MHB Recasting/Reducing ODEs of order n to first order

[nacho-man]

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 doesn't 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]

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-man]

This is my lecturer's solution.

What even man

 

[MarkFL]

The may or may not help you, but if I was going to solve the second one, I would note the characteristic equation is:

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

which has the roots:

$$r=1,\,\frac{-1\pm\sqrt{3}i}{2}$$

Hence, the general solution will take the form:

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

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

 

[zzephod]

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 doesn't 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 \({\bf{Y}}(0)=({\bf{Y}}_1(0),{\bf{Y}}_2(0)) =(y'(0),y(0))\)

.

 

[Prove It]

Please disregard my last post, I misread the question.

 

[nacho-man]

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