Question

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

Short answer

The solution of initial value problem using graphing utility is$y=\frac{3}{4}-\ln x+\frac{1}{4}{x}^2$

Step-by-step solution

Deriving the given solution by differential equation

Substituting in the equation

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

$Divideby{x}^{m-1}$

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

$y_c=c_1+c_2\ln x$

This is the complementary solution$y_1=1,y_2=\ln x$

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Solving the coefficient and find wronskian

Start solving by variation of parameters, the coefficient of$y\text{'}\text{'}$should be 1 divide the equation by x ,

$y\text{'}\text{'}+\frac{y\text{'}}{x}=1,f\left(x\right)=1$

We set the Wronskian

Solve the integrals and find partial solution

Solving this integral by parts

$$\begin{gathered} u_1^{\text{'}}=\frac{W_1}{W} \\ =-x\ln x \\ {u}_1=\int -x\ln xdx \end{gathered}$$

$\begin{array}{rcl}{u}_1& =& \frac{1}{4}{x}^2-\frac{1}{2}{x}^2\ln x\\ u_2^{\text{'}}& =& \frac{W_2}{W}\\ & =& x\\ u_2& =& \frac{x^2}{2}\\ y_p& =& u_1{y}_1+u_2{y}_2\\ & & \end{array}$

$$\begin{gathered} y_p=\left(\frac{1}{4}{x}^2-\frac{1}{2}{x}^2\ln x\right)\times 1+\frac{x^2}{2}\times \ln x \\ {y}_p=\frac{1}{4}{x}^2-\frac{1}{2}{x}^2\ln x+\frac{1}{2}{x}^2\ln x \end{gathered}$$

This is the particular solution

$y_p=\frac{1}{4}{x}^2$

$y=y_c+y_p$

This is the general solution

$y=c_1+c_2\ln x+\frac{1}{4}{x}^2$

Deriving the given solution using initial value

We now find y' and apply the initial value conditions

$y\text{'}=\frac{c_2}{x}+\frac{1}{2}x$

Solving the given two equation by applying initial value $\text{y}\left(\text{1}\right)\text{= 1}$ and $\text{y'}\left(\text{1}\right)\text{=}-\frac{\text{1}}{\text{2}}$

$$\begin{gathered} c_1+0+\frac{1}{4}=1 \\ {c}_2+\frac{1}{2}=-\frac{1}{2} \end{gathered}$$

We get

$$\begin{gathered} c_1=\frac{3}{4} \\ {c}_2=-1 \end{gathered}$$

Therefore $y=\frac{3}{4}-\ln x+\frac{1}{4}{x}^2$

Draw the graph using graphing utility

Using a graphing utility to graph the solution