Question

Differential Equation: y'=10^(x+y)

Answer 1

\[y=-\frac{ \ln(c-10^x) }{\ln10 }\] that's the answer

Answer 2

@paki ?

Answer 3

hmm let's try

Answer 4

\[y'=10^{x+y}=10^x 10^y \Longrightarrow \frac{dy}{dx}=10^x10^y\]
separation method leads to:\[\int \frac{dy}{10^y}=\int 10^xdx\]

Answer 5

try that!

Answer 6

I did!

Answer 7

it is best if you write is : \[\int 10^{-y}dy=\int 10^xdx\]

Answer 8

hmm so what's the problem then?

Answer 9

that I'm not smart enough to see what you see....

Answer 10

not our professor nor the TA speak much the language tbh

Answer 11

well i'm not smart either, i don't see anything at this point

Answer 12

so it's almost sign language

Answer 13

from here it's 10^-y/ln10 right?

Answer 14

hmm i don't memorize the antiderivatives
so my way would be writing that as natural exp which i know how to handle

Answer 15

the same goes for the right hand side

Answer 16

you mean lne^10?

Answer 17

i meant \[10^{-y}=e^{\ln (10^{-y})}=e^{-y\ln10}\]

Answer 18

now i can do it easily with substitution

Answer 19

that's change of base

Answer 20

right!

Answer 21

\[e^\ln(10^-x)=e^-\ln(10^x)\]

Answer 22

so we would have \[\int e^{-y\ln10}dy=\int e^{x\ln10}x\]

Answer 23

that suppoed to be dx not x

Answer 24

times x?

Answer 25

no dx

Answer 26

oh ok

Answer 27

gotta go keep trying :)

Answer 28

you should get that answer

Answer 29

wait shouldn't I first take off the exponent?

Answer 30

@xapproachesinfinity ?

Answer 31

@freckles ?

Answer 32

am I right?

Answer 33

can I just take the exponent off?

Answer 34

@zepdrix ?

Answer 35

Please I'm trying this problem since yesterday

Answer 36

\[\int\limits_{}^{} a^x dx=\frac{a^x}{\ln(a)}+C ,a \in (0,1) \cup (1,\infty)\]

Answer 37

\[\int\limits_{}^{}10^{x}dx=\frac{10^x}{\ln(10)}+C\]

Answer 38

\[\int\limits_{}^{}10^{-y} dy \\ \text{ you could sub the } -y=u \\ -dy=du\]

Answer 39

but here it's \[\int\limits e^(-yln10)\]

Answer 40

sorry @freckles I'm not getting it

Answer 41

@IrishBoy123 what am I missing here?

Answer 42

can you integrate
e^x dx ?

you get back e^x

what about integral e^(ax) dx where a is a constant ?

Answer 43

1/a

Answer 44

your problem is
\[ \int e^{ax} \ dx\]
except you are using y instead of x, and a= -ln(10)

Answer 45

but it's ln, not e.......

Answer 46

sorry I just don't follow the solutions, did you continue from where @xo_kansasprincess_xo finished?

Answer 47

@phi

Answer 48

@freckles I swear I don't get it

Answer 49

and I ever thought about a problem for 2 days in a row

Answer 50

never*

Answer 51

I assume you can get to here
\[ \int e^{-y\ln10}dy=\int e^{x\ln10} dx \]
which we can write
\[ \int e^{(-\ln10) y}dy=\int e^{(\ln10)\ x} dx \]

Answer 52

ln(10) is just a number (a constant) and so is -ln(10)

Answer 53

\[ \text{ Let } y=a^x \\ \ln(y)=\ln(a^x) \\ \ln(y)=x \ln(a) \\ \frac{y’}{y}=\ln(a) \\ y’=y \ln(a) \\ \text{ recall } y=a^x \\ \text{ so we have} y’=a^x \ln(a) \\ \frac{dy}{dx}=a^x \ln(a) \\ dy=a^x \ln(a) dx \\ \text{ integrating both sides we have } \\ y+C=\int a^x \ln(a) dx \\ \text{ now } ln(a) \text{ is a constant multiple of that integral so we can bring it outside the integral } \\ y+C=\ln(a) \int a^x dx \\ \text{ now divide both sides by} \ln(a) \\ \frac{y+C}{\ln(a)}= \int a^x dx \\ \text{ recall again } y=a^x \\ \frac{a^x+C}{\ln(a)}=\int a^x dx \\ \frac{a^x}{\ln(a)}+\frac{C}{\ln(a)}=\int a^x dx \\ \frac{a^x}{\ln(a)}+K=\int a^x dx \\ \text{ where } K \text{ represents some contant } \]

Answer 54

if you can integrate e^ax dx then you can do both sides of your problem

Answer 55

\[\int\limits 10^{-y}dy=\int\limits 10^xdx \]
you can apply that formula I wrote on both sides of here

or you can rewrite it and use a different method in which @phi is talking of

Answer 56

but either way you will still get the same result on both sides

Answer 57

still I have e at the bottom

Answer 58

I don't see an e

Answer 59

unless you choose to write it that way

Answer 60

if you do the integral

Answer 61

after integrating , you can change the e^-ln10 y back to 10^-y

Answer 62

here is example
\[\int\limits_{}^{}5^{-v} dv \\ \text{ \let } u=-v \\ du=-dv \\ -du=dv \\ \int\limits 5^{u} (-du)=- \int\limits 5^u du =-\frac{5^u}{\ln(5)}+C=-\frac{5^{-v}}{\ln(5)}+C\]

Answer 63

I know that @freckles ..........

Answer 64

then what is the problem

Answer 65

oh!

Answer 66

\[y'=10^{x+y} \\ \frac{dy}{dx}=10^x 10^y \\ \frac{dy}{10^{y}}=10^x dx \\ 10^{-y} dy=10^x dx \\ \text{ integrate both sides }\]

Answer 67

and the 10? how do I expres the equation with y then?

Answer 68

I get then \[-10^y=10^x+\ln10c\]

Answer 69

just C, because a constant (ln10) times an arbitrary constant is still just an arbitrary constant.

Answer 70

and the exponent should be -y (not +y)

Answer 71

oh so 10^-y?

Answer 72

-10^-y = 10^x + C

Answer 73

you can also do it that other way...
\[\int\limits_{}^{}5^{-v} dv=\int\limits e^{\ln(5^{-v})} dv \text{ since} \ln(x) \text{ and } e^x \text{ are inverses } \\ =\int\limits e^{-v \ln(5)} dv \text{ by power rule for } \log \\ \text{ then do a sub } \\ \text{ \let } u=-\ln(5) v \\ du=-\ln(5) dv \\ \frac{-1}{\ln(5)} du =dv \\ \text{ so we have } \int\limits e^{-v \ln(5)} dv=\int\limits e^{u} \frac{-1}{\ln(5)} du=\frac{-1}{\ln(5)}e^u+C \\ =\frac{-1}{\ln(5)}e^{-\ln(5) v}+C\]

\[=\frac{-1}{\ln(5)}e^{\ln(5^{-v})} +C \text{ by power rule again } \\ =\frac{-1}{\ln(5)}5^{-v}+C \text{ since again } \ln(x) \text{ and } e^x \text{ are inverse functions }\]
just wanted to show you the other way in which was spoke here to integrate that one example

Answer 74

oh ok

Answer 75

and i know you are past that point
but I wanted you to see both ways
so you can pick your favorite

Answer 76

thanks! but I didn't understand how to keep on from here

Answer 77

-10^-y = 10^x + C

How about write it as
10^-y = C- 10^x
now take the ln of both sides

Answer 78

in other words, multiply both sides by -1
notice that -C can be written as C

Answer 79

\[y=-\ln(lnC-10^x)\]

Answer 80

I know C is constant though I try to show it as much as similar to the answer

Answer 81

10^-y = C- 10^x

ln (10^-y ) = ln ( C- 10^x )

Answer 82

can you continue?

Answer 83

I'm an idiot, that's ll I have to say

Answer 84

of course

Answer 85

it's the middle of the night here, and other excuses

Answer 86

thank you all so much!!!!!!!!!!!!!!!!!!!!!!!!!!11

Answer 87

you were both very helpful, I'll give @freckles 'cause he's got less than you @phi but really thank you so much! I wish I could be helpful to you someday...

Answer 88

good night fellas!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!111

Answer 89

Well I will become phi's fan since I already gave a medal to x approaches infinity guy.

Answer 90

you helped me for such a long time, I don't take it for granted, you've brightened my day