Differential Equation, Solve for function

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The discussion revolves around solving the differential equation \(\frac{dx}{dt} + ax = Asin(ωt)\) with initial condition \(x(0) = b\). The Laplace transform is applied, leading to an expression for \(X(s)\), but confusion arises regarding the equivalence of two derived expressions. It is suggested that partial fraction decomposition could clarify the relationship between the terms, although this method is noted to be complex. An alternative approach using the method of undetermined coefficients is proposed, suggesting a solution of the form \(x(t) = Be^{-at} + C\cos(ωt) + D\sin(ωt)\). The conversation highlights the challenges of the Laplace method and explores simpler solution techniques.
GreenPrint
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Homework Statement



Given

\frac{dx}{dt} + ax = Asin(ωt), x(0) = b

Solve for x(t)

Homework Equations


The Attempt at a Solution



I take the Laplace transform of both sides and get

sX(s) - x(0) + aX(s) = \frac{Aω}{s^{2} + ω^{2}}
X(s) = \frac{b}{s + a} + \frac{Aω}{(s^{2} + ω^{2})(s+1)}

The solution then state that the expression above is equal to

(b + \frac{ωA}{a^{2}+ω^{2}})\frac{1}{s + a} + \frac{aA}{a^{2} + ω^{2}}\frac{ω}{s^{2} + ω^{2}} - \frac{ωA}{a^{2} + ω^{2}}\frac{s}{s^{2} + ω^{2}}

I don't know how the two expressions are equal to each other or were \frac{aA}{a^{2} + ω^{2}}\frac{ω}{s^{2} + ω^{2}} - \frac{ωA}{a^{2} + ω^{2}}\frac{s}{s^{2} + ω^{2}} came from. I believe that \frac{b}{s + a} + \frac{Aω}{(s^{2} + ω^{2})(s+a)} = (b + \frac{ωA}{a^{2}+ω^{2}})\frac{1}{s + a} as it's just factoring. So then if this is true than \frac{aA}{a^{2} + ω^{2}}\frac{ω}{s^{2} + ω^{2}} = \frac{ωA}{a^{2} + ω^{2}}\frac{s}{s^{2} + ω^{2}}, which I don't see how that's true.

Thanks for any help.
 
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GreenPrint said:

Homework Statement



Given

\frac{dx}{dt} + ax = Asin(ωt), x(0) = b

Solve for x(t)

Homework Equations





The Attempt at a Solution



I take the Laplace transform of both sides and get

sX(s) - x(0) + aX(s) = \frac{Aω}{s^{2} + ω^{2}}
X(s) = \frac{b}{s + a} + \frac{Aω}{(s^{2} + ω^{2})\color{red}{(s+a)}}

The solution then state that the expression above is equal to...

I think you meant ##s+a## where I have corrected in red. I didn't bother to work through your next calculations. This is going to be messy even without any mistakes. What the author surely did was use the standard procedure of partial fractions to break up that last fraction:$$
\frac{Aω}{(s^2 + ω^2)(s+a)}= \frac C {s+a} + \frac{Ds+E}{s^2+\omega^2}$$Use your calculus partial fraction methods to solve for ##C,D,E##. They will be a bit messy and if neither you nor the author made any mistakes it should work. That last fraction is easy enough to take the inverse transform. I would inverse just as it stands first, then substitute the values for ##C,D##, and ##E## in your answer.

I should comment that LaPlace transforms is an terrible way to work this problem. I'm assuming you were asked to do it that way or I would have a rather different hint.
 
Last edited:
How would you recommend solving this problem then, if there's a much easier way to solve this than I would like to know how that can be done. Resolving that into partial fractions is very difficult to do.
 
Another way of inverting the second term is to use convolution, especially with the s+a term there in the denominator.
 
GreenPrint said:
How would you recommend solving this problem then, if there's a much easier way to solve this than I would like to know how that can be done. Resolving that into partial fractions is very difficult to do.

The method of undertermined coefficients suggests a solution of the form
<br /> x(t) = Be^{-at} + C\cos \omega t + D \sin \omega t<br />
for some constants B, C and D.
 
GreenPrint said:
How would you recommend solving this problem then, if there's a much easier way to solve this than I would like to know how that can be done. Resolving that into partial fractions is very difficult to do.

It's a constant coefficient equation. I would have recommended the same as pasmith did above.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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