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Elimination of arbitrary constants.

bergausstein

Active member
Jul 30, 2013
191
1. eliminate B and $\alpha$ from the relation

$\displaystyle x=B cos(\omega+\alpha)$

in which $\omega$ is a parameter(not to be eliminated).

I first took the two derivatives of x with respect to t:

$\displaystyle \frac{dx}{dt}=-\omega B\sin(\omega+\alpha)$

$\displaystyle \frac{d^2x}{dt^2}=-{\omega}^2 B\cos(\omega+\alpha)$

2.) Eliminate $c_1$ and $c_2$ from the relation

$\displaystyle y={c_1}\sin(x)+{c_2}\cos(x)+x^2$


can you help me what to do next? I just dont understand how my book explained the steps because it's brief. please show me the steps on eliminating the constants.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Let's look at the first problem. I believe what you are given is:

\(\displaystyle x=\beta\cos(\omega t+\alpha)\)

and you will find:

\(\displaystyle \frac{d^2x}{dt^2}=-\beta\omega^2\cos(\omega t+\alpha)\)

Now, can you see how to replace part of the right side using the original equation, effectively eliminating the parameters $\beta$ and $\alpha$?
 

bergausstein

Active member
Jul 30, 2013
191
this is what I tried I add the two eqn.

the result is,

$\displaystyle \frac{d^2x}{dt^2}+x+w^2=0$ is this correct?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What I have in mind is the following:

\(\displaystyle \frac{d^2x}{dt^2}=-\beta\omega^2\cos(\omega t+\alpha)=-\omega^2\left(\beta\cos(\omega t+\alpha) \right)\)

Now, using the first equation, can you see that you may substitute for the expression \(\displaystyle \beta\cos(\omega t+\alpha)\)?
 

bergausstein

Active member
Jul 30, 2013
191
now I get it, substituting x for $\beta\cos(\omega t+\alpha)$ in the second eqn.

I have,

$\displaystyle \frac{d^2x}{dt^2}=-{\omega}^2 x$

$\displaystyle \frac{d^2x}{dt^2}+w^2x=0$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
now I get it, substituting x for $\beta\cos(\omega t+\alpha)$ in the second eqn.

I have,

$\displaystyle \frac{d^2x}{dt^2}=-{\omega}^2 x$

$\displaystyle \frac{d^2x}{dt^2}+w^2x=0$
Yes, good work! Now the second problem is worked very similarly. Can you give it a try?
 

bergausstein

Active member
Jul 30, 2013
191
Since the second prob has two constants, I assume that it is a solution to a 2nd order DE

$\displaystyle y={c_1}\sin(x)+{c_2}\cos(x)+x^2$ ---a

I'll take the derivative twice,

$\displaystyle y'={c_1}\cos(x)-{c_2}\sin(x)+2x$----b

$\displaystyle y''=-{c_1}\sin(x)-{c_2}\cos(x)+2$----c

I noticed that from c

$\displaystyle y''=-\left({c_1}\sin(x)+{c_2}\cos(x)\right)+2$

which can also be written as

$\displaystyle y''=x^2-y+2$ is this correct?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Since the second prob has two constants, I assume that it is a solution to a 2nd order DE

$\displaystyle y={c_1}\sin(x)+{c_2}\cos(x)+x^2$ ---a

I'll take the derivative twice,

$\displaystyle y'={c_1}\cos(x)-{c_2}\sin(x)+2x$----b

$\displaystyle y''=-{c_1}\sin(x)-{c_2}\cos(x)+2$----c

I noticed that from c

$\displaystyle y''=-\left({c_1}\sin(x)+{c_2}\cos(x)\right)+2$

which can also be written as

$\displaystyle y''-y-2=0$ is this correct?
That's almost correct. Let's go back to here:

\(\displaystyle y''=-\left(c_1\sin(x)+c_2\cos(x)\right)+2\)

Now, the original equation lets us write:

\(\displaystyle c_1\sin(x)+c_2\cos(x)=y-x^2\)

Now, continue...:D

edit: your edit of your post is correct, although the 2nd order ODE would traditionally be written as:

\(\displaystyle y''+y=x^2+2\)
 

bergausstein

Active member
Jul 30, 2013
191
Edited

Since the second prob has two constants, I assume that it is a solution to a 2nd order DE

$\displaystyle y={c_1}\sin(x)+{c_2}\cos(x)+x^2$ ---a

I'll take the derivative twice,

$\displaystyle y'={c_1}\cos(x)-{c_2}\sin(x)+2x$----b

$\displaystyle y''=-{c_1}\sin(x)-{c_2}\cos(x)+2$----c

I noticed that from c

$\displaystyle y''=-\left({c_1}\sin(x)+{c_2}\cos(x)\right)+2$

which can also be written as

$\displaystyle {c_1}\sin(x)+{c_2}\cos(x)=2-y''$ -----d

substituting d to a I have,

$y=2-y''+x^2$ or $y''=x^2-y+2$ edited