Question

Solve the given differential equation.${\mathit{x}}^{\mathbf{3}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution for the given differential equation is $y=c_{1}x+c_2\left(\text{ln}x\right)x+c_3(\text{ln}x{)}^{2}x$.

Step-by-step solution

Determine the general solution by solving the given equation

Consider the given differential equation,$x^{3}y\text{'}\text{'}+xy\text{'}-y=0$

Hence, by using the substitution $y=x^m$we obtain,

$y=x^m$

$y\text{'}=m{x}^{m-1}$

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

$y\text{'}\text{'}\text{'}=m(m-1)(m-2)x^{m-3}$

Determine the general solution by solving the given equation

Thus, substitute the values of $y,y\text{'},y\text{'}\text{'},y\text{'}\text{'}\text{'}$in the equation,

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

Hence, we obtain the auxiliary solution $m^3-3m^2+3m-1$.

Simplify the equation as:

$\begin{array}{rcl}{m}^3-3m^2+3m-1& =& 0\\ {\left(m-1\right)}^3& =& 0\\ m& =& 1,1,1\\ & & \end{array}$

Since, we obtain the same real roots, the general solution is $y=c_1{x}^{m_1}+c_2\left(\text{ln}x\right)x^{m_2}+c_3(\text{ln}x{)}^2{x}^{m_3}$.

Therefore, the solution for the given differential equation is$y=c_{1}x+c_2\left(\text{ln}x\right)x+c_3(\text{ln}x{)}^{2}x$ .