Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

UnkleRhaukus

  • one year ago

\[y'(x)-y\tan(x)=2\sin(x)h(x)\] [where \(h(t)\) is the heaviside unit step function ]

  • This Question is Closed
  1. Outkast3r09
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    are you like majoring in Math with study in Differential Equations?

  2. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ \newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd \begin{align*} y'(x)-y\tan(x)&=2\sin(x)h(x)\\ \\ \mu(x)=e^{\int{-\tan(x)}\dd x}\\ =e^{\int\frac{-\sin(x)\dd x}{\cos(x)}}\\ =e^{\ln\cos(x)}\\ =\cos(x)\\ \\ \left(y\cos(x)\right)'=&2\cos(x)\sin(x)h(x)\\ y\cos(x)&=2\int\cos(x)\sin(x)h(x)\dd x\\ &=\int\sin(2x)h(x)\dd x\\ \\ \end{align*} \]

  3. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im not sure what i should do with the h(x) in the indefinite integral

  4. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Im majoring in nuclear science technologies.

  5. Outkast3r09
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i haven't seen this stuff lol. Perhaps integrate as a function

  6. Outkast3r09
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i mean sin(2x). i think one of the two... still haven't remembered those 100%

  7. Outkast3r09
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh i see you already did that lol

  8. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[h(x)=\begin{cases}0&x<0\\1&x>0\end{cases}\]

  9. Outkast3r09
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it says h(t)

  10. Outkast3r09
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why do you have h(x)

  11. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whatever

  12. Outkast3r09
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol i'd have to see notes on how to do this... otherwise im not much help

  13. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can i take the step function out of the integral ?

  14. Outkast3r09
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I dont think so =/ only constants can come out

  15. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i know how to integrate definite integrals with the step function , but i dont know a about indefinite integrals.

  16. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @abb0t

  17. sirm3d
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    it's a two-part DE. \[y'-y \tan x = 0, \qquad x < 0\\y' - y \tan x = 2 \sin x, \quad x \geq 0\]

  18. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    something like this?\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} y\cos(x)&=2\int\cos(x)\sin(x)h(x)\dd x\\ &=\int\sin(2x)h(x)\dd x\\ \\ &= \begin{cases} \int\sin(2x)\dd x &x\geq0\\ 0 &x<0 \end{cases}\\ &= \begin{cases} \frac{-\cos(2x)}{2}+c &x\geq0\\ 0 &x<0 \end{cases}\\ y&= \begin{cases} \frac{-\cos(2x)}{2\cos(x)}+c &x\geq0\\ 0 &x<0 \end{cases}\\ &= \begin{cases} \frac{1-\cos^2(x))}{2\cos(x)}+c &x\geq0\\ 0 &x<0 \end{cases}\\ &= \begin{cases} \frac12\big(\sec(x)-\cos(x)\big)+c &x\geq0\\ 0 &x<0 \end{cases}\\ \end{align*}\]

  19. sirm3d
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \[y=\begin{cases}-\frac{\cos 2x}{2\cos x}+C_1 &, x\geq 0\\\sec x+c_2 & ,x<0\end{cases}\]

  20. sirm3d
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \[y=\begin{cases}\frac{-(1/2)\cos 2x}{\cos x}+c_1&=\frac{1+\sin ^2 x-(3/2)}{\cos x}+C_1\\\sec x +C_2& =\frac{1}{\cos x}+C_2\end{cases}\] \[y=\begin{cases}\frac{1}{\cos x}+\frac{\sin ^2 x-(3/2)}{\cos x}+C_1\\\frac{1}{\cos x}+C_2\end{cases}\] \[\Large y = \frac{1}{\cos x} + \frac{\sin^2 x - (3/2)}{\cos x}h(x) + C\]

  21. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \begin{align*} y'(x)-y\tan(x)&=2\sin(x)h(x)\\ &=\begin{cases}0,&x<0\qquad \text{I}\\2\sin(x),&x\geq0\qquad \text{II}\end{cases}\\ \qquad \text{I for } x<0\\ \\ y'-y\tan(x)&=0\\ \de yx&=y\tan(x)\\ \frac{\dd y}{y}&=\tan(x)\dd x\\ \ln y&=\int\tan(x)\dd x\\ \ln y&=-\ln \cos(x)+c_1\\ y&=k\sec(x)\\ \\ \qquad \text{II for } x\geq0&\\ \\ y'(x)-y\tan(x)&=2\sin(x)\\ \\ \mu(x)=e^{\int{-\tan(x)}\dd x}\\ =e^{\int\frac{-\sin(x)\dd x}{\cos(x)}}\\ =e^{\ln\cos(x)}\\ =\cos(x)\\ \\ \left(y\cos(x)\right)' &=2\cos(x)\sin(x)\\ &=\sin(2x)\\ y\cos(x) &=\int\sin(2x)\dd x\\ &=\frac{-\cos(2x)}{2}+c_2\\ y &=\frac{-\cos(2x)}{2\cos(x)}+c_2\\ &=\frac{1-2\cos^2(x))}{2\cos(x)}+c_2\\ y&=\frac{\sec(x)}2-\cos(x)+c_2\\ \\ \end{align*}\]

  22. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i don't think im getting notifications on OS lately cuz i didn't even get the mention. i just ran across this question right now o.0

  23. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \begin{align*} y'(x)-y\tan(x)&=2\sin(x)h(x) &=\begin{cases}0,&x<0\qquad \text{I}\\2\sin(x),&x\geq0\qquad \text{II}\end{cases}\\ \qquad \text{I for } x<0\\ \\ y'-y\tan(x)&=0\\ \de yx&=y\tan(x)\\ \frac{\dd y}{y}&=\tan(x)\dd x\\ \ln y&=\int\tan(x)\dd x\\ \ln y&=-\ln \cos(x)+c_1\\ y&=k\sec(x)\\ \\ \qquad \text{II for } x\geq0&\\ \\ y'(x)-y\tan(x)&=2\sin(x)\\ \\ \mu(x)=e^{\int{-\tan(x)}\dd x}\\ =e^{\int\frac{-\sin(x)\dd x}{\cos(x)}}\\ =e^{\ln\cos(x)}\\ =\cos(x)\\ \\ \big(y\cos(x)\big)' &=2\cos(x)\sin(x)\\ &=\sin(2x)\\ y\cos(x) &=\int\sin(2x)\dd x\\ &=\frac{-\cos(2x)}{2}+c_2\\ y &=\frac{-\cos(2x)}{2\cos(x)}+\frac{c_2}{\cos(x)}\\ &=\frac{1-2\cos^2(x))}{2\cos(x)}+c_2\sec(x)\\ &=\frac{\sec(x)}2-\cos(x)+c_2\sec(x)\\ &=\frac{(1+2c_2)\sec(x)}2-\cos(x)\\ &=c_3\sec(x)-\cos(x)\\ \end{align*} \]

  24. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so \[y(x)=k\sec(x)+\big(c_3\sec(x)-\cos(x)\big)h(x)\]\[=\big(k+c_3h(x)\big)\sec(x)-\cos(x)h(x)\] right?

  25. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    close, but I think you want the k sec(x) term to go away for x≥0, right? one way to do this is multiply it by h(-x)

  26. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh yeah i didn't think about that , also the solution to a 1st order DE should only have one constant right?

  27. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I don't know how many constants should be in this problem (I only know enough to be dangerous). But if that is the case, it would be nice, because you could write your solution as x<0 c1 sec(x) x≥0 c2 sec(x) - cos(x) if c1 must equal c2 this can be written as c1 sec(x) - cos(x) h(x)

  28. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ah cool, that looks good, however i think i made a mistake somewhere in solving y′(x)−ytan(x)=2sin(x) because it dosent agree with the wolf http://www.wolframalpha.com/input/?i=y%27-y*tanx%3D2*sin+x

  29. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I think a multiple term trig function can take many equivalent forms 1/2* sec(x) - cos(x) = -1/2* cos(2x)*sec(x) http://www.wolframalpha.com/input/?i=plot+1%2F2*sec%28x%29+-cos%28x%29%2C-1%2F2*+cos%282x%29*sec%28x%29

  30. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I think it comes down to what the constants are for the two branches x<0 and x≥0 we could write x< 0 c1 sec(x) x≥0 c2 sec(x) + 0.5 sec(x) - cos(x) and we could combine the sec(x) and assign a different constant...

  31. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Cosine of zero is equal to one so c_1 sec(x) - cos(x) h(x) is not continuous at the boundary. \[ \begin{align*} &y=\begin{cases}k\sec(x)&x<0\qquad \text{(I)}\\c\sec(x)-\cos(x)&x\geq0 \qquad \text{(II)}\end{cases}\\ \\ \\ k\sec(0)=c\sec(0)-\cos(0)&\\ k=c-1\qquad\qquad \\ \\ &y(x)=(k+ch(x))\sec(x)-\cos(x)h(x) \end{align*} \]

  32. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ah this is better \[\begin{align*} y&=\begin{cases}k\sec(x)&x<0\qquad \text{(I)}\\c\sec(x)-\cos(x)&x\geq0 \qquad \text{(II)}\end{cases}\\ \\ \\ k\sec(0)=c\sec(0)-\cos(0)\\ k=c-1\qquad\qquad\quad\\ \\ \\ y&=k\sec(x)+\big(\sec x-\cos(x)\big )h(x)\\ &=k\sec(x)+\left(\frac1{\cos(x)}-\frac{\cos^2(x)}{\cos(x)}\right)h(x)\\ &=k\sec(x)+\left(\frac{\sin^2(x)}{\cos(x)}\right)h(x)\\ &=k\sec(x)+\sin(x)\tan(x)h(x)\color{red}\checkmark\\ \end{align*}\]

  33. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.