Derivative of Laplace Transforms

[vanceEE]

Homework Statement

$$xy'' + xy = 0$$

Homework Equations

y'(0) = 0, y(0) = 1

The Attempt at a Solution

$$ L[xy''] + L[xy] = 0$$
$$-L[-xy''] - L[-xy] = 0$$
$$-\frac{d}{dp}L[y''] - \frac{dY}{dp} = 0 $$
$$-\frac{d}{dp}(-y'(0)-py(0) + p^2Y) - \frac{dY}{dp} = 0 $$
$$\frac{d}{dp}(-p + p^2Y) + \frac{dY}{dp} = 0 $$

How would I evaluate $p^2Y$? (does $\frac{d}{dp}(p^2Y) = 2pY$ or $2pY + p^2*\frac{dY}{dp}?$)
Should I end up with $$\frac{dY}{dp} = \frac{1-2pY}{p^2+1}$$?

 

[LCKurtz]

Homework Statement

$$xy'' + xy = 0$$

So divide by $x$ and you have $y''+y=0$.

 

[vanceEE]

So divide by $x$ and you have $y''+y=0$.

Yes, I'm aware that this problem can be solved using a simpler method but I need to solve using derivatives of the laplace transforms, in other words, leaving the x's.

 

[Dick]

Yes, I'm aware that this problem can be solved using a simpler method but I need to solve using derivatives of the laplace transforms, in other words, leaving the x's.

Ok, then you have the correct ODE for Y. This method of turning an easy problem into a hard one should work if you keep pushing forward. Now solve for Y(p).

 

[LCKurtz]

Yes, I'm aware that this problem can be solved using a simpler method but I need to solve using derivatives of the laplace transforms, in other words, leaving the x's.

OK, I will have another look below.

Homework Statement

$$xy'' + xy = 0$$

Homework Equations

y'(0) = 0, y(0) = 1

The Attempt at a Solution

$$ L[xy''] + L[xy] = 0$$
$$-L[-xy''] - L[-xy] = 0$$
$$-\frac{d}{dp}L[y''] - \frac{dY}{dp} = 0 $$
$$-\frac{d}{dp}(-y'(0)-py(0) + p^2Y) - \frac{dY}{dp} = 0 $$
$$\frac{d}{dp}(-p + p^2Y) + \frac{dY}{dp} = 0 $$

How would I evaluate $p^2Y$? (does $\frac{d}{dp}(p^2Y) = 2pY$ or $2pY + p^2*\frac{dY}{dp}?$)
Should I end up with $$\frac{dY}{dp} = \frac{1-2pY}{p^2+1}$$?

You have to use the product rule, so the second one. If you do that you should be able to write your result as$$
(p^2+1)Y' + 2pY = 1$$which is a first order linear DE for $Y$. You almost have it. You wind up with a first order differential equation in $Y$ instead of the original second order DE.

 

[vanceEE]

Ok, then you have the correct ODE for Y. This method of turning an easy problem into a hard one should work if you keep pushing forward. Now solve for Y(p).

Please check for any errors:

$\frac{dY}{dp} = \frac{1-2pY}{p^2+1}$

$(2pY-1)dp + (p^2+1)dY \equiv 0$

$\frac{∂M}{∂Y} = 2p = \frac{∂N}{∂p}$

$∫\frac{∂f}{∂p}dp = ∫(2pY-1)dp$

$f(p,Y) = p^2Y-p+\phi(Y)$

$\phi'(Y) = 1$

$\phi(Y) = Y$

$p^2Y-p+Y = C$

$Y(p^2+1) = C + p$

$Y = \frac{C}{p^2+1} + \frac{p}{p^2+1}$

$L[y] = C(\frac{1}{p^2+1}) + (\frac{p}{p^2+1})$

$L[y] = L[Csinx+cosx]$

$y = Csinx + cosx$

$1 = Csin0 + cos0$

$∴ y = cos x$

 

[LCKurtz]

OK, I will have another look below.



You have to use the product rule, so the second one. If you do that you should be able to write your result as$$
(p^2+1)Y' + 2pY = 1$$which is a first order linear DE for $Y$. You almost have it. You wind up with a first order differential equation in $Y$ instead of the original second order DE.

It would have been shorter to note the above equation is already exact so$$
[(p^2+1)Y]' = 1$$so $(p^2+1)Y = p+C$.

Please check for any errors:

$\frac{dY}{dp} = \frac{1-2pY}{p^2+1}$

$(2pY-1)dp + (p^2+1)dY \equiv 0$

$\frac{∂M}{∂Y} = 2p = \frac{∂N}{∂p}$

$∫\frac{∂f}{∂p}dp = ∫(2pY-1)dp$

$f(p,Y) = p^2Y-p+\phi(Y)$

$\phi'(y) = 1$

$\phi(y) = y$

$p^2Y-p+Y = C$

$Y(p^2+1) = C + p$

$Y = \frac{C}{p^2+1} + \frac{p}{p^2+1}$

$L[y] = C(\frac{1}{p^2+1}) + (\frac{p}{p^2+1})$

$L[y] = L[Csinx+cosx]$

$y = Csinx + cosx$

$1 = Csin0 + cos0$

$∴ y = cos x$

I don't see how that second to last equation implies $C=0$. Of course, the original boundary conditions do. I'm not sure how else to get $C=0$. Other than that it looks OK.

 

[vanceEE]

I don't see how that second to last equation implies $C=0$. Of course, the original boundary conditions do. I'm not sure how else to get $C=0$. Other than that it looks OK.

The last equation does not imply that C =0, y(0) MUST equal 1, therefore since sin(0) = 0 whatever value C holds does not matter. Since Csin0 = 0 and cos0 = 1.

 

[Dick]

I don't see how that second to last equation implies $C=0$. Of course, the original boundary conditions do. I'm not sure how else to get $C=0$. Other than that it looks OK.

The last equation does not imply that C =0, y(0) MUST equal 1, therefore since sin(0) = 0 whatever value C holds does not matter. Since Csin0 = 0 and cos0 = 1.

Use the boundary condition y'(0)=0.

 

[LCKurtz]

But if $C$ does not matter, you have gotten the solution $y = C\sin x + \cos x$ for any $C$, which is not correct. The boundary conditions are supposed to be built into the LT method. Somehow that has been missed in your solution. I don't right off see where. But you shouldn't have to go back to the original problem and reapply the boundary conditions to get $C=0$. It might be an inherent problem caused whenever you use the $L(tf(t))$ formula, which introduces a derivative in the transform.

 

[vanceEE]

It would have been shorter to note the above equation is already exact \

That's why I solved for Y(p) using the method of exact equations. I'm sure I could've used an integrating factor since it's first order linear, I just favor this method even though it can get kind of lengthy sometimes

 

[vanceEE]

But if $C$ does not matter, you have gotten the solution $y = C\sin x + \cos x$ for any $C$, which is not correct.

Not true, since y = Csinx + cosx and y(0) = 1 then 1 = 0*C + cos(0), since cos (0) $\equiv$ 1 and sin(0) $\equiv$ 0, this MUST be our solution since it holds true that C*0 + cos (0) = y(0). We do not need Csinx in our equation, hence we do not even need to find C. cos(0) = 1 = y(0).

 

[LCKurtz]

That's why I solved for Y(p) using the method of exact equations. I'm sure I could've used an integrating factor since it's first order linear, I just favor this method even though it can get kind of lengthy sometimes

No, I meant it was already in the exact form in that it was a derivative of a product as it stands. You just integrate it in one step. But it doesn't matter, your steps were OK too.

 

[Dick]

That's why I solved for Y(p) using the method of exact equations. I'm sure I could've used an integrating factor since it's first order linear, I just favor this method even though it can get kind of lengthy sometimes

Sure, but $y = C\sin x + \cos x$ isn't a solution to the original problem for nonzero C. You solved another differential equation for dY/dp which gave you an extra constant.

 

[LCKurtz]

Not true, since y = Csinx + cosx and y(0) = 1 then 1 = 0*C + cos(1), since cos (0) $\equiv$ 1 and sin(0) $\equiv$ 0, this MUST be our solution since it holds true that C*0 + cos (0) = y(0). We do not need Csinx in our equation, hence we do not even need to find C. cos(0) = 1 = y(0).

No, you are misunderstanding this. The only way you can get $C=0$ is the other boundary condition $y'(0)=0$. If you do not do that you don't have the correct solution. You can't just say you don't need that term. That term is in your solution unless you can prove $C=0$ and it is incorrect otherwise.

 

[vanceEE]

No, you are misunderstanding this. The only way you can get $C=0$ is the other boundary condition $y'(0)=0$. If you do not do that you don't have the correct solution. You can't just say you don't need that term. That term is in your solution unless you can prove $C=0$ and it is incorrect otherwise.

I'm am not implying that C = 0. I am stating the fact that C*sin(0) $\equiv$ 0 and cos(0) satisfies the condition that y(0) = 1.

 

[LCKurtz]

I'm am not implying that C = 0. I am stating the fact that C*sin(0) $\equiv$ 0 and cos(0) satisfies the condition that y(0) = 1.

But you need to imply C=0 or your solution is incorrect. But since I'm now just repeating myself, I've said all I want to in this thread.

 

[vanceEE]

But you need to imply C=0 or your solution is incorrect. But since I'm now just repeating myself, I've said all I want to in this thread.

C*sin(0) $\equiv$ 0 satisfies any implications. C*sin(0) does not exist in the solution, point blank. Simple intuition should tell one that a constant multiplied by 0 is 0 and hence negligible to the equation; the value of C has no effect on the equation because sin(x) is not in the solution, this easily seen by plugging in the coordinate points of the particular solution. If sin(x) doesn't exist in our particular solution, neither does C.

 

[Dick]

C*sin(0) $\equiv$ 0 satisfies any implications. C*sin(0) does not exist in the solution, point blank. Simple intuition should tell one that a constant multiplied by 0 is 0 and hence negligible to the equation; the value of C has no effect on the equation because sin(x) is not in the solution, this easily seen by plugging in the coordinate points of the particular solution. If sin(x) doesn't exist in our particular solution, neither does C.

Csin(x) doesn't satisfy all implications. It doesn't satisfy y'(0)=0. That's what rules it out. Not that y(0)=0 says Csin(0)=0 which has no implications for C whatsoever.

 

[vanceEE]

Csin(x) MUST equal zero for this equation to be true; if y(x) is satisfied then y'(x) is also satisfied. $$y(x) = Csin(x) + cos(x)$$
$$Csin(x) \equiv 0$$
$$y'(x) = \frac{d}{dx}(Csin(x)) - sin(x)$$
$$y'(x) = \frac{d}{dx}(0) - sin(x)$$
$$y'(0) = -sin(0) = 0$$

I understand what you both are saying and this can go on forever, the particular solution is y(x) = cos(x) no if ands or buts, you feel that I need to solve for C; I feel that it's unnecessary in this particular case, big deal; if I had a C in my particular solution I would obviously need to state the value of any constant of integration that exists in my solution. But anyways, thanks for your help on initially helping me with the transform. Goodnight! :thumbs:

 

[Dick]

$y(x) = Csin(x) + cos(x)$ says $y'(x) = Ccos(x) - sin(x)$. So $y'(0)=Ccos(0)=C$. Not $y'(x)=-sin(x)$. Yeah, I'm done too.