Solution of IVP

[ChuckNoise]

Hi

I am struggeling to wrap my head around the following attached example.
Why is the solution to equation 2
c₂e^t+c₁²/3(e^t-e^-2t)
and not simply
c₂e^t+(c₁e^-t)² ??

Thanks

 

[yoscar04]

You forgot to write down your IVP.

 

[ChuckNoise]

You forgot to write down your IVP.

It is an example from my textbook. A picture should be atached.
It's not the whole example it just starts out by stating the solution and i can't wrap my head around the solution

 

[HallsofIvy]

You say "consider the non-linear system (1)" but don't show what that is!

Is it \(\displaystyle \frac{d\begin{bmatrix}x_1 \\ x_2\end{bmatrix}}{dt}= \begin{bmatrix}-x_1 \\ x_1^2+ x_2 \end{bmatrix}\)?

If so then write it as two separate differential equations:
\(\displaystyle \frac{dx_1}{dt}= -x_1\) and
\(\displaystyle \frac{dx_2}{dt}= x_1^2+ x^2\).

The first can be written as \(\displaystyle \frac{dx_1}{x_1}= -dt\) and integrated:
\(\displaystyle ln(x_1)= -t+ C\) for some constant C. Taking the exponential of both sides, \(\displaystyle x_1= C'e^{-t}$ where $C_1= e^C \) is another constant.

Then the second equation can be written \(\displaystyle \frac{dx_2}{t}= x_1^2+ x_2= C'^2e^{-2t}+ x_2\). Fortunately that is now linear and we can write it as \(\displaystyle \frac{dx_2}{dt}- x_2= C'^2e^{-2t} \). The "associated homogeneous equation" is \(\displaystyle \frac{dx_2}{dt}- x_2= 0\) which has "characteristic equation" $r- 1= 0$ and "characteristc value $r= 1$. The general solution to the associated homogeneous equation is \(\displaystyle x_2= De^{t}\). To find a solution to the entire equation, try something of the form \(\displaystyle x_2= Ae^{-2t}\). Then \(\displaystyle \frac{dy}{dt}= -2Ae^{-2t}\) so the equation becomes \(\displaystyle -2Ae^{-2t}+ Ae^{-2t}= -Ae^{-2t}= C'^2e^{-2t}\) so \(\displaystyle A= -C'^2/2\).​

​

That gives \(\displaystyle x_2= De^t+ (C'^2/2)e^{-2t}\).​

​

So we have \(\displaystyle x_1= C'e^{-t}\) and \(\displaystyle x_2= De^t+ (C'^2/2)e^{-2t}\). The initial condition \(\displaystyle x_1(0)= C'= c_1\) and \(\displaystyle x_2= D+ C'^2/2= D+ c_1^2/2= c_2\) so \(\displaystyle D= c_2- c_1^2/2 \).​

 

[ChuckNoise]

You say "consider the non-linear system (1)" but don't show what that is!

Is it \(\displaystyle \frac{d\begin{bmatrix}x_1 \\ x_2\end{bmatrix}}{dt}= \begin{bmatrix}-x_1 \\ x_1^2+ x_2 \end{bmatrix}\)?

If so then write it as two separate differential equations:
\(\displaystyle \frac{dx_1}{dt}= -x_1\) and
\(\displaystyle \frac{dx_2}{dt}= x_1^2+ x^2\).

The first can be written as \(\displaystyle \frac{dx_1}{x_1}= -dt\) and integrated:
\(\displaystyle ln(x_1)= -t+ C\) for some constant C. Taking the exponential of both sides, \(\displaystyle x_1= C'e^{-t}$ where $C_1= e^C \) is another constant.

Then the second equation can be written \(\displaystyle \frac{dx_2}{t}= x_1^2+ x_2= C'^2e^{-2t}+ x_2\). Fortunately that is now linear and we can write it as \(\displaystyle \frac{dx_2}{dt}- x_2= C'^2e^{-2t} \). The "associated homogeneous equation" is \(\displaystyle \frac{dx_2}{dt}- x_2= 0\) which has "characteristic equation" $r- 1= 0$ and "characteristc value $r= 1$. The general solution to the associated homogeneous equation is \(\displaystyle x_2= De^{t}\). To find a solution to the entire equation, try something of the form \(\displaystyle x_2= Ae^{-2t}\). Then \(\displaystyle \frac{dy}{dt}= -2Ae^{-2t}\) so the equation becomes \(\displaystyle -2Ae^{-2t}+ Ae^{-2t}= -Ae^{-2t}= C'^2e^{-2t}\) so \(\displaystyle A= -C'^2/2\).​

​

That gives \(\displaystyle x_2= De^t+ (C'^2/2)e^{-2t}\).​

​

So we have \(\displaystyle x_1= C'e^{-t}\) and \(\displaystyle x_2= De^t+ (C'^2/2)e^{-2t}\). The initial condition \(\displaystyle x_1(0)= C'= c_1\) and \(\displaystyle x_2= D+ C'^2/2= D+ c_1^2/2= c_2\) so \(\displaystyle D= c_2- c_1^2/2 \).​

Ohh sorry! I get it now! The non-linear system (1) is simply x'=f(x). I attached an image with it.

But i still don't get how the solution becomes c₂e^-t+c₁²/3(e^t-e^-2t)

But to be honest i got a pretty hard time deciphering what you wrote

*EDIT* As soon as your post got into the "Reply" text it suddently looked different. I'll have another look at it now

*EDIT**EDIT* Nope! Still don't see it! How is the solution you wrote equal to c₂e^-t+c₁²/3(e^t-e^-2t) ?? I just can't see it

 

[HallsofIvy]

The point is that since the first equation, \(\displaystyle \frac{dx_1}{dt}= -x_1\) does not involve \(\displaystyle x_2\), we can solve it immediately- \(\displaystyle x_1= Ce^{-t}\). Then we can replace \(\displaystyle x_1^2\) in the second equation by \(\displaystyle C^2e^{-2t}\) so it is no longer non-linear!

\(\displaystyle \frac{dx_2}{dt}= x_1^2+ x_2= C^2e^{-2t}+ x_2\). That is equivalent to the linear, non-homogeneous, equation, \(\displaystyle \frac{dx_2}{dt}- x_2= C^2e^{-2t}\). The "associated homogeneous equation" is \(\displaystyle \frac{dx_2}{dt}- x_2= 0\) so \(\displaystyle \frac{dx_2}{dt}= x_2\) and the general solution to the associated homogeneous equation is \(\displaystyle x_2= De^t\) for D any constant.

We look for a solution to the entire equation of the for \(\displaystyle x_2= Ae^{-2t}\). Then \(\displaystyle \frac{dx_2}{dt}= -2Ae^{-2t}\) so \(\displaystyle \frac{dx_2}{dt}- x_2= -2Ae^{-2t}- Ae^{-2t}= -3Ae^{-2t}= C^2e^{-2t}\). We must have \(\displaystyle A= -\frac{C^2}{3}\).

The general solution to the second equation is \(\displaystyle x_2(t)= De^t- \frac{C^2}{3}e^{-2t}\).

So we have \(\displaystyle x_1(t)= Ce^{-t}\), \(\displaystyle x_2= De^t- \frac{C^2}{3}e^{-2t}\)

 

[ChuckNoise]

The point is that since the first equation, \(\displaystyle \frac{dx_1}{dt}= -x_1\) does not involve \(\displaystyle x_2\), we can solve it immediately- \(\displaystyle x_1= Ce^{-t}\). Then we can replace \(\displaystyle x_1^2\) in the second equation by \(\displaystyle C^2e^{-2t}\) so it is no longer non-linear!

\(\displaystyle \frac{dx_2}{dt}= x_1^2+ x_2= C^2e^{-2t}+ x_2\). That is equivalent to the linear, non-homogeneous, equation, \(\displaystyle \frac{dx_2}{dt}- x_2= C^2e^{-2t}\). The "associated homogeneous equation" is \(\displaystyle \frac{dx_2}{dt}- x_2= 0\) so \(\displaystyle \frac{dx_2}{dt}= x_2\) and the general solution to the associated homogeneous equation is \(\displaystyle x_2= De^t\) for D any constant.

We look for a solution to the entire equation of the for \(\displaystyle x_2= Ae^{-2t}\). Then \(\displaystyle \frac{dx_2}{dt}= -2Ae^{-2t}\) so \(\displaystyle \frac{dx_2}{dt}- x_2= -2Ae^{-2t}- Ae^{-2t}= -3Ae^{-2t}= C^2e^{-2t}\). We must have \(\displaystyle A= -\frac{C^2}{3}\).

The general solution to the second equation is \(\displaystyle x_2(t)= De^t- \frac{C^2}{3}e^{-2t}\).

So we have \(\displaystyle x_1(t)= Ce^{-t}\), \(\displaystyle x_2= De^t- \frac{C^2}{3}e^{-2t}\)

Thank you for the detailed version, I got it now!