Diff eq problem appears simple but is deadly

1. Sep 30, 2010

ngm01

1. Given i. dx/dt + y = 0; x(0) = 1 : ii. dy/dt + x = 0; y(0) = 0 Solve for the solns of x and y

Although these are first order dif eq's I worked them as leading to a second order undetermined coefficients problem

First I wrote them as
i. x' + y = 0 and ii. y' + x = 0

Eqn. ii leads to y' = -x and this to y'' = -x' => -y" = x' substitute back in i. for -y" + y = 0

=> -y" + 0y' + y = 0 now to solve i in it's new form of a 2nd order eqn.
I began by writing as y" -0y'-y=0 and if std form is y''+ay'+by then a=0, b=-1 and a^2 = 0 4b = -4 => a^2 >4b and soln is form of x=C1e^-r1x + C2e^r2x the values for r1 and r2 are calculated from y" + 0y' + y = 0 or r^2 - 1 =0 so r = +/- 1. the soln then becomes x = C1e^x + C2e^-x. Now applying initial value conditions x(0) = 1 => 1 = C1 + C2

I then complete the other half of the problem similarly and end with 0 = C1 + C2

and I'm stuck. Can someone please point out where I'm going wrong?

thanks

2. Sep 30, 2010

Staff: Mentor

You're on the right track but the clump of stuff above is kind of a mess. You're solving y'' - y = 0, so the solution would be y = c1et + c2e-t.

x and y are dependent variables and t is the independent variable.

Edit: Misread the initial condition for y.
Use your initial condition y(0) = 0 to get an equation involving c1 and c2.

After doing that, solve for x, which you show as being equal to -y'. When you get x, use the initial condition x(0) = 1 to get another equation involving c1 and c2.

Last edited: Sep 30, 2010
3. Sep 30, 2010

JThompson

My first reaction to this part

"the soln then becomes x = C1e^x + C2e^-x. Now applying initial value conditions x(0) = 1 => 1 = C1 + C2"

is that you were solving from y, but switched to x in the end. Also your independent variable should be t (since the differential is dy/dt). It looked good up to here. You should have

y=C1e^t + C2e^-t , with y(0)=0

4. Sep 30, 2010

ngm01

Hi sorry for the text jumble...I've managed to pdf and attach my working. Can ayone figure out where I'm going wrong?. Thanks

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• dydx problem.pdf
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5. Sep 30, 2010

HallsofIvy

You have correctly determined that $y= C_1e^t+ C_2e^{-t}$. From the initial condition, $y(0)= C_1+ C_2= 0$. You then go back and solve the corresponding differential equation for x and get the same thing, $x= C_1e^t+ C_2e^{-t}$, for which the initial condition gives $C_1+ C_2= 1$ and, obviously, those cannot both be true. The problem is that you have no reason to think that the "$C_1$" and "$C_2$" constants in the two different solution are the same! But if you write $y= C_1e^t+ C_2e^{-t}$ and $x= D_1e^t+ D_2e^{-t}$ you have four unknown constants and only two conditions to determine them!

Answer: once you have $y= C_1e^{t}+ C_2e^{-t}$, do not solve another differential equation for x. Instead use the second equation, dy/dt+ x= 0 to find $x= -dy/dt= C_1e^{t}- C_2e^t$.

Now you have $y(0)= C_1+ C_2= 0$ and $x(0)= C_1- C_2= 1$ to solve for $C_1$ and $C_2$.

6. Sep 30, 2010

Staff: Mentor

You have y = c1et + c2e-t, and with the initial condition, you get 0 = c1 + c2.

Since y' = - x or equivalently, x = -y', differentiate the equation you have for y and take its negative to get your equation for x. Substitute in your initial condition x(0) = 1 to get a second equation for the constants c1 and c2.

7. Sep 30, 2010

ngm01

Thanks for the responses...as a follow up to HallsofIvy - the crux of the solution was deciding how to proceed after solving the first diff eq. - I'm still not clear on what led you to determine that solving the second diff eq for x was not the right way to proceed. Would you have made the same decision w/o knowing that the solution of the dx/dt eqn led to a impossible answer?. In other words would you have to work the problem to a dead end before realizing that another avenue was needed?. Cheers

8. Sep 30, 2010

JThompson

The primary reason for not solving the second differential equation is that it's unnecessary, and trying to do so will create more work/problems. The purpose of doing so would be to find "x=something", but in the problem statement you are given x=-dy/dt. You already know y, so (in general) differentiating it isn't much work.

9. Sep 30, 2010

ngm01

Aha!... I never thought to look at it in that context... just saw the eqn. and started solving "a mile a minute"...thanks for your insight!... Cheers