      # Wrong Solution Of Some Differential Equations

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### #1 acid111

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Posted 22 April 2007 - 07:29 PM

Hi, i'm german and i hope you can understand all what i'm going to write ok, i tried to check a solution, that i calculated by hand, with the classpad
but the classpad showed me an other solution as i calculated.
But i know that my solution is the right one here is the differential equation:

y'' + 2y' - 8y = 3e^(2x)

the right solution is: y= C1*e^(2x) + C2*e^(-4x) + 0,5x * e^(2x)

but the classpad shows: y= C1*e^(2x) + C2*e^(-4x) + 0,5x * e^(2x) - 1/12 * e^(2x)

why does the classpad calculate a wrong solution ?
Can you try to calculate this equation with your classpad, maybe its only my classpad shall i better send this problem to casio ?

thx

### #2 PAP

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Posted 23 April 2007 - 11:01 AM

here is the differential equation: y'' + 2y' - 8y = 3e^(2x)
the right solution is: y= C1*e^(2x) + C2*e^(-4x) + 0,5x * e^(2x)
but the classpad shows: y= C1*e^(2x) + C2*e^(-4x) + 0,5x * e^(2x) - 1/12 * e^(2x)
why does the classpad calculate a wrong solution ?

Hello, and welcome to this forum The solution you get is not wrong; in fact, it is equivalent to your "by hand" solution:
C1*e^(2x) + C2*e^(-4x) + 0,5x*e^(2x) - 1/12*e^(2x) = (C1-1/12)*e^(2x)+C2*e^(-4x)+0.5xe^(2x)
Now, since C1 is an arbitrary constant, C1-1/12 is also arbitrary, so these two solutions are equivalent. ClassPad returns a slightly different solution probably due to the way it uses to solve the differential equation. I admit that the solution obtained "by hand" is more compact, but ClassPad's solution is correct as well. Btw, Mathematica's solution is identical to ClassPad's solution.

### #3 acid111

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Posted 23 April 2007 - 03:05 PM

yes, you are right !
i haven's seen it, hmm ... ok, now i think that i can trust the solutions of the classpad...  thank you !

### #4 The_AFX_Master

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Posted 23 April 2007 - 05:14 PM

I managed to solve the equation via Laplace transform on the OS 3, and i got the same result. then i guess that the calc uses laplace transforms to solve linear ODE's

### #5 PAP

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Posted 24 April 2007 - 06:49 AM

i guess that the calc uses laplace transforms to solve linear ODE's

Hmmm, not necessarily. The differential equation is linear to the dependent variable y, so it may use other methods to solve it analytically. Anyway, Laplace transforms can be used as well.

### #6 Guest_Logicman112_*

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Posted 23 February 2011 - 06:05 AM

If we have a linear system with the following differential equation:
d2y/dt2+7*dy/dt+12*y=dx/dt+5*x
So H(s) = (s+5)/((s+4)*(s+3)) = -1/(s+4)+2/(s+3)--> h(t) = (-e^(-4*t)+2*e^(-3*t))*u(t)
(Please calculate the unit impulse response by Laplace transform and verify the result by yourself)
Why this answer does not satisfy the differential equation?
The right answer is:
h(t) = (-2*e^(-4*t)+3*e^(-3*t))*u(t)

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