- #1

karush

Gold Member

MHB

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$$\begin{array}{lrll}

\textit {Given }\\

&y'+(\cot x)y&=2 \csc x \qquad y(\pi/2)=1 &_{(1)}\\

\textit {Find u(x) }\\

&\displaystyle u(x)&=\exp\int \cot x \, dx\\

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

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

\textit{multiply thru w/ $\sin x$} \\

&\sin x y' +\cos x y&=1 &_{(3)}\\

\textit{rewrite:}\\

&(\sin{x} y)'&=1 &_{(4)}\\

\textit{Integrate }\\

&\displaystyle \sin{x} y

&=\displaystyle\int \, dx\\

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

\textit{divide thru by $\sin{x}$}\\

&\displaystyle y

&=x\csc x+\displaystyle\frac{c}{\sin x} &_{(6)}\\

\textit{So if }\\

&\displaystyle y(\pi/2)

&=\displaystyle\frac{\pi}{2}(1)+c=1

% +\displaystyle\frac{c}{\sin \frac{\pi}{2}}=1\\

\\ \\

&&c=\displaystyle 1-\frac{\pi}{2} &_{(7)}\\

\textit{finally W|A says}\\

%&y&=\color{red}{\displaystyle\frac{2x+1-\pi}{\sin{x}},

%\quad 0<x<\pi}

&y&=\color{red}{\displaystyle y c_1 \csc{x}

+ 2 x \csc{x},

\quad 0<x<\pi} &_{(8)}\\

\end{array}$$

ok I didnt quite get the last 3 steps red is W|A answer

$$\begin{array}{lrll}

\textit {Given }\\

&y'+(\cot x)y&=2 \csc x \qquad y(\pi/2)=1 &_{(1)}\\

\textit {Find u(x) }\\

&\displaystyle u(x)&=\exp\int \cot x \, dx\\

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

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

\textit{multiply thru w/ $\sin x$} \\

&\sin x y' +\cos x y&=1 &_{(3)}\\

\textit{rewrite:}\\

&(\sin{x} y)'&=1 &_{(4)}\\

\textit{Integrate }\\

&\displaystyle \sin{x} y

&=\displaystyle\int \, dx\\

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

\textit{divide thru by $\sin{x}$}\\

&\displaystyle y

&=x\csc x+\displaystyle\frac{c}{\sin x} &_{(6)}\\

\textit{So if }\\

&\displaystyle y(\pi/2)

&=\displaystyle\frac{\pi}{2}(1)+c=1

% +\displaystyle\frac{c}{\sin \frac{\pi}{2}}=1\\

\\ \\

&&c=\displaystyle 1-\frac{\pi}{2} &_{(7)}\\

\textit{finally W|A says}\\

%&y&=\color{red}{\displaystyle\frac{2x+1-\pi}{\sin{x}},

%\quad 0<x<\pi}

&y&=\color{red}{\displaystyle y c_1 \csc{x}

+ 2 x \csc{x},

\quad 0<x<\pi} &_{(8)}\\

\end{array}$$

ok I didnt quite get the last 3 steps red is W|A answer

Last edited: