# Wrong Solution Of Some Differential Equations

Started by
acid111
, Apr 22 2007 07:29 PM

5 replies to this topic

### #1

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

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

Posted 23 April 2007 - 11:01 AM

Hello, and welcome to this forumhere 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 ?

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

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 !

i haven's seen it, hmm ...

ok, now i think that i can trust the solutions of the classpad...

thank you !

### #4

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

Posted 24 April 2007 - 06:49 AM

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

### #6 Guest_Logicman112_*

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)

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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