Question

solve the given differential equation by variation of parameters.

$\mathbf{24}\mathbf{.}\mathbf{}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}\mathbf{+}\mathbf{1}}$

Short answer

The solution for the given differential equation for variation of parameters

$$\begin{gathered} y=c_1{x}^{-1}+c_{2}x-\frac{1}{2}-\frac{1}{2}x\ln x \\ +\frac{1}{2}x\ln (x+1)-\frac{1}{2x}\ln (x+1) \end{gathered}$$

Step-by-step solution

Define Cauchy – Euler:

The Cauchy – Euler equation or Euler equation is simply a standard equation with a variable coefficient. It is sometimes called the equidimensional equation.

Find a homogenous differential equation:

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

Divide$\frac{1}{x^2}$on both sides,

$$\begin{gathered} y\text{'}\text{'}+\frac{1}{x}y\text{'}-\frac{1}{x^2}y=\frac{1}{(x+1)} \\ y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=f\left(x\right) \end{gathered}$$

Let's consider,

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

Substitute in the above equation,

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

Where x^m≠ 0,

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

m₁ = -1 and m₂= 1

Use Case 1: Distinct real roots

Then the general solution becomes,

$y=c_1{x}^{m_1}+c_2{x}^{m_2}$

Substitute the value of m₁ and m₂

$y_c=c_1{x}^{-1}+c_{2}x\to \left(1\right)$

Use the Wronskian method:

Where, Y = X and $y_2=x^{-1}$

role="math" localid="1667890416906" $$\begin{gathered} W=\left|x{x}^{-1} \\ \frac{d}{dx}\left[x\right]\frac{d}{dx}\left[x^{-1}\right]\right| \\ W=\left|x{x}^{-1} \\ 1-x^{-2}\right| \\ W=-x^{-1}-\left(x^{-1}\right) \\ W=-2x^{-1}=-\frac{2}{x} \\ {W}_1=\left|0x^{-1} \\ \frac{1}{x^2(x+1)}-x^{-2}\right| \\ {W}_1=-\frac{1}{x^2(x+1)}·x^{-1} \\ {W}_1=-\frac{1}{x^3(x+1)} \end{gathered}$$

And, W₂

$$\begin{gathered} W_2=\left|x0 \\ 1\frac{1}{x^2(x+1)}\right| \\ {W}_2=\frac{1}{x^2(x+1)}·x \\ {W}_2=\frac{1}{x(x+1)} \end{gathered}$$

Calculate

$$\begin{gathered} u_1^{\text{'}}=\frac{W_1}{W} \\ {u}_1^{\text{'}}=-\frac{1}{x^3(x+1)}·\left(-\frac{x}{2}\right) \\ {u}_1^{\text{'}}=\frac{1}{2x^2(x+1)} \\ {u}_1=\int \frac{1}{2x^2(x+1)}dx \end{gathered}$$

Taking integrand in partial form,

$$\begin{gathered} \frac{1}{2x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \\ \frac{1}{2x^2(x+1)}=\frac{Ax(x+1)}{x^2(x+1)}+\frac{\left(B\right)(x+1)}{x^2(x+1)}+\frac{C{x}^2}{x^2(x+1)} \\ \frac{1}{2}=Ax(x+1)+B(x+1)+C{x}^2 \\ \frac{1}{2}=A{x}^2+Ax+Bx+B+C{x}^2 \\ \frac{1}{2}=(A+C)x^2+(A+B)x+B \end{gathered}$$

Comparing the equation and taking,

$B=\frac{1}{2}$

Then,

$$\begin{gathered} A+B=0 \\ A=-B \\ A=-\frac{1}{2} \end{gathered}$$

For C,

$$\begin{gathered} A+C=0 \\ A=-C \\ C=\frac{1}{2} \end{gathered}$$

Substitute the values in the above equation

$\frac{1}{2x^2(x+1)}=-\frac{1}{2x}+\frac{1}{2x^2}+\frac{1}{2(x+1)}$

Find y:

Use integration for$u_1^{\text{'}}$

$$\begin{gathered} u_1=\int -\frac{1}{2x}+\frac{1}{2x^2}+\frac{1}{2(x+1)}dx \\ {u}_1=-\frac{1}{2}\ln x-\frac{1}{2x}+\frac{1}{2}\ln (x+1) \end{gathered}$$

Similarly for, $u_2^{\text{'}}$

$$\begin{gathered} u_2^{\text{'}}=\frac{W_2}{W} \\ {u}_2^{\text{'}}=\frac{1}{x(x+1)}·\left(-\frac{x}{2}\right) \\ {u}_2^{\text{'}}=-\frac{1}{2(x+1)} \end{gathered}$$

From,$u_1=-\int \frac{1}{2(x+1)}dx$

$u_2=-\frac{1}{2}\ln (x+1)$

Find

role="math" localid="1667891608790" $$\begin{gathered} y_p=\left[-\frac{1}{2}\ln x-\frac{1}{2x}+\frac{1}{2}\ln (x+1)\right]·\left(x\right)+\left[-\frac{1}{2}\ln (x+1)\right]·\left(x^{-1}\right) \\ {y}_p=-\frac{1}{2}x\ln x-\frac{1}{2}+\frac{1}{2}x\ln (x+1)-\frac{1}{2x}\ln (x+1) \\ {y}_p=-\frac{1}{2}x\ln x-\frac{1}{2}+\frac{1}{2}x\ln (x+1)-\frac{1}{2x}\ln (x+1) \\ {y}_p=-\frac{1}{2}-\frac{1}{2}x\ln x+\frac{1}{2}x\ln (x+1)-\frac{1}{2x}\ln (x+1) \\ y=y_c+y_p \end{gathered}$$

The solution is$y=c_1{x}^{-1}+c_{2}x-\frac{1}{2}-\frac{1}{2}x\ln x+\frac{1}{2}x\ln (x+1)-\frac{1}{2x}\ln (x+1)$