Thread: Verifying the solution of a differential equation.

1. 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

2. Re: General solution of differential equation

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:

$x'(t)=-\frac{1}{2}Ce^{-t/2}$

which we see is only a solution if $C=1$ (that is the arbitary constant is in the wrong place)

The correct solution is: $x(t)=C+e^{-t/2}$

CB

3. Re: General solution of differential equation

@CB:
Can you explain why wolphram alpha comes to the solution:
dx&#47;dt&#61;-x&#47;2 - Wolfram|Alpha

Is that similar, or?

4. Re: General solution of differential equation

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:

$x'(t)=-\frac{1}{2}Ce^{-t/2}$

which we see is only a solution if $C=1$ (that is the arbitary constant is in the wrong place)

The correct solution is: $x(t)=C+e^{-t/2}$

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?

5. Re: General solution of differential equation

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 $\ln(C)=C \Leftrightarrow e^{C}=C$, which is not true, sou I would say:
$\ln(x)=\ln(e^{\frac{-t}{2}})+\ln(K)$ where K is a new constant, in fact: $K=e^{C}$.

Or you just can write:
$\ln(x)=\ln(e^{\frac{-t}{2}})+\ln(e^{C})$

6. Re: General solution of differential equation

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?

\displaystyle \begin{align*} \ln{|x|} &= -\frac{t}{2} + c \\ |x| &= e^{-\frac{t}{2} + c} \\ |x| &= e^ce^{-\frac{t}{2}} \\ x &= \pm e^ce^{-\frac{t}{2}} \\ x &= Ce^{-\frac{t}{2}}\textrm{ where }C = \pm e^c\end{align*}

7. Re: General solution of differential equation

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

8. Re: General solution of differential equation

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?

\displaystyle \begin{align*} \ln{|x|} &= -\frac{t}{2} + c \\ |x| &= e^{-\frac{t}{2} + c} \\ |x| &= e^ce^{-\frac{t}{2}} \\ x &= \pm e^ce^{-\frac{t}{2}} \\ x &= Ce^{-\frac{t}{2}}\textrm{ where }C = \pm e^c\end{align*}
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??

9. Re: General solution of differential equation

I thinkt that's correct.