Question

In Problems 1–4 the given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem.

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{x}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{x}\mathbf{}\mathbf{In}\mathbf{}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)} \\ {\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}\text{'}}\mathbf{-}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

Short answer

A solution of the initial-value problem is $\mathrm{y}=3\mathrm{x}-4\mathrm{x}\mathrm{In}\mathrm{x}$.

Step-by-step solution

Find the value of the first constant of the general solution

The initial conditions are:

$$\begin{gathered} {\mathrm{y}}_1\left(1\right)=3 \\ \mathrm{y}\text{'}\left(1\right)=-1 \end{gathered}$$

The point lying on the general solution $\mathrm{y}={\mathrm{c}}_1\mathrm{x}+{\mathrm{c}}_2\mathrm{x}$is $\left(\mathrm{x},\mathrm{y}\right)=\left(1,3\right)$.

Substituting the value in the general solution,

$$\begin{gathered} 3={\mathrm{c}}_1\left(1\right)+{\mathrm{c}}_2\left(1\right)\mathrm{In}\left(1\right) \\ 3={\mathrm{c}}_1+{\mathrm{c}}_2\times \left(0\right) \\ {\mathrm{c}}_1=3 \end{gathered}$$

Find the value of the second constant of the general solution

Determining the first derivative of the general solution of the differential equation,

$$\begin{gathered} \mathrm{y}\text{'}={\mathrm{c}}_1+{\mathrm{c}}_2\mathrm{In}\mathrm{x}+{\mathrm{c}}_2\mathrm{x}\times \frac{1}{\mathrm{x}} \\ ={\mathrm{c}}_1+{\mathrm{c}}_2\mathrm{In}\mathrm{x}+{\mathrm{c}}_2 \\ ={\mathrm{c}}_1+{\mathrm{c}}_2\left(\mathrm{In}\mathrm{x}+1\right) \end{gathered}$$

Substituting the point value as $\left(\mathrm{x},\mathrm{y}\text{'}\right)=\left(1,-1\right)$on this solution,

$$\begin{gathered} -1={\mathrm{c}}_1+{\mathrm{c}}_2\left(\mathrm{In}\left(1\right)+1\right) \\ -1={\mathrm{c}}_1+{\mathrm{c}}_2\left(0+1\right) \\ -1={\mathrm{c}}_1+{\mathrm{c}}_2 \end{gathered}$$

Substituting the value of ${\mathrm{c}}_1=3$,

$$\begin{gathered} -1=3+{\mathrm{c}}_2 \\ {\mathrm{c}}_2=-4 \end{gathered}$$

Find the general solution of the differential equation

The general solution is $\mathrm{y}={\mathrm{c}}_1\mathrm{x}+{\mathrm{c}}_2\mathrm{x}\mathrm{In}\mathrm{x}$

Substituting the following values:

$$\begin{gathered} {\mathrm{c}}_1=3 \\ {\mathrm{c}}_2=-4 \end{gathered}$$

The solution member of family with given initial conditions will be:

localid="1667910840608" $$\begin{gathered} \mathrm{y}=3\mathrm{x}+\left(-4\right)\mathrm{x}\mathrm{In}\mathrm{x} \\ \mathrm{y}=3\mathrm{x}-4\mathrm{x}\mathrm{In}\mathrm{x} \end{gathered}$$

localid="1668488772194" $\mathbf{Hence}\mathbf{}\mathbf{the}\mathbf{}\mathbf{solution}\mathbf{}\mathbf{is}\mathbf{}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{}\mathit{I}\mathit{n}\mathbf{}\mathbf{x}$