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UnkleRhaukus

  • 2 years ago

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

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  1. Outkast3r09
    • 2 years ago
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    are you like majoring in Math with study in Differential Equations?

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

  3. UnkleRhaukus
    • 2 years ago
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    im not sure what i should do with the h(x) in the indefinite integral

  4. UnkleRhaukus
    • 2 years ago
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    Im majoring in nuclear science technologies.

  5. Outkast3r09
    • 2 years ago
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    i haven't seen this stuff lol. Perhaps integrate as a function

  6. Outkast3r09
    • 2 years ago
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    i mean sin(2x). i think one of the two... still haven't remembered those 100%

  7. Outkast3r09
    • 2 years ago
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    oh i see you already did that lol

  8. UnkleRhaukus
    • 2 years ago
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    \[h(x)=\begin{cases}0&x<0\\1&x>0\end{cases}\]

  9. Outkast3r09
    • 2 years ago
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    it says h(t)

  10. Outkast3r09
    • 2 years ago
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    why do you have h(x)

  11. UnkleRhaukus
    • 2 years ago
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    whatever

  12. Outkast3r09
    • 2 years ago
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    lol i'd have to see notes on how to do this... otherwise im not much help

  13. UnkleRhaukus
    • 2 years ago
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    can i take the step function out of the integral ?

  14. Outkast3r09
    • 2 years ago
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    I dont think so =/ only constants can come out

  15. UnkleRhaukus
    • 2 years ago
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    i know how to integrate definite integrals with the step function , but i dont know a about indefinite integrals.

  16. UnkleRhaukus
    • 2 years ago
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    @abb0t

  17. sirm3d
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    \[y=\begin{cases}-\frac{\cos 2x}{2\cos x}+C_1 &, x\geq 0\\\sec x+c_2 & ,x<0\end{cases}\]

  20. sirm3d
    • 2 years ago
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    \[y=\begin{cases}\frac{-(1/2)\cos 2x}{\cos x}+c_1&=\frac{1+\sin ^2 x-(3/2)}{\cos x}+C_1\\\sec x +C_2& =\frac{1}{\cos x}+C_2\end{cases}\] \[y=\begin{cases}\frac{1}{\cos x}+\frac{\sin ^2 x-(3/2)}{\cos x}+C_1\\\frac{1}{\cos x}+C_2\end{cases}\] \[\Large y = \frac{1}{\cos x} + \frac{\sin^2 x - (3/2)}{\cos x}h(x) + C\]

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

  22. abb0t
    • 2 years ago
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    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
    • 2 years ago
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    \[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \begin{align*} y'(x)-y\tan(x)&=2\sin(x)h(x) &=\begin{cases}0,&x<0\qquad \text{I}\\2\sin(x),&x\geq0\qquad \text{II}\end{cases}\\ \qquad \text{I for } x<0\\ \\ y'-y\tan(x)&=0\\ \de yx&=y\tan(x)\\ \frac{\dd y}{y}&=\tan(x)\dd x\\ \ln y&=\int\tan(x)\dd x\\ \ln y&=-\ln \cos(x)+c_1\\ y&=k\sec(x)\\ \\ \qquad \text{II for } x\geq0&\\ \\ y'(x)-y\tan(x)&=2\sin(x)\\ \\ \mu(x)=e^{\int{-\tan(x)}\dd x}\\ =e^{\int\frac{-\sin(x)\dd x}{\cos(x)}}\\ =e^{\ln\cos(x)}\\ =\cos(x)\\ \\ \big(y\cos(x)\big)' &=2\cos(x)\sin(x)\\ &=\sin(2x)\\ y\cos(x) &=\int\sin(2x)\dd x\\ &=\frac{-\cos(2x)}{2}+c_2\\ y &=\frac{-\cos(2x)}{2\cos(x)}+\frac{c_2}{\cos(x)}\\ &=\frac{1-2\cos^2(x))}{2\cos(x)}+c_2\sec(x)\\ &=\frac{\sec(x)}2-\cos(x)+c_2\sec(x)\\ &=\frac{(1+2c_2)\sec(x)}2-\cos(x)\\ &=c_3\sec(x)-\cos(x)\\ \end{align*} \]

  24. UnkleRhaukus
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    oh yeah i didn't think about that , also the solution to a 1st order DE should only have one constant right?

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

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