# Solution of IVP

#### ChuckNoise

##### New member
Hi

I am struggeling to wrap my head around the following attached example.
Why is the solution to equation 2
c2et+c12/3(et-e-2t)
and not simply
c2et+(c1e-t)2 ??

Thanks

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#### yoscar04

##### Full Member
You forgot to write down your IVP.

#### ChuckNoise

##### New member
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

##### Elite Member
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 $$\displaystyle D= c_2- c_1^2/2$$.​

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#### ChuckNoise

##### New member
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(et-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(et-e-2t) ?? I just can't see it

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