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!