Solve the following equations with the given initial conditions:
1. y ' - 2 y = 0, y(1)= - 3
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p(t)=-2
P(t)=-2t
=e^(2t)
y(1)=-3 so C=-3
y=Ce^(-P(t))
y=-3e^(2t)
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2. t^2 y ' - 9 y = 0, y(1) = 2.
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y’-(9/t^2)y=0
p(t)=(-9/t^2)
P(t)=(9/t)
y(1)=C so C=2
=2e^(-9/t)
@& & y(t) = C e^(9 / t), so
y(1) = C e^(9/1) = C e^9,
y(1) = 1 so C e^9 = 1 and C = 1 / e^9 = e^-9.
The solution is
y(t) = e^-9 * e^(9/t),
or
y(t) = e^(9 / t - 9),
which could be expressed as
y(t) = e^( 9 ( 1/t - 1) ).*@
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3. (t^2 + t) y' + (2t + 1) y = 0, y(0) = 1.
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I don’t know if you do the same process as the y(1) problems, but that’s what I did:
y’+[(2t+1)/(t^2+t)]y=0
y’+(3/t^2)y=0
p(t)=(3/t^2)
P(t)=(-3/t)
y(0)=C so C=1
=1e^(3/t)
@& Dividing through by t^2 + t you get
y ' + (2 t + 1) / (t^2 + t) y = 0.
p(t) = (2 t + 1) / (t^2 + t),
so you have to integrate p(t).
The substitution u = t^2 + t yields du = 2 t + 1, so the integral becomes
integral(du / u),
which is equal to ln | u | .
Substituting t^2 + t for u we get
ln (t^2 + t).
Now e^(-integral ( p(t) ) ) = e^- (ln ( t^2 + t) = 1 / (t^2 + t), so the solution is
y = c * 1 / (t^2 + t) = c / (t^2 + t).*@
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4. y ' + sin(3 t) y = 0, y(0) = 2.
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p(t)=sin(3t)
P(t)=[-cos(3t)/3]
y(0)=C so C=2
=2e^(cos(3t)/3)
@& The solution is
y(t) = C e^(-cos(t)).
y(0) = C e^(-cos(0 ) = C e^(-1) = C / e,
so
C / e = 1 and
C = e.
y(t) = e * e^(-cos(t) ) = = e^(-cos(t) + 1).*@
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5. Match each equation with one of the direction fields shown below, and explain why you chose as you did.
y ' - t^2 y = 0
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Graph E, because on each of these, I found p(t) then integrated that to get P(t) and then put it in the exponential function, for example, then graphed each “y” function:
p(t)=(-t^2)
P(t)=(-t^3/3)
y=e^(t^3/3)
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y ' - y = 0
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Graph A
p(t)=-1
P(t)=(-t)
y=e^(t)
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y' - y / t = 0
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Graph C
p(t)=(-1/t)
P(t)=(-ln(abs (t)))
y=abs (t)
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y ' - t y = 0
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Graph B
p(t)=(-t)
P(t)=(-t^2/2)
y=e^(t^2/2)
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y ' + t y = 0
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Graph F
p(t)=t
P(t)=(t^2/2)
y=e^(-t^2/2)
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6. The graph of y ' + b y = 0 passes through the points (1, 2) and (3, 8). What is the value of b?
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I graphed the points and connected them. I’m not sure how to go about solving this, but since p(t)=b would you integrate that and get P(t)=(b^2/2)?! How do I use my 2 points?!
@& The variable of integration is t, not b, so integrating b wouldn't give you b^2 / 2.
Your solution is y(t) = C e^(- b t).
Substituting we get the equations
1 = C e^(-b) and
8 = C e^(-3 b).
We solve these two simultaneous equations for C and b.
Dividing the second equation by the first we get
8 = e^(-2 b)
so that
-2 b = ln(8) and
b = -ln(8) / 2.
Substituting this into the first equation we get
1 = C e^(ln(8) / 2)
e^(ln(8) / 2) = e^(ln(8)) ^(1/2) = 8^(1/2) = sqrt(8) so
1 = C * sqrt(8) .
Therefore
C = 1 / sqrt(8) = 1 / (2 sqrt(2) ) = sqrt(2) / 4.
The solution is
y = (sqrt(2) / 4) e^(-ln(8) / 2 * t)
= sqrt(2) / 4 * (1/8) ^ ( t / 2).
*@
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7. The equation y ' - y = 2 is first-order linear, but is not homogeneous.
If we let w(t) = y(t) - 2, then:
What is w ' ?
What is y(t) in terms of w(t)?
What therefore is the equation y ' - y = 2, written in terms of the function w and its derivative w ' ?
Now solve the equation and check your solution:
Solve this new equation in terms of w.
Substitute y + 2 for w and get the solution in terms of y.
Check to be sure this function is indeed a solution to the equation.
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If w(t)=y(t)+2
w’(t)=y’(t)+0
Solving for y(t): y(t)=w(t)-2
The equation y’-y=2 would be w(t)-(w(t)-2)=0
Is this a good start?!
@& Good start.
The substitution should have been w(t) = y(t) - 2.
With this substitution
w ' (t) = y ' (t) - 0 = y ' (t),
and
y(t) = w(t) + 2
so the equation
y ' - y = 2
becomes
w ' + w + 2 = 2, or just
w ' + w = 0.
The solution is w(t) = C e^(-t), so
y(t) = w(t) - 2 = C e^(-t) - 2.
You can substitute this into the original equation and see that it works.*@
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8. The graph below is a solution of the equation y ' - b y = 0 with initial condition y(0) = y_0. What are the values of y_0 and b?
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p(t)=(-b)
mu(t)=e^(-b^2/2)
(e^(-b^2/2)*y’) – (e^(-b^2/2))=0
Would I integrate both sides wrt “b” then solve for y?
@& The variable of integration is t, not b.
The integral of -b with respect to t is - b t + c.
The solution that follows is
y = y0 e^( b t).
The y-intercerpt of this equation is y_0, and can be read from the graph of the solution.
y ' = b * y_0 e^(b t),
so the t = 0 slope of the graph is
y ' (0) = b * y_0.
Having observed the value of y_0 from the graph, you can estimate the slope at t = 0, set the result equal to b * y_0, and solve for b.*@
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self-critique #$&*
#$&* self-critique
self-critique rating
@& You need to go through the details of writing out the solution and substituting known values in order to evaluate constants.*@
@& See my other notes as well, and let me know if you have questions.*@