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course Mth 279
February 6 around 3:25pm. Not sure how well I did, but I attempted them all!
Section 2.3Solve each equation:
1. y ' + y = 3
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p(t)= 1 so P(t)=t and mu(t)=e^(t)
e^ty’ + e^ty = 3e^t
e^ty = 3e^t + C
y= 2e^t + C
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@& The integrating factor is e^(1 * t) = e^t.
(y e^t) ' = y ' e^t + y e^t, so the equation
e^t y ' + e^t y = 3 e^t is
(y e^t) ' = 3 e^t
with solution
y e^t = 3e^t + C.
Solving for y we get
y = (3 e^t + C) / e^t
= 3 e^t / e^t + C / e^t
= 3 + C e^(-t).
You were find up to the last step.
Be sure you understand how this one works algebraically in that last step.*@
2. y ' + t y = 3 t
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p(t)=t so P(t)=t^2/2 and mu(t)=e^(t^2/2)
e^(t^2/2)y’ + t e^(t^2/2)y = 3t e^(t^2/2)
e^(t^2/2)y = 3e^(t^2/2)
y=2 e^(t^2/2) + C
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@& The integrating factor is e^(t^2 / 2).
(y e^(t^2 / 2) ) ' = 2. e^(t^2 / 2) y ' + t e^(t^2 / 2) y
so the equation is
(y e^(t^2 / 2) ) ' = 3 t
which we integrate to get the solution
y e^(t^2 / 2) = 3 t^2 / 2 + C.
Solving for y we have
y = (3 t^2 / 2 + C) / e^(t^2 / 2)
= 3 /2 + C e^(-t^2 / 2).*@
3. y ' - 4 y = sin(2 t)
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p(t)=-4 so P(t)=-4t and mu(t)=e^(-4t)
e^(-4t)y’ -4 e^(-4t)y = sin(2t) e^(-4t)
y= e^(4t) INT (e^(-4t)*sin(2t) dt + C e^(4t)
y= e^(4t)*([(- e^(-4t)*cos(2t))/10]-[ (e^(-4t)*sin(2t))/5] + Ce^(4t)?!
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@& The integrating factor is e^(-4 t).
(y e^(-4 t) ) ' = e^(-4 t) y ' - 4 e^(-4 t) y
so the equation is
(y e^(-4 t) ) ' = sin(2 t)
which we integrate to get the solution
y e^(-4 t) = -1/2 cos(2 t) + C.
Solving for y we have
y = -1/2 cos(2 t) / e^(-4 t) + C / e^(-4 t)
= (-1/2 cos(2 t) + C) e^(4 t).*@
4. y ' + y = e^t, y (0) = 2
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p(t)=1 so P(t)=t and mu(t)=e^(t)
e^(t)y’ + e^(t)y = e^(2t)
e^(t)y = (e^(2t))/2 OR e^(t)y = ½ e^(2t)
y=( e^(2t)/2) – (e^(t)/1) + C
y(0)=2 so:
y=( e^(2t)/2) – (e^(t)/1) + 2
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@& The integrating factor is e^(t).
(y e^( t) ) ' = e^( t) y ' + e^( t) y
so the equation, multiplied by the integrating factor, is
(y e^( t) ) ' = e^(2t)
which we integrate to get the solution
y e^( t) = 1/2 e^(2t) + C
Solving for y we have
y = 1/2 e^(2t) / e^( t) + C / e^( t)
= e^t + C e^(-t).*@
5. y ' + 3 y = 3 + 2 t + e^t, y(1) = e^2
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p(t)=3 so P(t)=3t and mu(t)=e^(3t)
e^(3t)y’ + 3e^(3t)y = (3+2t+e^t)* e^(3t)
e^(3t)y = (e^(4t)/4) + (2t/3 + 7/9)*e^(3t)
y = (e^(4t)/4) + (2t/3 - 2/9)*e^(3t) + C
How do I do y(1)=e^2 on this problem? I’m used to doing y(0)?!
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@& The integrating factor is e^(3 t).
(y e^(3 t) ) ' = e^( 3t) y ' + e^( 3 t) y
so the equation, multiplied by the integrating factor, is
(y e^( 3 t) ) ' = 3 e^(3 t) + 2 t e^(3 t) + e^t * e^(3 t)
which we integrate to get the solution
y e^( 3 t) = e^(3 t) + 2/3 t e^(3 t) - 2 e^(3 t) / 9 + 1/4 e^(4 t). + C
Solving for y we have
y = 1 + 2 t / 3 - 2/9 + 1/4 e^t + C e^(-3 t).
y(1) = 1 + 2/3 - 2/9 + 1/4 e^1 + C e^-3
1 + 2/3 - 2/9 + 1/4 e^1 + C e^-3 = e^2
can be solved for C.*@
6. The general solution to the equation y ' + p(t) y = g(t) is y = C e^(-t^2) + 1, t > 0. What are the functions p(t) and g(t)?
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You would have to work this problem backwards from “y=” in order to find p(t) and g(t).
You would have to get an exponential on the left side with the y, then take the derivatives of each side, then see what is in front of y’ and that would be your mu(t) and you would know your g(t) function, which would be on the right side, without the mu(t) multiplied by it. Since mu(t) is “e” raised to the something, that something would be your P(t) and then you would need to take the derivative of that to get your p(t) function.
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@& The general solution to
y ' + p(t) y = g(t)
is found by using the integrating factor e^(p(t)) to obtain
e^(p(t) ) y ' + e^(p(t) ) y = e^(p(t)) g(t),
which is
(e^(p(t)) y) ' = g(t) e^(p(t))
Integrating both sides this is
e^(p(t)) y = integral(g(t) e^(p(t)) dt) + C
so that
y = ( integral(g(t) e^(p(t)) dt) + C ) * e^(-p(t)).
Now in the given solution the term C e^(-t^2) matches C e^(-p(t)), so that p(t) = -t^2.
The other term in the given solution is 1, so that
integral(g(t) e^(p(t)) dt) * e^(-p(t)) = 1
implying that
integral(g(t) e^(p(t)) dt) = e^(p(t)) = e^(-t^2).
It follows that g(t) e^p(t) is the derivative of e^(-t^2 ). Thus
g(t) e^p(t) = - 2 t e^(-t^2).
Since p(t) = -t^2, it follows that g(t) = - 2 t.
*@