Thread: Verifying the solution of a differential equation.

I have the following differential equation >> dx/dt = -1/2 (x)
it says i need to verify that x(t)=Ce^-t/2

i have done:

rearrange >> 1/x dx = -1/2 dt
integrate >> ln x = -t/2 + C
take log to simplify >> ln x = ln e^-t/2 + ln C
>> ln x = ln (Ce^-t/2)
which finally gives >> x = Ce^-t/2

is this correct or am i doing it wrong??

It then goes on to say find the particular solution that satisfies x(0) = 2 and sketch on a direction field...so:

input the condition >> 2 = Ce^-0/2
simplify >> 2 = Ce^0
>> 2 = C(1)
therefore >> 2 = C
and the particular solution is >> x = 2e^-t/2

again is this correct? And how would i go about sketching this?

Thanks

Originally Posted by NFS1

I have the following differential equation >> dx/dt = -1/2 (x)
it says i need to verify that x(t)=Ce^-t/2

With a question like this, where you are asked to verify that a proposed solution is in fact a solution you just need to show that the given solution satisfies the ODE.

So differentiate to get:



which we see is only a solution if (that is the arbitary constant is in the wrong place)

The correct solution is:

CB

@CB:
Can you explain why wolphram alpha comes to the solution:
dx/dt=-x/2 - Wolfram|Alpha

Is that similar, or?

Originally Posted by CaptainBlack

With a question like this, where you are asked to verify that a proposed solution is in fact a solution you just need to show that the given solution satisfies the ODE.

So differentiate to get:



which we see is only a solution if (that is the arbitary constant is in the wrong place)

The correct solution is:

CB

The way i have done is incorrect then? I have looked through my uni notes and thought the way i have done it would be correct. Also when i have substituted the intial condition of x(0)=2 is that correct?

Originally Posted by NFS1

rearrange >> 1/x dx = -1/2 dt
integrate >> ln x = -t/2 + C
take log to simplify >> ln x = ln e^-t/2 + ln C

I would change C into (for example) K, because in this step you show that , which is not true, sou I would say:
where K is a new constant, in fact: .

Or you just can write:

Originally Posted by NFS1

I have the following differential equation >> dx/dt = -1/2 (x)
it says i need to verify that x(t)=Ce^-t/2

i have done:

rearrange >> 1/x dx = -1/2 dt
integrate >> ln x = -t/2 + C
take log to simplify >> ln x = ln e^-t/2 + ln C
>> ln x = ln (Ce^-t/2)
which finally gives >> x = Ce^-t/2

is this correct or am i doing it wrong??

It then goes on to say find the particular solution that satisfies x(0) = 2 and sketch on a direction field...so:

input the condition >> 2 = Ce^-0/2
simplify >> 2 = Ce^0
>> 2 = C(1)
therefore >> 2 = C
and the particular solution is >> x = 2e^-t/2

again is this correct? And how would i go about sketching this?

Thanks

First of all, don't forget that the integral of 1/x is ln|x|, the modulus is important.

Second, why not just exponentiate both sides to simplify?

Originally Posted by NFS1

The way i have done is incorrect then? I have looked through my uni notes and thought the way i have done it would be correct. Also when i have substituted the intial condition of x(0)=2 is that correct?

It is a question of technique, if asked to veryfy something you can just check that it is a solution, you do not need to solve the problem from scratch.

CB

Originally Posted by Prove It

First of all, don't forget that the integral of 1/x is ln|x|, the modulus is important.

Second, why not just exponentiate both sides to simplify?

Thankyou mate. Makes alot more sense. So am i writing in say the particular solution is x=2e^-t/2 if the initial condition is x(0)=2??

I thinkt that's correct.