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Recasting/Reducing ODEs of order n to first order

nacho

Active member
Sep 10, 2013
156
Could someone please provide a worked solution for me. I think that is the only way I will understand this. It was covered very vaguely in our lectures and my notes start talking about vectors and using co-domain notation which is very frustrating!

1. $y''(x) = x + y'(x) + e^{y(x)}$ with $y(0)=0, y'(0)=1$


I know MHB doesnt endorse just handing out solutions, so I will try and attempt the second problem myself if someone may help me with the first. I really need to learn this for my exam in 7 days.

2. $y'''(x) = y(x)$ with $y(1) = 4, y'(1)=4, y''(1)=0$
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
The second one is easy if you substitute [tex]\displaystyle \begin{align*} u(x) = y'(x) \end{align*}[/tex], making a second order linear DE with constant coefficients [tex]\displaystyle \begin{align*} u''(x) = u(x) \end{align*}[/tex].
 

nacho

Active member
Sep 10, 2013
156
This is my lecturer's solution.

What even man
 

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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The may or may not help you, but if I was going to solve the second one, I would note the characteristic equation is:

\(\displaystyle r^3-1=(r-1)\left(r^2+r+1 \right)=0\)

which has the roots:

\(\displaystyle r=1,\,\frac{-1\pm\sqrt{3}i}{2}\)

Hence, the general solution will take the form:

\(\displaystyle y(x)=c_1e^x+e^{-\frac{1}{2}x}\left(c_2\cos\left(\frac{\sqrt{3}}{2}x \right)+c_3\sin\left(\frac{\sqrt{3}}{2}x \right) \right)\)

Now it is a matter of differentiating to get a 3X3 linear system in the 3 parameters.
 

zzephod

Well-known member
Feb 3, 2013
134
Could someone please provide a worked solution for me. I think that is the only way I will understand this. It was covered very vaguely in our lectures and my notes start talking about vectors and using co-domain notation which is very frustrating!

1. $y''(x) = x + y'(x) + e^{y(x)}$ with $y(0)=0, y'(0)=1$


I know MHB doesnt endorse just handing out solutions, so I will try and attempt the second problem myself if someone may help me with the first. I really need to learn this for my exam in 7 days.

2. $y'''(x) = y(x)$ with $y(1) = 4, y'(1)=4, y''(1)=0$
For a second order ODE that can be written in the form: \(y''(x)=f(y'(x),y(x),x)\) we reduce this to a first order system by introducting the state vector \( {\bf{Y}}(x)=(y'(x),y(x))\). Now if we differentiate this we get:

\[ {\bf{Y}}'(x)=(y''(x),y'(x))\]
Now we may substitute \(y''(x)\) from the original equation into this to get:
\[ {\bf{Y}}'(x)=(f(y'(x),y(x),x),y'(x))\]
Then using \(y'(x)={\bf{Y}}_1\) and \(y(x)={\bf{Y}}_2\) we get:
\[ {\bf{Y}}'(x)=(f({\bf{Y}}_1,{\bf{Y}}_2,x),{\bf{Y}}_1)\]
with inotial condition \({\bf{Y}}(0)=({\bf{Y}}_1(0),{\bf{Y}}_2(0)) =(y'(0),y(0))\)

.
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Please disregard my last post, I misread the question.
 

nacho

Active member
Sep 10, 2013
156
For the longest of time, our lecturer made it out to be that recasting and reduction of order were the same thing.

now that I know that they are not, everything is much more clearer.

Thanks for the help guys/gals