Question

\[y'(x)-y\tan(x)=2\sin(x)h(x)\]

[where \(h(t)\) is the heaviside unit step function ]

Answer 1

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

Answer 2

\[
\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*}
\]

Answer 3

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

Answer 4

Im majoring in nuclear science technologies.

Answer 5

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

Answer 6

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

Answer 7

oh i see you already did that lol

Answer 8

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

Answer 9

it says h(t)

Answer 10

why do you have h(x)

Answer 11

whatever

Answer 12

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

Answer 13

can i take the step function out of the integral ?

Answer 14

I dont think so =/ only constants can come out

Answer 15

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

Answer 16

@abb0t

Answer 17

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

Answer 18

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*}\]

Answer 19

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

Answer 20

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

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

\[\Large y = \frac{1}{\cos x} + \frac{\sin^2 x - (3/2)}{\cos x}h(x) + C\]

Answer 21

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

Answer 22

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

Answer 23

\[\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*}
\]

Answer 24

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?

Answer 25

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)

Answer 26

oh yeah i didn't think about that ,

also the solution to a 1st order DE should only have one constant right?

Answer 27

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)

Answer 28

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

Answer 29

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

Answer 30

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...

Answer 31

Cosine of zero is equal to one
so c_1 sec(x) - cos(x) h(x) is not continuous at the boundary.


\[
\begin{align*}
&y=\begin{cases}k\sec(x)&x<0\qquad \text{(I)}\\c\sec(x)-\cos(x)&x\geq0 \qquad \text{(II)}\end{cases}\\
\\
\\
k\sec(0)=c\sec(0)-\cos(0)&\\
k=c-1\qquad\qquad
\\
\\
&y(x)=(k+ch(x))\sec(x)-\cos(x)h(x)
\end{align*}
\]

Answer 32

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