Question

solve the given initial-value problem. Use a graphing utility to graph the solution curve.${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{2}$

Short answer

The solution for the given initial –value problem is$y=\cos \left(\ln x\right)+2\sin \left(\ln x\right)$ .

Step-by-step solution

Define Initial-value:

The initial-value, or y-intercept, is the output value when the line function input is zero. It is the y value of the point where the line crosses the y axis.

Find homogenous differential equation:

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

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 above equation,

$m(m-1)x^m+m{x}^m+x^m=0$

Divide$x^m$on both side,

$m(m-1)+m+1=0$

So,$x^m\ne 0$

$$\begin{gathered} m^2+1=0 \\ m=\pm i \end{gathered}$$

So the equation is $y=c_1\cos \left(\ln x\right)+c_2\sin \left(\ln x\right)$

Find y using initial value conditions:

$y\text{'}=\frac{-c_1\sin \left(\ln x\right)}{x}+\frac{c_2\cos \left(\ln x\right)}{x}$

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

$$\begin{gathered} c_1+0=1 \\ 0+c_2=2 \end{gathered}$$

So the values are$c_1=1$and $c_2=2$

The solution is $y=\cos \left(\ln x\right)+2\sin \left(\ln x\right)$.

using graphing utility draw the graph of  y=cos(lnx)+2sin(lnx)