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Finding Differential Equation

bergausstein

Active member
Jul 30, 2013
191
find the desired equation.

a.) $\displaystyle y=c_1+c_2e^{3x}$

taking two derivatives

$\displaystyle \frac{dy}{dx}=3c_2e^{3x}$

$\displaystyle \frac{d^2y}{dx^2}=9c_2e^{3x}$

b.) $\displaystyle y=c_1e^{ax}\cos(bx)+c_2e^{ax}\sin(bx)$ a and b are parameters.

can you help me continue with the problems. thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
a) Differentiating with respect to $x$, we obtain:

\(\displaystyle y'=3c_2e^{3x}\)

But, from the original equation, we have:

\(\displaystyle c_2e^{3x}=y-c_1\)

So substitute and differentiate again. What do you find?
 

bergausstein

Active member
Jul 30, 2013
191
what I see is

$\displaystyle y'=3(y-c_1)$ now when I differentiate it again I get $\displaystyle y''=3y'$

so, the D.E is $y''-3y'=0$ is this correct?

but I want to know other technique to eliminate the constant using the 2 derivatives and the function in my OP. can you help me with that?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
what I see is

$\displaystyle y'=3(y-c_1)$ now when I differentiate it again I get $\displaystyle y''=3y'$

so, the D.E is $y''-3y'=0$ is this correct?

but I want to know other technique to eliminate the constant using the 2 derivatives and the function in my OP. can you help me with that?
Yes, that is correct. The technique I described is to me the most straightforward way to approach this problem. If you differentiate twice, you will still wind up substituting, and then you'll have to differentiate again to eliminate the remaining parameter.
 

bergausstein

Active member
Jul 30, 2013
191
I noticed something. If I use the 2nd derivative

$ \displaystyle y''=9c_2e^{3x}$

since $\displaystyle c_2e^{3x}=y-c_1$ I will have $\displaystyle y''=9(y-c_1)$

so $y''-9y'=0$ is this also correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I noticed something. If I use the 2nd derivative

$ \displaystyle y''=9c_2e^{3x}$

since $\displaystyle c_2e^{3x}=y-c_1$ I will have $\displaystyle y''=9(y-c_1)$

so $y''-9y'=0$ is this also correct?
No, you did not differentiate the left side of the equation.
 

bergausstein

Active member
Jul 30, 2013
191
yes It must be $y'''-9y'=0$

how bout the second problem?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
yes It must be $y'''-9y'=0$

how bout the second problem?
Actually in the second problem, the parameters are $c_1$ and $c_2$. These are what you wish to eliminate.

I think I would write it in the equivalent two-parameter form to make differentiation simpler:

\(\displaystyle y=c_1e^{ax}\sin\left(bx+c_2 \right)\)

And, let's take Ackbach's advice and write it as:

\(\displaystyle c_1=\frac{y}{e^{ax}\sin\left(bx+c_2 \right)}\)

What do you get when you differentiate with respect to $x$?
 

bergausstein

Active member
Jul 30, 2013
191
how did you get this?

\(\displaystyle y=c_1e^{ax}\sin\left(bx+c_2 \right)\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
how did you get this?

\(\displaystyle y=c_1e^{ax}\sin\left(bx+c_2 \right)\)
Through the use of a linear combination trigonometric identity. For example:

\(\displaystyle \sin(x)+\cos(x)=\sqrt{2}\sin\left(x+\frac{\pi}{4} \right)\)
 

bergausstein

Active member
Jul 30, 2013
191
is there another method? :confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
is there another method? :confused:
Yes, certainly there are many ways to proceed, however, I am trying to show you how to work the problem in a manner that will ease the amount of computation required.