Question

\[y′+y=g(x)\]
find the solution

Answer 1

\[y'+y=g(x)\]\[g(x) =\begin{cases}x & 0\leq x <1\\0 & 1\leq x\end{cases}\qquad\qquad y(0)=0\]

Answer 2

\[\mu (x)=e^{\int\text dx}=e^x\]

\[\left(ye^x\right)'=e^xg(x)\]\[ye^x=\int e^xg(x)\text dx\]\[=\int_0^1 xe^x\text dx\]\[=\left.xe^x\right|_0^1-\int_0^1 e^x\text dx\]\[=e-\left.e^x\right|_0^1\]\[=e-e+1\]\[=1\]

Answer 3

\[ye^x=1\]
\[y=e^{-x}\]
\[y(0)=e^{-0}=1\neq0\] ?

Answer 4

where did i go wrong?

Answer 5

@mukushla what do you think ?

Answer 6

it will be
ye^x=1 + c
c is the constant

Answer 7

when u substitute y(0) i.e. x=0
u need to find the constant and then plug in the value of c in
ye^x =1 + c
this will be your final answer

Answer 8

or maybe

\[\large ye^x=\int_{c}^{1} xe^x dx \]

Answer 9

\[\left(ye^x\right)'=e^xg(x)\]\[ye^x+c=\int_0^1 xe^x\text dx\]\[ye^x+c=1\]\[ye^x=1-c\]\[y=e^{-x}-ce^{-x}\]
\[y(x)=e^{-x}-ce^{-x}\]\[y(0)=1-c=0\]\[c=1\]\[y(x)=e^{-x}-e^{-x}=0\]?

Answer 10

\[ ye^x=\int_{c}^{1} xe^x dx\]\[=\left.xe^x\right|_c^1-\left.e^x\right|_c^1\]\[=e-ce^c-e+e^c\]\[ye^x=e^c-ce^c\]\[y(x)=e^x(e^{c}-ce^{c})\]\[y(0)=e^{c}-ce^{c})=0\]\[e^c(1-c)=0\]\[c=1\]\[y(x)=e^x(e-ce)\]\[y(x)=e^{x+1}(1-1)=0\]?

Answer 11

@UnkleRhaukus

how about laplace transform?

Answer 12

what answer do you get

Answer 13

after applying laplace we have\[\large Y(s)+sY(s)-y(0)=\int_{0}^{1} xe^{-sx} dx\]

Answer 14

and \[Y(s)=\frac{1}{(1+s)s^2}-\frac{e^{-s}}{s^2}\]since \(y(0)=0\)

Answer 15

hm

Answer 16

we can calculate inverse of it but wolfram says

\[y=-(x-1) H(x-1)+x+e^{-x}-1\]

Answer 17

im wondering how we can obtain it with other methods....!!!

Answer 18

\[y'+y=xu(x-1)=(x-1)u_1+u_1\]\[sY(s)-y(0)+Y(s)=\frac{e^{-s}}{s^2}+\frac{e^{-s}}s\]\[Y(s)(s+1)=e^{-s}\left(\frac1{s^2}+\frac1s\right)=e^{-s}\left({s+1\over s^2}\right)\]\[Y(s)={e^{-s}\over s^2}\implies y=(x-1)u_1\]what did I do wrong?

Answer 19

what is \(u_1\)

Answer 20

heavyside
H(x-1) maybe you know it as

Answer 21

i dont think im ment to use laplace for this question

Answer 22

why not, I think I successfully redefined the problem in terms of the heavyside...
what section of your book is this in?

Answer 23

this is the first assignment of a Differential Equations class

Answer 24

laplace trans forms dont come in till chapter 5, we havent covered everything form chapter 3 yet

Answer 25

do you know what the answer is supposed to be?

Answer 26

im not sure

Answer 27

oh wait, my heavyside thing is wrong
but I won't fix it since you said no laplace yet anyway

Answer 28

should have started with\[y'+y=x(1-u_1)\]that is the step function redone with the heaviside

Answer 29

but if not laplace allowed this does not help

Answer 30

@lalaly

Answer 31

well ... \((e^x y)'=e^x g(x)\) and we have 2 cases

\((I)\) \( 0\le x <1 \)

\[(e^x y)'=e^x g(x)=xe^x\]\[ye^x=xe^x-e^x+c\]\[y=x-1+ce^{-x}\]\[y(0)=0\]it gives \(c=1\)

\((II)\) \(x \ge1\)

\[(ye^x)'=0\]\[ye^x=b\]\[y=be^{-x}\]
so we can write

\[y =\begin{cases}x-1+e^{-x} & 0\leq x <1\\be^{-x} & 1\leq x\end{cases}\]if we assume that \(y\) is a continous function easily u can see \(b=1\) and the final answer:\[y =\begin{cases}x-1+e^{-x} & 0\leq x <1\\e^{-x} & 1\leq x\end{cases}=-(x-1)H(x-1)+x-1+e^{-x}\]

Answer 32

exactly th same result from laplace transform

Answer 33

@mukushla you are awesome.this is 100% correct. i was reviewing my DEqS book for this Problem was just about to go for this process as you did.the key Point was the requirement of continuity at x=1.

Answer 34

:)

Answer 35

hey that's brilliant #Mukushla !
i wasn't sure how to handle the piecewise function but now i see i must take cases,
thank you

Answer 36

yw :)

Answer 37

im a little bit unsure how i can assume the solution is continuos but every think else makes so much sense now

Answer 38

i feel i need another line or some words so something,

how do we know continuity of the solution at x=1?

Answer 39

i say y(1)=1, but that kinda comes outer nowhere

Answer 40

@eliassaab

sir..how we can say y is continous? is this come out from type of equation?

Answer 41

or it should be mentiond as a condition for problem?

Answer 42

quite different with what i got !!

Answer 43

y should be continuous and piecewise differentiable

Answer 44

so its obtainable from equation that y is continuous and piecewise differentiable...

Answer 45

Sorry the solution, that I posted above was for g[x]=1 for x between 0 and 1.
Here is the solution for the given problem
\[
y(x)=\left \{\begin{array}{cc}
0 & x\leq 0 \\
e^{-x} \left( 1-e^x+ x e^x\right) & 0<x\leq 1 \\
e^{-x} & x \ge 1 \\
\end{array}
\right .
\]

Answer 46

@mukushla your solution is correct.

Answer 47

oh...thank u sir..:)

Answer 48

if you take
\[
u(x)=e^{-x} \left(e^x
x-e^x+1\right)\\
u'(x) + u(x) =x\\
and\\
u(x)=e^{-x}\\
u'(x) + u(x) =0
\]

Answer 49

yeah...actually im stuck with my first question...i want just a little explanation about continuity if u dont mind me asking this again...

Answer 50

\[u(1)= e^{-1}
\]
on both side of 1.

Answer 51

should i say something about the limit of y(x) as x approaches 1, ?