MHB Elimination of arbitrary constants.

[bergausstein]

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 don't understand how my book explained the steps because it's brief. please show me the steps on eliminating the constants.

 

[MarkFL]

Let's look at the first problem. I believe what you are given is:

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

and you will find:

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

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

 

[bergausstein]

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?

 

[MarkFL]

What I have in mind is the following:

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

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

 

[bergausstein]

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]

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]

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?

 

[MarkFL]

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:

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

Now, the original equation let's us write:

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

Now, continue...:D

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

$$y''+y=x^2+2$$

 

[bergausstein]

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