MHB Integrating Factor Method for Solving y' + y = e^{-2t}

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The discussion focuses on solving the first-order linear differential equation y' + y = e^{-2t} using the integrating factor method. The associated homogeneous equation is y' = -y, leading to the solution y = C'e^{-t}. For the non-homogeneous part, a particular solution is assumed in the form y = Ae^{-2t}, which is determined by substituting into the original equation. After calculations, the coefficient A is found to be -1/2, resulting in the general solution y(t) = Ce^{-t} - (1/2)e^{-2t}. The thread emphasizes the validity of both the integrating factor method and the method of undetermined coefficients for solving this type of differential equation.
hiyum
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\[ y'=-y+e^{(-2)t} \]
 
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hiyum said:
\[ y'=-y+e^{(-2)t} \]
As in the other thread:
[math]y' + y = e^{-2t}[/math]

How do you find the integrating factor here?

-Dan
 
Since this, while a first order differential equation, is also linear we can also separate it. The "associated homogeneous equation" is y'= -y which we can write as \(\frac{dy}{dx}= -y\) and separate as \(\frac{dy}{y}= -dt\). Integrating, \(ln{y}= -t+ C\). Taking the exponential of both sides, \(y=C'e^{-t}\) where \(C'= e^C\).

Since the "non-homogeneous part", \(e^{-2t}\), is of the type of function we expect as a solution to a "linear differential equation with constant coeffcients" we try a solution of the form \(y= Ae^{-2t}\) (this is the "method of undetrmined coefficients". A is the coefficient to be determined.

If \(y= Ae^{-2t}\) then \(y'= -2Ae^{-2t}\) and putting those into the differential equation, \(-2Ae^{-2t}= -Ae^{-2t}+ e^{-2t}\). \(-Ae^{-2t}= e^{-2t}\). Dividing by \(e^{-2t}\) we have -A= 1 so A= -1/2.

The general solution to this differential equation is \(y(t)= Ce^{-t}- \frac{1}{2}e^{-2t}\).
 
topsquark said:
As in the other thread:
[math]y' + y = e^{-2t}[/math]

How do you find the integrating factor here?

-Dan
Though you could try the integrating factor approach as I showed you in the other Forum.

-Dan
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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