The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

PC4 - Differential Equations - dx/dt = e^(x+t)

Hi, I'm stuck on this question, I can't seem to get to the final answer given in the book, any pointers would be much appreciated. Q: Find the general solution of this differential equation giving answer in explicit form: dx/dt = e^(x+t), answer given in the book is x = -ln |A - e^t| So far I've tried separating the variables so I get: integral e^-x dx = integral e^t dt, which i think gives -e^-x = e^t + A, and then solving for x gives: x = ln (e^t + A). Can anyone show me where i'm going wrong or how I can get to the correct answer? Thanks...
Original post by gsr929
Hi, I'm stuck on this question, I can't seem to get to the final answer given in the book, any pointers would be much appreciated. Q: Find the general solution of this differential equation giving answer in explicit form: dx/dt = e^(x+t), answer given in the book is x = -ln |A - e^t| So far I've tried separating the variables so I get: integral e^-x dx = integral e^t dt, which i think gives -e^-x = e^t + A, and then solving for x gives: x = ln (e^t + A). Can anyone show me where i'm going wrong or how I can get to the correct answer? Thanks...


You have to multiply through by -1 to get rid of the negative in front of e to the minus x :smile:

ln⁡(−1)\ln(-1)

isn't real, remember :smile:
Reply 2
Avatar for davros
davros
16
Original post by gsr929
integral e^-x dx = integral e^t dt, which i think gives -e^-x = e^t + A, and then solving for x gives: x = ln (e^t + A). Can anyone show me where i'm going wrong or how I can get to the correct answer? Thanks...


You seem to have dropped a crucial minus sign when you solved for x!
Reply 3
Avatar for Hasufel
Hasufel
16
somethimes it`s best with seperable equations to put them in the form:

∫f(x)dx+∫g(t)dt=c\int f(x)dx+\int g(t)dt=c or:

∫e−xdx−∫etdt=k\int e^{-x}dx-\int e^{t}dt=k

giving: −e−x−et=k-e^{-x}-e^{t}=k

OR:


e−x=A−ete^{-x}=A-e^{t} where A=-k
(edited 11 years ago)
Reply 4
Avatar for gsr929
gsr929
OP
Thanks everyone for your help - I think the problem was me not realising that i should use a different lettered constant (e.g. 'k') up to the point just after I've multiplied both sides by -1 to get the LHS from -e^-x to e^-x, and then I just let the RHS '-k' constant = A to get the required form of the answer. I was using the constant 'A' from straight after the integration, so then having to later say let -A = A to get the book form of the answer is what was confusing me. Thanks again :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you