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DIFFERENTIAL EQUATIONS- WHY IS THE CONSTANT +ln C??

Can someone please explain to me why on some differential equations instead of the constant being +C, it is +InC?

thanks

Reply 1

visesh

They are both constants. Using lnC makes manupilation easier.

Reply 2

melbourne

Ah i see

Another point, i was trying to do this question from my C4 book

EX 6J Q1a

Find the general solutions of the following differential equations. Leave you answer in the form of y=f(x)

dy/dx = (1+y)(1-2x)

I cant do this and the answer in the book is y=Ae^(x-x^2) -1

How on earth they got that i dont know!!!!

Reply 3

visesh

What bit are you stuck on?

Have you separated the variables and formed:
INT(1/(1+y)) dy = INT (1-2x) dx

solve that, and let the constant on the LHS be lnA

Reply 4

yazan_l

dy/dx = (1+y)(1-2x)
1/(1+y) dy = 1-2x dx
int 1/(1+y) dy = int 1-2x dx
ln|1+y| = x-x² + c
1+y = e^(x-x² + c)
1 + y = e^c . e^(x-x²)
e^c = A
y = A e^(x-x²)-1

Reply 5

m:)ckel

dy/dx = (1+y)(1-2x)
INT [1/(1+y)] dy = INT (1-2x) dx

ln(1+y) = x - x^2 + c

1+y = e^(x-x^2).e^c

Call e^c, A:

y = Ae^(x-x^2) - 1

{oops}

Reply 6

melbourne

mockel

dy/dx = (1+y)(1-2x)
INT [1/(1+y)] dy = INT (1-2x) dx

ln(1+y) = x - x^2 + c

1+y = e^(x-x^2).e^c

Call e^c, A:

y = Ae^(x-x^2) - 1

{oops}

oh thanks guys!!!

Reply 7

melbourne

so whats the rule of thumb for when to use C and when to use ln C for the constant? is it just if u have ln anyway?

Reply 8

El Stevo

if your lhs integral is a ln, then your constant is a ln...

Reply 9

melbourne

El Stevo

if your lhs integral is a ln, then your constant is a ln...

yeh dats what i was trying to say hehe

thanks el stevo

Reply 10

yazan_l

first put it C, if you thought that the side where C contains lnx for example, and u think that you can join lnx and C together , remove C and put lnC
example:
int dy = int (2x+1)/(x²+x+1) dx
this will be
y = ln|x²+x+1| + K
you can see here that you can simplify this more by removing K and putting lnC where K=lnC
so it's y = ln|x²+x+1| + lnC
you know that lna + lnb = ln(ab)
so y = ln|x²+x+1| + lnC
becomes : y = ln|C(x²+x+1)|

now in our case:

dy/dx = (1+y)(1-2x)
1/(1+y) dy = 1-2x dx
int 1/(1+y) dy = int 1-2x dx
ln|1+y| = x-x² + c
1+y = e^(x-x² + c)
1 + y = e^c . e^(x-x²)
e^c = A
y = A e^(x-x²)-1

the side where C is , doesnt have ln, so it isnt worth it to make it lnK. ( although if u did that , you will get the same answer! as lnK, C, A ... all are constants )

Reply 11

melbourne

how would i go about ...

dy/dx = 2e^(x-y) ?

Reply 12

dvs

Just use:
e^(x-y) = e^x/e^y

Reply 13

melbourne

dvs

Just use:
e^(x-y) = e^x/e^y

hehe dats what i thought

Reply 14

melbourne

hehe dats what i thought

nah im lost, ive got:

2 dy/dx = e^x / e^y

e^y dy/dx = e^x /2

I e^y dy = I e^x /2 dx

e^y =1/2 e^x + C

:frown:

Reply 15

yazan_l

nah im lost, ive got:

2 dy/dx = e^x / e^y

e^y dy/dx = e^x /2

I e^y dy = I e^x /2 dx

e^y =1/2 e^x + C

that's right
if u want it y=f(x) just take ln of both sides:

e^y =1/2 e^x + C
y = ln|1/2 e^x + C|

if it's not the same as the answer of the book maybe it's becase u moved the "2" from the y side to the x side, the book might had done it this way:

2 dy/dx = e^x / e^y
I 2e^y dy = I e^x dx
2e^y = e^x + C
y + ln2 = ln|e^x + C|
y = ln|e^x + C| - ln2
y = ln|(e^x + C)/2|
y = ln|1/2 e^x + C/2|
y = ln|1/2 e^x + K|
which is the same as the first one