Question

Homogeneous equation:

(a/b)y'(t)+y(t) = x(t)
the answer is y(t) = Ae^(-b/a)t

Huh?

Answer 1

i know its a first order differential

Answer 2

but where does the "A" and minus sign come from?

Answer 3

The equation in the general form in which you've written it is inhomogeneous. The homogeneous version of that equation is x(t) = 0. I.e.,

(a/b)y'(t)+y(t) = 0

Now this equation is separable. Separate and solve.

Answer 4

I.e.,
dy/dx = -b/a . y
dy/y = -b/a dx

=> ln y = -(b/a)x + C

You take it from here.

Answer 5

jamesj... i dont get this step:

dy/dx = -b/a . y

Answer 6

(a/b) y' + y = 0
(a/b) y' = -y
y' = -(b/a)y
dy/dx = -(b/a)y

Answer 7

Hence
\[ \int \frac{dy}{y} = \int -(b/a) dx \ \ \ \implies \ln y = -(b/a)x + C \]

Answer 8

oke... y = e^-(b/2) + C

but the "A" where does that one come from

Answer 9

\[ \implies y = \exp( -(b/a)x + C ) \]
Hence y = ...

Answer 10

\[\frac{a}{b}y'+y=t\]

now I want the coefficient of y' to be 1
so I will multiply both sides by b/a

\[y'+\frac{b}{a}y=t \frac{b}{a}\]

now i will multiply both sides by v

\[vy'+\frac{b}{a} v y=t v \frac{b}{a}\]

I want to choose this v such that it satisifies the following equation

\[v'=\frac{b}{a}v\]

I want to choose it this way so that i can write

\[vy'+v'y=t v \frac{b}{a}\]

so then I can write

\[(vy)'=t v \frac{b}{a}\]

so what is this v well we need to solve
\[v'=\frac{b}{a}v\]
we can do this by separation of variables

\[\frac{dv}{dt} =\frac{b}{a} v\]

\[\frac{1}{v} dv =\frac{b}{a} dt\]

now integrate both sides

\[\ln|v|=\frac{b}{a} t+k \text{ let k=0 }\]

let v>0 so we can just write

\[\ln(v)=\frac{b}{a}t\]

\[v=e^{\frac{b}{a}t}\]

so we have

\[(e^{ \frac{b}{a}t}y)'=t e^{ \frac{b}{a} t}\frac{b}{a}\]

now integrating both sides gives

\[e^{ t \frac{b}{a}} y+c_1=\int\limits_{}^{}\frac{b}{a} t e^{\frac{b}{a} t} dt\]

so now we have to figure out how to do

\[\frac{b}{a}\int\limits_{}^{}t e^{\frac{ b}{a} t} dt =\frac{b}{a} [ t \frac{a}{b}e^{\frac{b}{a} t} -\int\limits_{}^{}\frac{a}{b} e^{\frac{ b}{a} t} dt]\]
\[=te^{\frac{b}{a} t} -\frac{a}{b}e^{\frac{b}{a} t} +c_2\]

so we have
\[e^{ t \frac{b}{a}} y+c_1=te^{\frac{b}{a} t} -\frac{a}{b}e^{\frac{b}{a} t} +c_2\]


now we only need to really write one constant since constants are closed under addition

so we have

\[e^{t \frac{b}{a} } y=te^{ \frac{b}{a} t}-\frac{a}{b} e^{\frac{b}{a}t}+C\]

Answer 11

did i make a mistake somewhere?

Answer 12

Your equation
y = e^-(b/2) + C

is wrong.

\[ y = e^{-(b/a)x + C} = e^C.e^{-(b/a)x} = Ae^{-(b/a)x} \]
where A = e^C

Answer 13

no wait i think it looks good :)

Answer 14

@myin, you solved a different equation. The homogeneous equation is

(a/b) y' + y = 0.

Answer 15

owhh i forgot the (...)

Answer 16

oh

Answer 17

oops i wrote it down wrong altogether

Answer 18

jamesj...

the \[\int\limits_{}^{} -(b/a)dx\]

isnt it

\[-(b/a) (x + c)\]

Answer 19

thats right

Answer 20

you can also write it as -bx/a+c

Answer 21

It can be, in which case you're just scaling the constant c. It makes no difference to the final answer.

Answer 22

The important thing is
(a/b)y' + y = 0

is a linear equation. Hence if y1 is a solution of the equation, then Ay1 is also a solution for any constant A.

Answer 23

ahhh oke :)

Answer 24

Hence for our purposes we could have supposed c = 0. Then we would have found a (i.e., one) solution
\[ y = e^{-(b/a)x} \]

Then the general solution of the equation is
\[ y = Ae^{-(b/a)x} \]
for an arbitrary constant A.

Answer 25

It's very good practice though to not suppose c = 0, as a way to remind yourself there is a degree of freedom in the general solution. It is also essential when solving a number of inhomogeneous equations that the c stay in expressions.

Answer 26

you mean x(t)?

Answer 27

For example, in this equation is x(t) were not identically zero, then when working through the solution, you want to keep the c 'alive' in the algebra.

Answer 28

but there are other sorts of inhomogeneous equations of course.

Answer 29

oh i see what c you are talking about

Answer 30

what about the particular solution... for example..

(a/b)y'+y = 1

Answer 31

i love my way above

Answer 32

In that case the particular solution is yp(x) = 1.

Answer 33

could you explain why it is 1?

Answer 34

Substitute and see: it works.

Answer 35

get the coeificent of y' to be 1
and find integrating factor

Answer 36

but do what james said

Answer 37

i like to make things extra long

Answer 38

In fact, for any constant K, the particular solution of
(a/b)y' + y = K
is yp(x) = K

Answer 39

oke, thats clear, but what do you mean with substitute... what do i have to substitute?

Answer 40

K=1

Answer 41

owh wait... for an example..

if it was y'+y = x

yp = ax+b right?

Answer 42

then i have to get the first derivative

Answer 43

and substitute it in y'+y

Answer 44

I mean the function yp(x) = K. Substitute it into the equation and you see it indeed satisfies
(a/b)y' + y = K

Answer 45

For the method of solving

(a/b)y' + y = x(t)

in general, watch this:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-7-first-order-linear-with-constant-coefficients/

Answer 46

oke jamesj.. im gonna watch the video :) thanks <3

Answer 47

Actually, I now realize this is slightly too far along. Instead, start at lecture 3, here:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/

Answer 48

\[y'+\frac{b}{a}y=x \frac{b}{a}\]

\[vy'+v \frac{b}{a}y=v x \frac{b}{a}\]

we want to choose v (v>0) such that
\[v'=v \frac{b}{a} => \frac{dv}{dt}=v \frac{b}{a} => \frac{1}{v} dv=\frac{b}{a} dt \]
now integrating both sides gives

\[\ln(v)=\frac{b}{a} t +k \text{ let k=0} => \ln(v)=\frac{b}{a} t => v=e^{\frac{b}{a} t}\]

we chose this v this way so that we could write

vy'+v'y=(vy)'

so we have

\[(e^{\frac{ b}{a} t} y)'=e^{\frac{b}{a} t} x \frac{b}{a}\]

\[(e^{\frac{ b}{ a} t} y)'=\frac{b}{a} x e^{\frac{b}{a} t}\]

integrating both sides we get

\[e^{\frac{b}{a}t} y=\int\limits_{}^{}\frac{b}{a}xe^{\frac{b}{a} t} d t+C\]

\[y=\frac{b}{a e^{\frac{b}{a} t}} \int\limits_{}^{} x e^{\frac{b}{a} t} dt +C\]

i made a formula for the general case :)

Answer 49

this is what i meant to do way above but i wrote t instead of x

Answer 50

ahh okee... thnks myininaya :)

Answer 51

do you understand what i did?
is this consider to advance?

Answer 52

too*

Answer 53

@myin: carefully in your last expression, as the exponential multiplies both the integral and the constant.

Answer 54

oops

Answer 55

james is write the last expression i should have wrote was
\[y=\frac{b}{a e^{\frac{b}{a} t}} ( \int\limits\limits_{}^{} x e^{\frac{b}{a} t} dt +C )\]

Answer 56

its a bit complex... but i think i got it :)

Answer 57

above the constant wasn't being multiplied by b/a
so it doesn't matter the way that i wrote it since b/a *C is still a constant

Answer 58

but still the constant needs to be divided by that exponential guy

Answer 59

its hard for some to use the product rule backwards
and thats what i was trying to do above

Answer 60

fg'+f'g
=
(fg)'

Answer 61

yeahh.. indeed.. the product rule backwards is hard >.<

Answer 62

so what if i asked you to write this as (something)'

\[\frac{-1}{x^2} \cdot e^{-x^2}-2e^{-x^2}\]

do you think you could do it?

Answer 63

let me think

Answer 64

maybe that was a hard one to start off with
i should have chose an easier one like
xy'+y

Answer 65

oeehh this one is hard

(x^-1*e^(x^2))'

Answer 66

almost that x^2 should have negative in front it

Answer 67

but excellent job

Answer 68

yeaaaahh ^^ yeahh dammit i forgot that one -.-

Answer 69

that is useful for solving
things in this form
y'+p(x)y=q(x)
this is called a linear differential equation

Answer 70

some might require you to use the formula
but it is also good understand how the formula is derived

Answer 71

i mean they have a formula for y'+p(x)y=q(x)

Answer 72

i don't remember it i just derive it every single time

Answer 73

using the long process above

Answer 74

I forget the formula all the time, but one can rederive it in a minute.

Anyway, it's all in the lecture 3 I linked to earlier:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/

Answer 75

sometimes you can manipulate a nonlinear differential equation so you can use the process i have above

Answer 76

i like doing that

Answer 77

i have only done it once though lol