# Recasting/Reducing ODEs of order n to first order

#### nacho

##### Active member
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
The second one is easy if you substitute \displaystyle \begin{align*} u(x) = y'(x) \end{align*}, making a second order linear DE with constant coefficients \displaystyle \begin{align*} u''(x) = u(x) \end{align*}.

#### nacho

##### Active member
This is my lecturer's solution.

What even man

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#### MarkFL

Staff member
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
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 $${\bf{Y}}(0)=({\bf{Y}}_1(0),{\bf{Y}}_2(0)) =(y'(0),y(0))$$

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Last edited:

MHB Math Helper

#### nacho

##### Active member
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