Query_05

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course MTH-279

06/21 around 5:54AM.

Query 05 Differential Equations*********************************************

Question: 3.2.6. Solve y ' + e^y t = e^y sin(t) with initial condition y(0) = 0.

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Your solution:

a)To find the implicit and explicit solution we first see we have a ivp problem y'+e^y t=e^y sin(t),with the intial condition,y(0)=0.I see this is a seperable differential equation.

It has the form n(y)*y'+m(t)=0 now I must rewrite the equation:

y'+e^y t=e^y sin(t)

y'e^-y+t=sin(t)

y'e^-y+t-sin(t)=0

n(y)=e^-y and m(t)=t-sin(t)

now by taking anti derivatives:

integral sign(e^-y dy/dt+t-sin(t))dt=integral sign 0dt

integral sign e^-y dy/dt dt+integral sign t-sin(t)dt=C

INTEGRAL sign e^-y dy+integral sign t-sin(t)dt=C

-e^-y+t^2/2+cos(t)=C

-e^-y+t^2/2+cos(t)=C

BY APPLYING THE intial condition y(0)=0 then I can solve for C

-e^-0+0^2/2+cos(0)=C

-1+0+1=C

0=C

IMPLICIT soln:-e^-y+t^2/2+cos(t)=0

now can find the explicit solution by solving for y

-e^-y+t^2/2+cos(t)=0

t^2/2+cos(t)=e^-y

ln(1/2t^2+cos(t))=-y

-ln(1/2t^2+cos(t))=y

explct soln:y(t)=-ln(1/2t^2+cos(t))

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Good.

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Confidence rating: 2

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Given Solution:

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Self-critique (if necessary):

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Question: 3.2.10. Solve 3 y^2 y ' + 2 t = 1 with initial condition y(0) = -1.

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Your solution:

a)3y^2 y'+2t=1 y(-1)=-1

n(y)*y'+m(t)=0

3y^2 y'+2t=1

3y^2 y'+2t-1=0

n(y)=3y^2 and m(t)=2t-1

now by taking anti derivative

integral sign(3y^2 dy/dt+2t-1)dt=integral sign 0dt

integral sign 3y^2 dy+integral sign 2t-integral sign 1dt=C

3 y^3/3+2 t^2/2-t=C

y^3+t^2-t=C

y^3+t^2-t=C

by applying intial conditions:y(-1)=-1 and solving for C

(-1)^3+(-1)^2-(-1)=C

-1+1+1=C

1=C

implicit soln:y^3+t^2-t=1

now I can use it to find the expl. soln

y^3+t^2-t=1

y^3=1+t-t^2

y=3 radical symbol 1+t-t^2

b)interval of existence:-infinity

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Good solution.

The initial condition given here is y(0)= -1, which leads to solution

y = (-1 + t - t^2)^(1/3).

The condition you used is y(-1) = -1, and that leads to the value of C you used in your solution.

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Confidence rating: 2

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Given Solution:

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Self-critique (if necessary):

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Question: 3.2.18. State a problem whose implicit solution is given by y^3 + t^2 + sin(y) = 4, including a specific initial condition at t = 2.

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Your solution:

first we find the intial value

by pluging in t_0=2

y^3+2^2+siny=4

y^3+4+siny=4

y^3+siny=0

y^3=-siny

M(y)+N(t)=C

d/dt M(y)+d/dt N(t)=0

M(y)=y^3+sin(y) and N(t)=t^2

now we can calculate the differential eqn

d/dt(y^3+sin(y))+d/dt(t^2)=0

3y^2 y'+cos(y)y'+2t=0

(3y^2+cosy)y'+2t=0

ivp that satisfies intial conditions

(3y^2+cosy)y'+2t=0,y(2)=0

Confidence rating: 2

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Very good.

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

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Question: 3.2.24. Solve the equation y ' = (y^2 + 2 y + 1) sin(t) fand determine the t interval over which he solution exists.

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Your solution:

y ' = (y^2 + 2 y + 1) sin(t)

int(dy/(y^2 + 2 y + 1)) = int (sin(t))dt

int(y+1)^-2 dy = -cos(t) + c

-1/(y+1) = -cos(t) + c

-1 = (-cos(t))(y + 1)

y + 1 = -1/(-cos(t))

y = 1/(cos(t) + c) - 1

The Solution may exist everywhere but where:

cos(t) + c = 0

Confidence rating: 2

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Given Solution:

Self-critique (if necessary):

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Question: 3.2.28. Match the graphs of the solution curves with the equations y ' = - y^2, y ' = y^3 an dy ' = y ( 4 - y).

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Your solution:

y ' = - y^2

As y increases, y ' is a negative slope that increases by the square of y, making it graph C.

y ' = y^3

As y increases, y' is the cube root of y, so its slope is positive,

making it graph A.

y ' = y ( 4 - y)

As y increases, y' is the more complex equation above, but it’s good to note that by plugging in 0 or 4 for ‘y’ we get a slope of zero.

This matches our last graph which is B.

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You can also compare your analysis with the following:

For each function you need to analyze the behavior of the right-hand side as you move around in the y vs. t plane.

-y^2 is always negative unless y = 0, in which case -y^2 is zero, and the magnitude of y^2 increases as an increasing rate as you move away from y = 0.

y^3 is positive for positive y, negative for negative y, and zero for y = 0. The magnitude of y^2 increases as an increasing rate as you move away from y = 0.

y ( 4 - y_) is positive for 0 < y < 4, zero when y = 0 or y = 4, and negative everywhere else.

This results in the following slope-field behaviors:

y ' = - y^2 will result in a direction field that gets steeper as you move away from the t axis, since y^2 increases in magnitude at an increasing rate as you move away from t = 0.

On the t axis y is zero so the direction field is horizontal. For y not equal to zero, since y^2 is positive, - y^2 is always negative. So the field lines all have negative slopes, and get steeper as you move away from the t axis.

y ' = y^3 will again result in a direction field that gets steeper as you move away from the t axis, since y^3 increases in magnitude at an increasing rate as you move away from t = 0.

On the t axis y is zero so the direction field is horizontal.

Above the t axis y is positive so y^3 is positive, so the field lines all have positive slopes, and get steeper as you move away from the t axis..

Below the t axis y is negative so y^3 is negative, so the field lines all have negative slopes, and get steeper as you move away from the t axis..

y ' = y ( 4 - y) is positive between the t axis and the line y = 4, negative above that line. The closer you get to y = 4 the less steep the slopes.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

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&#Good work. See my notes and let me know if you have questions. &#