# Differential equations stuck trying to integrate 10^-u du (1 Viewer)

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#### dooogle

1. The problem statement, all variables and given/known data

trying to find the integral curve for ths equation get stuck at trying to integrate 10^-u
dy/dx=10^x+y

2. Relevant equations

3. The attempt at a solution

let u=x+y
so dy/dx=10^u
du/dx=1+dy/dx
=(10^u)-1=du/dx
du/dx=(10^u)-1
du/10^u=-dx

thanks in advance

Last edited:

#### Mark44

Mentor
1. The problem statement, all variables and given/known data

trying to find the integral curve for ths equation get stuck at trying to integrate 10^-u
dy/dx=10^x+y
What is the differential equation you're trying to solve?
When you write 10^x + y, I read that as (10^x) + y. Did you mean 10^(x + y)?
2. Relevant equations

3. The attempt at a solution

let u=x+y
so dy/dx=10^u
How does the line above follow from the previous line? It would be helpful if you included the differential equation you started from, which presumably does not include u.
du/dx=1+dy/dx
=(10^u)-1=du/dx
du/dx=(10^u)-1
du/10^u=-dx

thanks in advance

#### dooogle

hi the equation i started with was dy/dx=10^(x + y)

dy/dx=10^u
follows
u=x+y
since dy/dx=10^(x + y)

so replacing (x+y) with u gives 10^u

cheers

#### Mark44

Mentor
This equation is separable, so there is no need for a substitution.
10^(x + y) = 10^x * 10^y. After separation, the equation becomes
10^(-y)dy = 10^x dx

Both exponential expressions can be converted to ones with e raised to a power using this identity: a^b = (e^(ln a))^b = e^(b ln a)

#### dooogle

thanks for the help i have seperated the equation as stated getting:

e^-yln 10
= e^xln 10

which when i integrate gives a final solution of

10^y=10^x+c

which leads to y=x+c1 where c1=log10 c

does this sound ok to you

thanks very much for your help

#### Mark44

Mentor
thanks for the help i have seperated the equation as stated getting:

e^-yln 10
= e^xln 10
Please use parentheses! The left side should be written as e^(-y ln10), and similarly for the right side.
which when i integrate gives a final solution of

10^y=10^x+c
This isn't what I get.
which leads to y=x+c1 where c1=log10 c
Even if the equation 10^y = 10^x + c were correct, it doesn't result in y = x + c1.
does this sound ok to you

thanks very much for your help

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