UnkleRhaukus
  • UnkleRhaukus
\[y'(x)-y\tan(x)=2\sin(x)h(x)\] [where \(h(t)\) is the heaviside unit step function ]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
are you like majoring in Math with study in Differential Equations?
UnkleRhaukus
  • UnkleRhaukus
\[ \newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd \begin{align*} y'(x)-y\tan(x)&=2\sin(x)h(x)\\ \\ \mu(x)=e^{\int{-\tan(x)}\dd x}\\ =e^{\int\frac{-\sin(x)\dd x}{\cos(x)}}\\ =e^{\ln\cos(x)}\\ =\cos(x)\\ \\ \left(y\cos(x)\right)'=&2\cos(x)\sin(x)h(x)\\ y\cos(x)&=2\int\cos(x)\sin(x)h(x)\dd x\\ &=\int\sin(2x)h(x)\dd x\\ \\ \end{align*} \]
UnkleRhaukus
  • UnkleRhaukus
im not sure what i should do with the h(x) in the indefinite integral

Looking for something else?

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

More answers

UnkleRhaukus
  • UnkleRhaukus
Im majoring in nuclear science technologies.
anonymous
  • anonymous
i haven't seen this stuff lol. Perhaps integrate as a function
anonymous
  • anonymous
i mean sin(2x). i think one of the two... still haven't remembered those 100%
anonymous
  • anonymous
oh i see you already did that lol
UnkleRhaukus
  • UnkleRhaukus
\[h(x)=\begin{cases}0&x<0\\1&x>0\end{cases}\]
anonymous
  • anonymous
it says h(t)
anonymous
  • anonymous
why do you have h(x)
UnkleRhaukus
  • UnkleRhaukus
whatever
anonymous
  • anonymous
lol i'd have to see notes on how to do this... otherwise im not much help
UnkleRhaukus
  • UnkleRhaukus
can i take the step function out of the integral ?
anonymous
  • anonymous
I dont think so =/ only constants can come out
UnkleRhaukus
  • UnkleRhaukus
i know how to integrate definite integrals with the step function , but i dont know a about indefinite integrals.
UnkleRhaukus
  • UnkleRhaukus
@abb0t
sirm3d
  • sirm3d
it's a two-part DE. \[y'-y \tan x = 0, \qquad x < 0\\y' - y \tan x = 2 \sin x, \quad x \geq 0\]
UnkleRhaukus
  • UnkleRhaukus
something like this?\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} y\cos(x)&=2\int\cos(x)\sin(x)h(x)\dd x\\ &=\int\sin(2x)h(x)\dd x\\ \\ &= \begin{cases} \int\sin(2x)\dd x &x\geq0\\ 0 &x<0 \end{cases}\\ &= \begin{cases} \frac{-\cos(2x)}{2}+c &x\geq0\\ 0 &x<0 \end{cases}\\ y&= \begin{cases} \frac{-\cos(2x)}{2\cos(x)}+c &x\geq0\\ 0 &x<0 \end{cases}\\ &= \begin{cases} \frac{1-\cos^2(x))}{2\cos(x)}+c &x\geq0\\ 0 &x<0 \end{cases}\\ &= \begin{cases} \frac12\big(\sec(x)-\cos(x)\big)+c &x\geq0\\ 0 &x<0 \end{cases}\\ \end{align*}\]
sirm3d
  • sirm3d
\[y=\begin{cases}-\frac{\cos 2x}{2\cos x}+C_1 &, x\geq 0\\\sec x+c_2 & ,x<0\end{cases}\]
sirm3d
  • sirm3d
\[y=\begin{cases}\frac{-(1/2)\cos 2x}{\cos x}+c_1&=\frac{1+\sin ^2 x-(3/2)}{\cos x}+C_1\\\sec x +C_2& =\frac{1}{\cos x}+C_2\end{cases}\] \[y=\begin{cases}\frac{1}{\cos x}+\frac{\sin ^2 x-(3/2)}{\cos x}+C_1\\\frac{1}{\cos x}+C_2\end{cases}\] \[\Large y = \frac{1}{\cos x} + \frac{\sin^2 x - (3/2)}{\cos x}h(x) + C\]
UnkleRhaukus
  • UnkleRhaukus
\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \begin{align*} y'(x)-y\tan(x)&=2\sin(x)h(x)\\ &=\begin{cases}0,&x<0\qquad \text{I}\\2\sin(x),&x\geq0\qquad \text{II}\end{cases}\\ \qquad \text{I for } x<0\\ \\ y'-y\tan(x)&=0\\ \de yx&=y\tan(x)\\ \frac{\dd y}{y}&=\tan(x)\dd x\\ \ln y&=\int\tan(x)\dd x\\ \ln y&=-\ln \cos(x)+c_1\\ y&=k\sec(x)\\ \\ \qquad \text{II for } x\geq0&\\ \\ y'(x)-y\tan(x)&=2\sin(x)\\ \\ \mu(x)=e^{\int{-\tan(x)}\dd x}\\ =e^{\int\frac{-\sin(x)\dd x}{\cos(x)}}\\ =e^{\ln\cos(x)}\\ =\cos(x)\\ \\ \left(y\cos(x)\right)' &=2\cos(x)\sin(x)\\ &=\sin(2x)\\ y\cos(x) &=\int\sin(2x)\dd x\\ &=\frac{-\cos(2x)}{2}+c_2\\ y &=\frac{-\cos(2x)}{2\cos(x)}+c_2\\ &=\frac{1-2\cos^2(x))}{2\cos(x)}+c_2\\ y&=\frac{\sec(x)}2-\cos(x)+c_2\\ \\ \end{align*}\]
abb0t
  • abb0t
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
UnkleRhaukus
  • UnkleRhaukus
\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \begin{align*} y'(x)-y\tan(x)&=2\sin(x)h(x) &=\begin{cases}0,&x<0\qquad \text{I}\\2\sin(x),&x\geq0\qquad \text{II}\end{cases}\\ \qquad \text{I for } x<0\\ \\ y'-y\tan(x)&=0\\ \de yx&=y\tan(x)\\ \frac{\dd y}{y}&=\tan(x)\dd x\\ \ln y&=\int\tan(x)\dd x\\ \ln y&=-\ln \cos(x)+c_1\\ y&=k\sec(x)\\ \\ \qquad \text{II for } x\geq0&\\ \\ y'(x)-y\tan(x)&=2\sin(x)\\ \\ \mu(x)=e^{\int{-\tan(x)}\dd x}\\ =e^{\int\frac{-\sin(x)\dd x}{\cos(x)}}\\ =e^{\ln\cos(x)}\\ =\cos(x)\\ \\ \big(y\cos(x)\big)' &=2\cos(x)\sin(x)\\ &=\sin(2x)\\ y\cos(x) &=\int\sin(2x)\dd x\\ &=\frac{-\cos(2x)}{2}+c_2\\ y &=\frac{-\cos(2x)}{2\cos(x)}+\frac{c_2}{\cos(x)}\\ &=\frac{1-2\cos^2(x))}{2\cos(x)}+c_2\sec(x)\\ &=\frac{\sec(x)}2-\cos(x)+c_2\sec(x)\\ &=\frac{(1+2c_2)\sec(x)}2-\cos(x)\\ &=c_3\sec(x)-\cos(x)\\ \end{align*} \]
UnkleRhaukus
  • UnkleRhaukus
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
  • 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)
UnkleRhaukus
  • UnkleRhaukus
oh yeah i didn't think about that , also the solution to a 1st order DE should only have one constant right?
phi
  • 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)
UnkleRhaukus
  • UnkleRhaukus
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
  • 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
  • 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...
UnkleRhaukus
  • UnkleRhaukus
Cosine of zero is equal to one so c_1 sec(x) - cos(x) h(x) is not continuous at the boundary. \[ \begin{align*} &y=\begin{cases}k\sec(x)&x<0\qquad \text{(I)}\\c\sec(x)-\cos(x)&x\geq0 \qquad \text{(II)}\end{cases}\\ \\ \\ k\sec(0)=c\sec(0)-\cos(0)&\\ k=c-1\qquad\qquad \\ \\ &y(x)=(k+ch(x))\sec(x)-\cos(x)h(x) \end{align*} \]
UnkleRhaukus
  • UnkleRhaukus
Ah this is better \[\begin{align*} y&=\begin{cases}k\sec(x)&x<0\qquad \text{(I)}\\c\sec(x)-\cos(x)&x\geq0 \qquad \text{(II)}\end{cases}\\ \\ \\ k\sec(0)=c\sec(0)-\cos(0)\\ k=c-1\qquad\qquad\quad\\ \\ \\ y&=k\sec(x)+\big(\sec x-\cos(x)\big )h(x)\\ &=k\sec(x)+\left(\frac1{\cos(x)}-\frac{\cos^2(x)}{\cos(x)}\right)h(x)\\ &=k\sec(x)+\left(\frac{\sin^2(x)}{\cos(x)}\right)h(x)\\ &=k\sec(x)+\sin(x)\tan(x)h(x)\color{red}\checkmark\\ \end{align*}\]

Looking for something else?

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