# -aux.2.2.4 de y'+(\cot x)y&=2 \csc x; y(\pi/2)=a

• MHB
• karush
In summary, the given differential equation can be solved by finding the integrating factor u(x) and multiplying the equation through by sin x. After rewriting and integrating, the solution for y(x) can be found to be (2x+1-π) csc x, with the parameter c_1 determined from the initial value of y at π/2. The solution is equivalent to the one given by W|A, with the restriction that x cannot equal πk for any integer k, and the initial value being in the interval 0<x<π. The person discussing the solution also mentions carrying the 2 through in a step.
karush
Gold Member
MHB
652
$$\begin{array}{lrll} \textit {Given }\\ &y'+(\cot x)y&=2 \csc x \qquad y(\pi/2)=1 &_{(1)}\\ \textit {Find u(x) }\\ &\displaystyle u(x)&=\exp\int \cot x \, dx\\ &&\displaystyle=e^{\ln{\sin{x}}}\\ &&\displaystyle=\sin{x} &_{(2)}\\ \textit{multiply thru w/ \sin x} \\ &\sin x y' +\cos x y&=1 &_{(3)}\\ \textit{rewrite:}\\ &(\sin{x} y)'&=1 &_{(4)}\\ \textit{Integrate }\\ &\displaystyle \sin{x} y &=\displaystyle\int \, dx\\ &&=x+c &_{(5)}\\ \textit{divide thru by \sin{x}}\\ &\displaystyle y &=x\csc x+\displaystyle\frac{c}{\sin x} &_{(6)}\\ \textit{So if }\\ &\displaystyle y(\pi/2) &=\displaystyle\frac{\pi}{2}(1)+c=1 % +\displaystyle\frac{c}{\sin \frac{\pi}{2}}=1\\ \\ \\ &&c=\displaystyle 1-\frac{\pi}{2} &_{(7)}\\ \textit{finally W|A says}\\ %&y&=\color{red}{\displaystyle\frac{2x+1-\pi}{\sin{x}}, %\quad 0<x<\pi} &y&=\color{red}{\displaystyle y c_1 \csc{x} + 2 x \csc{x}, \quad 0<x<\pi} &_{(8)}\\ \end{array}$$

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

Last edited:
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:

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

mega mahalo again

I probably would have made if carried the 2 thru

## 5 Most Frequently Asked Questions About "-aux.2.2.4 de y'+(\cot x)y&=2 \csc x; y(\pi/2)=a"

1. What does the notation "-aux.2.2.4 de" mean in the given equation?

The notation "-aux.2.2.4 de" indicates that this is a differential equation. The "aux" stands for "auxiliary" and the numbers and letters after it represent the specific equation.

2. What do the terms "y', (\cot x)y, and \csc x" represent in the equation?

These terms represent the derivative of y with respect to x, the product of cotangent of x and y, and the cosecant of x, respectively.

3. How do I solve this differential equation?

To solve this equation, you can use various methods such as separation of variables, integrating factors, or substitution. The specific method will depend on the form of the equation and your personal preference.

4. What does the initial condition y(\pi/2)=a mean?

The initial condition y(\pi/2)=a means that the value of y at x=\pi/2 is equal to a. This is necessary in order to find a specific solution to the differential equation.

5. Can I use this equation to model a real-life situation?

Yes, this equation can be used to model various real-life situations such as radioactive decay, population growth, and chemical reactions. However, it is important to note that the specific values and parameters may need to be adjusted to accurately reflect the situation.

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