Question

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

$\mathbf{30}\mathbf{.}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{3}$

Short answer

$y=x^2+3x-2x\ln x$

Step-by-step solution

The second order differential equation

We have the second order differential equation

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

which is

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

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

$y\left(1\right)=4\text{and}{y}^{\text{'}}\left(1\right)=3,$

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

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

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

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

$$\begin{gathered} y^{\text{'}}=m{x}^{(m-1)}.......\left(2\right) \\ {y}^{\text{'}\text{'}}=m(m-1)x^{(m-2)}.......\left(3\right) \end{gathered}$$

After that, substitute with $y=e^{mx}$and equations (2) and (3) into the differential equation (1), then we obtain

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

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

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

Then the roots are

$m_{1,2}=1$

which are real and repeated.

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

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

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

$$\begin{gathered} W\left(x^3,x^{-2}\right)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right| \\ =\left|\begin{array}{cc}x& x\ln x\\ 1& \ln x+1\end{array}\right| \\ =\left(x\right)\times (\ln x+1)-\left(\ln x\right)\times \left(1\right) \\ =x\ln x+x-\ln x \\ =x \end{gathered}$$

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

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

Using the Green’s function

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

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

For the integration $\int \ln tdt$, we can made it using integration by parts as the following:

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

$$\begin{gathered} \int \ln tdt=s\times v-\int vds \\ =t\ln t-\int dt \\ =t\ln t-t \end{gathered}$$

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

$$\begin{gathered} y_p\left(x\right)=\ln x[t{]}_1^x-x[t\ln t-t{]}_1^x \\ =x\ln x\left[\right(x)-(1\left)\right]-x\left[\right(x\ln x-x)-(0-1\left)\right] \\ =x^2\ln x-x\ln x-x^2\ln x+x^2+x \\ =x^2+x-x\ln x...........\left(b\right) \end{gathered}$$

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

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

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

We have to apply the point$(x,y)=(1,4)$into equation (c), then we have

$$\begin{gathered} 4=c_1\left(1\right)+c_2\left(1\right)\left(\ln 1\right)+(1{)}^2 \\ 4=c_1+0+1 \\ {c}_1=3............\left(d\right) \end{gathered}$$

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

$y^{\text{'}}=c_1+c_2+c_2\ln x+2x$

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

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

Final proof

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

$c_1=3\text{and}{c}_2=-2$

Then, substitute with$c_1$ and$c_2$ into equation (c), then we obtain

$y=x^2+3x-2x\ln x$

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