Diff EQ w/ I.C. anyone see where i f'ed it up?

[mr_coffee]

Its me again! I'm just having a full day of Diff EQ! 90% of the time I'm wrong but I'm learning slowly! but here is another problem:
The directions say: Find the particular solution of the differential equation
\frac{dy}{dx} + 5 y = 7
satisfying the initial condition y(0)=0.
Answer: y = ?

I submited this answer, and it was wrong:
(e^x*e^(ln(7^(-1/5)))^-5)/5 - 7/5Here is my work:

 

[quantumdude]

The only mistake I see is that in the next to last step you divided the left side by -5 while dividing the right side by +5. Fix that and your answer should be correct.

 

[mr_coffee]

Thanks tom, but i submitted:
-(e^x*e^(ln(7^(-1/5)))^-5)/5 +7/5

and it still said it was wrong :/

 

[Integral]

$e^C$ is a constant. There is no need to solve for C in the manner you did. Just set $e^C = C_1 = C$ (The last since the name of the constant is immaterial). Now see what you get.

 

[HallsofIvy]

And why in the world are writing such a complicated thing as
(e^(ln(7^(-1/5)))^(-5)? That's just 1/7.

Back where you had "(-1/5)ln(-5y+ 7)= x+ C", I would have been inclined to write ln(-5y+ 7)= -5x+ C' (C' = -5C) then -5y+ 7= C"e^(-5x)
(C"= e^C') and finally y= C"' e^(-5x)- 7/5 (C"'= C/(-5)). (In fact, I would probably have just written C for each of C, C', C", C"' reminding myself that they different values of C.)

Now, if y(0)= Ce^(-5(0))- 7/5= C- 7/5= 0, then C= 7/5 and
y(x)= (7/5)(e^(-5x)- 1).

You said it still said it was wrong. Are you submitting this to an automatic checker? Those things are notoriously hardnosed about the answer being in exactly the right form.

 

[neurocomp2003]

your first answer may not have been wrong..if you were using a computerized answering system they may have been looking for the simplest form of the solution...
-(e^x*e^(ln(7^(-1/5)))^-5)/5 +7/5 simplify that. which you shouhld be able to do.

interesting that you solved for C before solving for the general why solution =]