Question

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

$\mathbf{27}\mathbf{.}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{x}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$

Short answer

$\mathrm{y}=4\mathrm{x}-2{\mathrm{x}}^2-\mathrm{xlnx}$

Step-by-step solution

The solve differential equation

$$\begin{gathered} m^2-3m+2=0 \\ (m-1)(m-2)=0 \end{gathered}$$We have the second order differential equation

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

which is

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

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

$y\left(1\right)=2\text{and}{y}^{\text{'}}\left(1\right)=-1$

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

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

$x^2{y}^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0$

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

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

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

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

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

$$\begin{gathered} m^2-3m+2=0 \\ (m-1)(m-2)=0 \end{gathered}$$

Then the roots are

$m_1=1\text{and}{m}_2=2$

which are real and distinct.

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

$y_c=k_{1}x+k_2{x}^2...........\left(a\right)$

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

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

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

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

Using the Green’s function

Then, since$f\left(t\right)=\frac{1}{t}$ and using the Green's function we have to find the particular solution$y_p\left(x\right)$ as

$$\begin{gathered} y_p\left(x\right)={\int }_{x_0}^{x}G(x,t)f\left(t\right)dt \\ ={\int }_{x_0}^x\left(x^2{t}^{-1}-x\right)\frac{1}{t}dt \\ =x^2{\int }_1^x\frac{1}{t^2}dt-x{\int }_1^x\frac{1}{t}dt \\ =x^2{\left[-\frac{1}{t}\right]}_1^x-x[\ln t{]}_1^x \\ =x^2\left[-\frac{1}{x}+\frac{1}{1}\right]-x[\ln x-\ln 1] \\ =-x+x^2-x\ln x.............(b) \end{gathered}$$

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

$$\begin{gathered} y=y_c+y_p \\ =k_{1}x+k_2{x}^2-x+x^2-x\ln x \\ =\left(k_1-1\right)x+\left(k_2+1\right)x^2-x\ln x \\ =c_{1}x+c_2{x}^2-x\ln x..............\left(c\right) \end{gathered}$$

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

We have to apply the point$(x,y)=(1,2)$ into equation (c), then we have

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

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

$$\begin{gathered} y^{\text{'}}=c_1+2c_{2}x-x\times \frac{1}{x}-\ln x \\ =c_1+2c_{2}x-1-\ln x \end{gathered}$$

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

$$\begin{gathered} -1=c_1+2c_2\left(1\right)-1-\ln 1 \\ -1=c_1+2c_2-1+0 \\ {c}_1+2c_2=0......\left(e\right) \end{gathered}$$

Final proof

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

$c_1=4\text{and}{c}_2=-2$

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

$y=4x-2x^2-x\ln x$

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