- #1

karush

Gold Member

MHB

- 3,269

- 5

$$xy^\prime+y=e^x, \qquad y(1)=1$$

$$\begin{array}{lrll}

\textit{Divide thru with x}\\

&\displaystyle y' +\frac{1}{x}y

&\displaystyle=\,\frac{e^x}{x} &_{(1)}\\

\textit {Find u(x)}\\

&\displaystyle u(x)

&\displaystyle=\exp\int\frac{1}{x}\,dx\\

&&=e^{\ln {x}}\\

&&=x &_{(2)}\\

\textit{Multiply thru with $x$} \\

&(xy)' +x'y&=e^x &_{(3)}\\

\textit{Rewrite:}\\

&(xy)'&=e^x &_{(4)}\\

\textit{Integrate }\\

&\displaystyle xy

&=\displaystyle\int e^x \, dx\\

&&=\displaystyle e^x+c &_{(5)}\\

\textit{Divide thru by $x$}\\

&\displaystyle y&=\displaystyle\frac{e^x}{x}+\frac{c}{e^x} &_{(6)}\\

\textit{So then if }\\

&\displaystyle y(1)&\displaystyle=e+\frac{c}{e}=1 &_{(7)}\\

\textit{with $c=?$ then }\\

&\displaystyle y

&=\color{red}{\displaystyle\frac{1}{x}(e^x + 1 - e)} &_{(8)}\\

\end{array}$$

ok (8) is the book answer but ? what wuld be c?

$\textit{State the interval in which the solution is valid. ?}\\$