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

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

Mathematics
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are you like majoring in Math with study in Differential Equations?
\[ \newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd \begin{align*} y'(x)-y\tan(x)&=2\sin(x)h(x)\\ \\ \mu(x)=e^{\int{-\tan(x)}\dd x}\\ =e^{\int\frac{-\sin(x)\dd x}{\cos(x)}}\\ =e^{\ln\cos(x)}\\ =\cos(x)\\ \\ \left(y\cos(x)\right)'=&2\cos(x)\sin(x)h(x)\\ y\cos(x)&=2\int\cos(x)\sin(x)h(x)\dd x\\ &=\int\sin(2x)h(x)\dd x\\ \\ \end{align*} \]
im not sure what i should do with the h(x) in the indefinite integral

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Im majoring in nuclear science technologies.
i haven't seen this stuff lol. Perhaps integrate as a function
i mean sin(2x). i think one of the two... still haven't remembered those 100%
oh i see you already did that lol
\[h(x)=\begin{cases}0&x<0\\1&x>0\end{cases}\]
it says h(t)
why do you have h(x)
whatever
lol i'd have to see notes on how to do this... otherwise im not much help
can i take the step function out of the integral ?
I dont think so =/ only constants can come out
i know how to integrate definite integrals with the step function , but i dont know a about indefinite integrals.
it's a two-part DE. \[y'-y \tan x = 0, \qquad x < 0\\y' - y \tan x = 2 \sin x, \quad x \geq 0\]
something like this?\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} y\cos(x)&=2\int\cos(x)\sin(x)h(x)\dd x\\ &=\int\sin(2x)h(x)\dd x\\ \\ &= \begin{cases} \int\sin(2x)\dd x &x\geq0\\ 0 &x<0 \end{cases}\\ &= \begin{cases} \frac{-\cos(2x)}{2}+c &x\geq0\\ 0 &x<0 \end{cases}\\ y&= \begin{cases} \frac{-\cos(2x)}{2\cos(x)}+c &x\geq0\\ 0 &x<0 \end{cases}\\ &= \begin{cases} \frac{1-\cos^2(x))}{2\cos(x)}+c &x\geq0\\ 0 &x<0 \end{cases}\\ &= \begin{cases} \frac12\big(\sec(x)-\cos(x)\big)+c &x\geq0\\ 0 &x<0 \end{cases}\\ \end{align*}\]
\[y=\begin{cases}-\frac{\cos 2x}{2\cos x}+C_1 &, x\geq 0\\\sec x+c_2 & ,x<0\end{cases}\]
\[y=\begin{cases}\frac{-(1/2)\cos 2x}{\cos x}+c_1&=\frac{1+\sin ^2 x-(3/2)}{\cos x}+C_1\\\sec x +C_2& =\frac{1}{\cos x}+C_2\end{cases}\] \[y=\begin{cases}\frac{1}{\cos x}+\frac{\sin ^2 x-(3/2)}{\cos x}+C_1\\\frac{1}{\cos x}+C_2\end{cases}\] \[\Large y = \frac{1}{\cos x} + \frac{\sin^2 x - (3/2)}{\cos x}h(x) + C\]
\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \begin{align*} y'(x)-y\tan(x)&=2\sin(x)h(x)\\ &=\begin{cases}0,&x<0\qquad \text{I}\\2\sin(x),&x\geq0\qquad \text{II}\end{cases}\\ \qquad \text{I for } x<0\\ \\ y'-y\tan(x)&=0\\ \de yx&=y\tan(x)\\ \frac{\dd y}{y}&=\tan(x)\dd x\\ \ln y&=\int\tan(x)\dd x\\ \ln y&=-\ln \cos(x)+c_1\\ y&=k\sec(x)\\ \\ \qquad \text{II for } x\geq0&\\ \\ y'(x)-y\tan(x)&=2\sin(x)\\ \\ \mu(x)=e^{\int{-\tan(x)}\dd x}\\ =e^{\int\frac{-\sin(x)\dd x}{\cos(x)}}\\ =e^{\ln\cos(x)}\\ =\cos(x)\\ \\ \left(y\cos(x)\right)' &=2\cos(x)\sin(x)\\ &=\sin(2x)\\ y\cos(x) &=\int\sin(2x)\dd x\\ &=\frac{-\cos(2x)}{2}+c_2\\ y &=\frac{-\cos(2x)}{2\cos(x)}+c_2\\ &=\frac{1-2\cos^2(x))}{2\cos(x)}+c_2\\ y&=\frac{\sec(x)}2-\cos(x)+c_2\\ \\ \end{align*}\]
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
\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \begin{align*} y'(x)-y\tan(x)&=2\sin(x)h(x) &=\begin{cases}0,&x<0\qquad \text{I}\\2\sin(x),&x\geq0\qquad \text{II}\end{cases}\\ \qquad \text{I for } x<0\\ \\ y'-y\tan(x)&=0\\ \de yx&=y\tan(x)\\ \frac{\dd y}{y}&=\tan(x)\dd x\\ \ln y&=\int\tan(x)\dd x\\ \ln y&=-\ln \cos(x)+c_1\\ y&=k\sec(x)\\ \\ \qquad \text{II for } x\geq0&\\ \\ y'(x)-y\tan(x)&=2\sin(x)\\ \\ \mu(x)=e^{\int{-\tan(x)}\dd x}\\ =e^{\int\frac{-\sin(x)\dd x}{\cos(x)}}\\ =e^{\ln\cos(x)}\\ =\cos(x)\\ \\ \big(y\cos(x)\big)' &=2\cos(x)\sin(x)\\ &=\sin(2x)\\ y\cos(x) &=\int\sin(2x)\dd x\\ &=\frac{-\cos(2x)}{2}+c_2\\ y &=\frac{-\cos(2x)}{2\cos(x)}+\frac{c_2}{\cos(x)}\\ &=\frac{1-2\cos^2(x))}{2\cos(x)}+c_2\sec(x)\\ &=\frac{\sec(x)}2-\cos(x)+c_2\sec(x)\\ &=\frac{(1+2c_2)\sec(x)}2-\cos(x)\\ &=c_3\sec(x)-\cos(x)\\ \end{align*} \]
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?
  • phi
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)
oh yeah i didn't think about that , also the solution to a 1st order DE should only have one constant right?
  • phi
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)
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
  • phi
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
  • phi
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...
Cosine of zero is equal to one so c_1 sec(x) - cos(x) h(x) is not continuous at the boundary. \[ \begin{align*} &y=\begin{cases}k\sec(x)&x<0\qquad \text{(I)}\\c\sec(x)-\cos(x)&x\geq0 \qquad \text{(II)}\end{cases}\\ \\ \\ k\sec(0)=c\sec(0)-\cos(0)&\\ k=c-1\qquad\qquad \\ \\ &y(x)=(k+ch(x))\sec(x)-\cos(x)h(x) \end{align*} \]
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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