Question

Consider the differential equation ${\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{)}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{)}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{3}}\mathbf{.}$

Verify that${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathit{x}$is one solution of the associated homogeneous equation. Then show that the method of reduction of order discussed in Section 4.2 leads to a second solution${\mathit{y}}_{\mathbf{2}}$ of the homogeneous equation as well as a particular solution ${\mathit{y}}_{\mathbf{p}}$of the nonhomogeneous equation. Form the general solution of the DE on the interval$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$

Short answer

The general solution of the DE on the interval$(0,\infty ).$

$y=c_{1}x+c_{2}x{e}^x-x^2$

Step-by-step solution

Define the particular solution of DE

A Particular Solution is a differential equation solution that is derived from the General Solution by assigning specific values to the random constants. Depending on the query, the requirements for finding the values of the random constants can be submitted to us as an Initial-Value Problem or Boundary Conditions

Obtain the second solution of the homogeneous differential equation

The second order differential equation$x^2{y}^{\text{'}\text{'}}-\left(x^2+2x\right)y^{\text{'}}+(x+2)y=x^3$

with a homogeneous differential equation$x^2{y}^{\text{'}\text{'}}-\left(x^2+2x\right)y^{\text{'}}+(x+2)y=0$

with$y_1=x$

we have to verify that it is a solution for this homogeneous differential equation as the following :

Differentiate this alleged solution $y_1=x$with respect to$X$then we have

$$\begin{gathered} y_1^{\text{'}}=1 \\ {y}_1^{"}=0 \end{gathered}$$

then we have

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

Then$y_1=x$ is a solution for the homogeneous differential equation for the given D.E.

Now obtain the second solution of the homogeneous differential equation using the method of reduction of order as the following :

Assume that$y=ux,$

then differentiate with respect to x, then we have

$$\begin{gathered} y^{\text{'}}=u+x{u}^{\text{'}} \\ {y}^{\text{'}\text{'}}=2u^{\text{'}}+x{u}^{\text{'}\text{'}} \end{gathered}$$

$$\begin{gathered} x^2\left(2u^{\text{'}}+x{u}^{\text{'}\text{'}}\right)-\left(x^2+2x\right)\left(u+x{u}^{\text{'}}\right)+(x+2)ux=0 \\ 2x^2{u}^{\text{'}}+x^3{u}^{\text{'}\text{'}}-x^{2}u-2xu-x^3{u}^{\text{'}}-2x^2{u}^{\text{'}}+x^{2}u+2xu=0 \\ {x}^3{u}^{\text{'}\text{'}}-x^3{u}^{\text{'}}=0 \\ {x}^3\left(u^{\text{'}\text{'}}-u^{\text{'}}\right)=0 \end{gathered}$$

Obtain the corresponding homogeneous solution of the given differential equation as.

After that, assume that$u=e^{mx},$

then differentiate with respect to x,

$$\begin{gathered} u^{\text{'}}=m{e}^{mx} \\ {u}^{\text{'}\text{'}}=m^2{e}^{mx} \end{gathered}$$

$\left(m^2-m\right)e^{mx}=0$

Since $e^{mx}$cannot be equal 0

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

The we have the roots as

$m_1=0\text{and}{m}_2=1$

Then we have

$u=c_1+c_2{e}^x$

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

$$\begin{gathered} y_c=\left(c_1+c_2{e}^x\right)x \\ =c_{1}x+c_{2}x{e}^x \end{gathered}$$

Find the particular solution of the DE.

Assumed that$y=ux$ and substituted in equation ,

$$\begin{gathered} x^3\left(u^{\text{'}\text{'}}-u^{\text{'}}\right)=x^3 \\ {u}^{\text{'}\text{'}}-u^{\text{'}}=1 \end{gathered}$$

Assume that $u_P=Ax,$

then differentiate with respect to x,

$$\begin{gathered} u^{\text{'}}=A \\ {u}^{\text{'}\text{'}}=0 \end{gathered}$$

$$\begin{gathered} 0-A=1 \\ A=-1 \end{gathered}$$

Then we have

$u_p=-x$

Then the particular solution becomes

$$\begin{gathered} y_p=u_p \\ x=-x\times x \\ =-x^2 \end{gathered}$$

The general solution is $y=c_{1}x+c_{2}x{e}^x-x^2$