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1st ODE > sum 1 check plz!

dxdt=ex+t;x(0)=−3 \frac {dx} {dt} = e^{x+t} ; x(0) = -3

So i did the usual mumbo jumbo by separating the variables and integrating both sides and got this :


−e−x=et+c -e^{-x} = e^t + c

Given that when x = 0 , t = -3 :

−e0=e−3+c -e^{0} = e^{-3} + c

So

c=−e0−e−3 c = -e^{0}- e^{-3}


c=−1−e−3 c = -1- e^{-3}


Now focussing on the above step, the software's working says that it is meant to be :

c=−1−e3 c = -1- e^3


Note the e+3 e^{+3} bit. How?

EDIT : Has it used the laws of logs to say that −e−x=ex -e^{-x} = e^x



Thanks!
Reply 1
Avatar for TheBigRC
TheBigRC
8
wizz_kid
dxdt=ex+t;x(0)=−3 \frac {dx} {dt} = e^{x+t} ; x(0) = -3

So i did the usual mumbo jumbo by separating the variables and integrating both sides and got this :


−e−x=et+c -e^{-x} = e^t + c

Given that when x = 0 , t = -3 :

−e0=e−3+c -e^{0} = e^{-3} + c

So

c=−e0−e−3 c = -e^{0}- e^{-3}


c=−1−e−3 c = -1- e^{-3}


Now focussing on the above step, the software's working says that it is meant to be :

c=−1−e3 c = -1- e^3


Note the e+3 e^{+3} bit. How?

EDIT : Has it used the laws of logs to say that −e−x=ex -e^{-x} = e^x



Thanks!


You have said that x(0)=-3
so x=-3 when t=0
if you follow through with this in mind you get the correct solution.
Reply 2
Avatar for wizz_kid
wizz_kid
OP
16
TheBigRC
You have said that x(0)=-3
so x=-3 when t=0
if you follow through with this in mind you get the correct solution.



O yea i got it the otherway round didnt i? Thanks!

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