MHB 2.2.5 de equation with y(1)=1/2

[karush]

$\textsf{Find the solution of the given initial value problem. State the interval in which the solution is valid.}$
$$xy^\prime+2y=x^2-x+1, \qquad y(1)=\frac{1}{2}$$
$\textit{rewrite}$
$$y' +\frac{2}{x}y=x-1+\frac{1}{x}$$
$\textit {Find u(x) }$
$$\displaystyle u(x)=\exp\int \frac{2}{x} \, dx =\ln e^{x^2}=x^2$$
$\textit{multiply thru with $u(x)=x^2$}$
$$x^2y' +(x^2)'y=x^2(x^2-x+1)$$
$\textsf{rewrite:}$
$$(x^2 y)'=x^4-x^3+x^2 \, \color{red}{\textit{I think the problem is here}}$$
$\textit{Integrate }$
$$\displaystyle x^2y=\int x^4-x^3+x^2 \, dx
=\frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}+c$$
$\textit{divide thru by $x^2$}$
$$\displaystyle y=\frac{x^3}{5}-\frac{x^2}{4}+\frac{x}{3}+\frac{c}{x^2}$$
$\textit{book answer(I couldn't get this) }$
$$y=\color{red}{\frac{1}{12}(3x^2-4x+6+\frac{1}{x^2})}$$

 

[Alone]

You have the ODE: $xy'+2y=x^2-x+1$ you should multiply both sides of the equation by $x$ and not x^2, that's where you got it wrong, your RHS is multiplied by x^2 and not x.

 

[karush]

$\textit{but isn't $u(x)=x^2$}$

 

[MarkFL]

$\textit{but isn't $u(x)=x^2$}$

Yes, but you used it on the original ODE, not the one in standard linear form. :)

 

[karush]

Ok here is my final flower arraignment

$\textit{Find the solution of the given initial value problem.} $
$\textit{State the interval in which the solution is valid.}$
\begin{array}{lrll}
\textit{Given:}\\
%\qquad&&&\qquad&\\
\displaystyle &xy^\prime+2y&\displaystyle =x^2-x+1,\quad y(1)=\frac{1}{2} &&&(1)\\
\textit{Divide thru by x}\\
&\displaystyle y' +\frac{2}{x}y&\displaystyle =x-1+\frac{1}{x}& &&(2)\\
\textit {Find u(x)}\\
&\displaystyle u(x)&\displaystyle =\exp\int \frac{2}{x} \, dx =\ln e^{x^2}=x^2&&&(3)\\
\textit{Multiply thru $x^2$}\\
&x^2y' +(x^2)'y&\displaystyle =x^2\left(x-1+\frac{1}{x}\right)&&&(4)\\
\textit{Rewrite:}\\
&(x^2 y)'&=x^3-x^2+x \, &&&(5)\\
\textit{Integrate}\\
&\displaystyle x^2y&\displaystyle =\int x^3-x^2+x \, dx &&&(6)\\
&&\displaystyle=\frac{x^4}{4}-\frac{x^3}{3}+\frac{x^2}{2}+c\\
\textit{Divide thru by $x^2$}\\
&\displaystyle y&\displaystyle=\frac{x^2}{4}-\frac{x}{3}+\frac{1}{2}+\frac{c}{x^2}&&&(7)\\
\textit{Solve for c}\\
&\displaystyle y(1)&\displaystyle=\frac{1}{4}-\frac{1}{3}+\frac{1}{2}+c=\frac{1}{2}&&&(8)\\
\textit{With $\displaystyle c=\frac{1}{12}$ then}\\
&y&=\color{red}{\displaystyle\frac{1}{12}\left(3x^2-4x+6+\frac{1}{x^2}\right)}&&&(9)
\end{array}

suggestions and insults welcome
actually I don't know what the interval is that valid?