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    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...
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    (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

    \ln(-1)

    isn't real, remember
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    (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!
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    somethimes it`s best with seperable equations to put them in the form:

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

    \int e^{-x}dx-\int e^{t}dt=k

    giving: -e^{-x}-e^{t}=k

    OR:


    e^{-x}=A-e^{t} where A=-k
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    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
 
 
 
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