The solution of a certain differential equation is of the form y(t) = a e^(2 t) + b e^(5 t),
where a and b are constants. The solution has initial conditions y(0) = 3 and y'(0) = 5 .
Find the solution by using the initial conditions to get linear equations for a and b.
y(t)= ?

Question

The solution of a certain differential equation is of the form y(t) = a e^(2 t) + b e^(5 t),
where a and b are constants. The solution has initial conditions y(0) = 3 and y'(0) = 5 .
Find the solution by using the initial conditions to get linear equations for a and b. 
y(t)= ?

turingtest · Answer

you never replied to my earlier efforts to help you...
where are you stuck?

anonymous · Answer

I thought I had it figured out but the answer didn't match

anonymous · Answer

I don't know where I went wrong

turingtest · Answer

$$y(t)=Ae^{2t}+Be^{5t}$$and we also need $y'(t)$
what did you get for that?

turingtest · Answer

$$y'(t)=2Ae^{2t}+5Be^{5t}$$so what system results from subbing in $t=0$ ?

anonymous · Answer

Yup that's what I used, e^0 = 1 so 2A + 5B

turingtest · Answer

well first off what did you get for a and b ?

anonymous · Answer

Just A + B right?

turingtest · Answer

=what?

anonymous · Answer

For y(t), then y'(t) = 2A +5B

turingtest · Answer

not quite$$y(t)=Ae^{2t}+Be^{5t}$$it's important to make the distinction between that and the given value$$y(0)=A+B=3$$only by using the second part entirely can we solve the system

turingtest · Answer

by "the second part" I mean both the x and y values of the initial conditions

turingtest · Answer

so we also have that$$y'(0)=2A+5B=?$$

anonymous · Answer

so y(0)=A+B=3 and y'(0)= 2A+5B=5

turingtest · Answer

good, now you only have to solve the system

anonymous · Answer

This is where I went wrong

turingtest · Answer

what were your final answers?

anonymous · Answer

A=1 B=4 ... I think I'm confusing this

turingtest · Answer

no that's not right
I think you may need a system solving review

turingtest · Answer

let's use elimination:
try subtracting
$$y'(0)-2[y(0)]$$

turingtest · Answer

what is the resulting equation?

anonymous · Answer

a= 8/3 b= 3-(8/3) does this work?

turingtest · Answer

no, I think you got B turned negative somehow and messed up A...

anonymous · Answer

b=1/3

turingtest · Answer

actually b=-1/3

turingtest · Answer

I suppose that's where you went wrong

turingtest · Answer

...are you not seeing it?

anonymous · Answer

I'm using a=8/3 and b=-1/3 and its still not working

turingtest · Answer

that's because $$A\neq\frac83$$

turingtest · Answer

use the first equation and remember the signs
that's what's getting you, those darn minus signs

turingtest · Answer

$$y'(0)=2A+5B=5$$$$-2y(0)=-2A-2B=-6$$so$$y'(0)-2y(0)\implies3B=-1\implies B=-\frac13$$now use the formula for $y(0)$ again...

turingtest · Answer

$$y(0)=A+B=3\implies A=3-B$$

anonymous · Answer

And isn't 3-(1/3) = 8/3

turingtest · Answer

but$$B\neq\frac13$$watch your signs!!!

anonymous · Answer

Ohhhh! My bad lol, so it's 10/3

turingtest · Answer

yeah :) and if these things are allowed you could have just solved the system using wolfram http://www.wolframalpha.com/input/?i=sovle%20x%2By%3D3%2C2x%2B5y%3D5&t=crmtb01 that becomes reasonable with larger systems

anonymous · Answer

That's where I went wrong :/ Thanks for your patience :-)

anonymous · Answer

Ahh thanks for the link, it'll be useful!

turingtest · Answer

...but you should know how to do a 2x2 system like this
you're welcome, anytime :)