MHB What Is the Differential Equation for Given Solutions?

[bergausstein]

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]

a) Differentiating with respect to $x$, we obtain:

$$y'=3c_2e^{3x}$$

But, from the original equation, we have:

$$c_2e^{3x}=y-c_1$$

So substitute and differentiate again. What do you find?

 

[bergausstein]

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]

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]

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]

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]

yes It must be $y'''-9y'=0$

how bout the second problem?

 

[MarkFL]

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:

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

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

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

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

 

[bergausstein]

how did you get this?

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

 

[MarkFL]

how did you get this?

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

Through the use of a linear combination trigonometric identity. For example:

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

 

[bergausstein]

is there another method?

 

[MarkFL]

is there another method?

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.