# [SOLVED]2.2.7 de with trig and y(1)=1

#### karush

##### Well-known member
$$\begin{array}{lrll} \textit {Given }\\ &y'+(\cot x)y&=2 \csc x \qquad y(\pi/2)=1 &_{(1)}\\ \textit {Find u(x) }\\ &\displaystyle u(x)&=\exp\int \cot x \, dx\\ &&\displaystyle=e^{\ln{\sin{x}}}\\ &&\displaystyle=\sin{x} &_{(2)}\\ \textit{multiply thru w/ \sin x} \\ &\sin x y' +\cos x y&=1 &_{(3)}\\ \textit{rewrite:}\\ &(\sin{x} y)'&=1 &_{(4)}\\ \textit{Integrate }\\ &\displaystyle \sin{x} y &=\displaystyle\int \, dx\\ &&=x+c &_{(5)}\\ \textit{divide thru by \sin{x}}\\ &\displaystyle y &=x\csc x+\displaystyle\frac{c}{\sin x} &_{(6)}\\ \textit{So if }\\ &\displaystyle y(\pi/2) &=\displaystyle\frac{\pi}{2}(1)+c=1 % +\displaystyle\frac{c}{\sin \frac{\pi}{2}}=1\\ \\ \\ &&c=\displaystyle 1-\frac{\pi}{2} &_{(7)}\\ \textit{finally W|A says}\\ %&y&=\color{red}{\displaystyle\frac{2x+1-\pi}{\sin{x}}, %\quad 0<x<\pi} &y&=\color{red}{\displaystyle y c_1 \csc{x} + 2 x \csc{x}, \quad 0<x<\pi} &_{(8)}\\ \end{array}$$

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

#### MarkFL

Staff member
When you multiply through by $\mu(x)$, you should have:

$$\displaystyle \frac{d}{dx}(\sin(x)y)=2$$

And then integrate:

$$\displaystyle \sin(x)y=2x+c_1$$

$$\displaystyle y(x)=2x\csc(x)+c_1\csc(x)$$

Determine parameter from initial values

$$\displaystyle y\left(\frac{\pi}{2}\right)=2\left(\frac{\pi}{2}\right)\csc\left(\frac{\pi}{2}\right)+c_1\csc\left(\frac{\pi}{2}\right)=\pi+c_1=1\implies c_1=1-\pi$$

And so the solution to the IVP is:

$$\displaystyle y(x)=2x\csc(x)+(1-\pi)\csc(x)=(2x+1-\pi)\csc(x)$$

This is equivalent to what W|A returns for me. In the original ODE, we have:

$$\displaystyle x\ne\pi k$$ where $$\displaystyle k\in\mathbb{Z}$$

And we find the initial value is in the interval:

$$\displaystyle 0<x<\pi$$

#### karush

##### Well-known member
mega mahalo again

I probably would have made if carried the 2 thru