and (b), we can obtain the general solution of the given differential equation as

$$\begin{gathered} y=y_c+y_p \\ =k_1{x}^3+k_2{x}^{-2}--\frac{1}{45}{x}^3-\frac{1}{6}\ln x+\frac{1}{36}-\frac{1}{20}{x}^{-2} \\ =\left(k_1-\frac{1}{45}\right)x^3+\left(k_2-\frac{1}{20}\right)x^{-2}-\frac{1}{6}\ln x+\frac{1}{36} \\ =c_1{x}^3+c_2{x}^{-2}+\frac{1}{36}-\frac{1}{6}\ln x.....\left(c\right) \end{gathered}$$

Finally, we have to apply the initial condition of the given differential equation as the following :

We have to apply the point$(x,y)=(1,1)$ into equation (c), then we have

$$\begin{gathered} 1=c_1(1{)}^3+c_2(1{)}^{-2}+\frac{1}{36}-\frac{1}{6}\ln 1 \\ 1=c_1+c_2+\frac{1}{36}-0 \\ {c}_1+c_2=\frac{35}{36}.........\left(d\right) \end{gathered}$$

After that, we have to take the first derivative for the general solution as

$y^{\text{'}}=3c_1{x}^2-2c_2{x}^{-3}-\frac{1}{6x}$

Then apply the point$\left(x,y^{\text{'}}\right)=(1,3)$ into this first derivative function as

$$\begin{gathered} 3=3c_1(1{)}^2-2c_2(1{)}^{-3}-\frac{1}{6\times 1} \\ 3=3c_1-2c_2-\frac{1}{6} \\ 3c_1-2c_2=\frac{19}{6}.......\left(e\right) \end{gathered}$$

Final proof 

Then by solving these two equations (d) and (e), we find that

$c_1=\frac{46}{45}\text{and}{c}_2=-\frac{1}{20}$

Then, substitute with$c_{1}and{c}_2$ into equation (c), then we obtain

$y=\frac{46}{45}{x}^3-\frac{1}{20}{x}^{-2}+\frac{1}{36}-\frac{1}{6}\ln x$

is the solution of the given differential equation at the given conditions.