Query 06

#$&*

course MTH 279

Query 06 Differential Equations*********************************************

Question: 3.3.4. If the equation is exact, solve the equation (6 t + y^3 ) y ' + 3 t^2 y = 0, with y(2) = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3t^2*y + (6t+y^3)dy/dt = 0 M = 3t^2*y; N = 6t+y^3 M_y = 3t^2; N_t = 6 The equation is not exact. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK

********************************************* Question: If the equation is exact, solve the equation (6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 = -t, with y(2) = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (6t + 3t^3)dy/dt +(6y+9/2t^2y^2+t) = 0 M = (6y+9/2t^2y^2+t) ; N = (6t+3t^3) M_y = 9yt^2+6 ; N_t = 9t^2+6 The equation is not exact. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK

********************************************* Question: 3.3.6. If the equation is exact, solve the equation y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2) with y(0) = pi. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (t cos(t y) + 2 y e^y^2) dy/dt + ( y cos(t y) + 1) = 0 M = y*cos(y*t)+1 ; N = t*cos(t*y) + 2y*e^(y^2) M_y = cos(t*y)-y*t*sin(t*y) ; N_t = cos(t*y)-t*y*sin(t*y) M_y = N_t The equation is exact. f(t,y) = ʃMdt = sin(t*y)+t + h(y) δf/δy = tcos(ty)+dh/dy tcos(ty) + dh/dy = tcos(ty)+2ye^(y^2) dh/dy = 2ye^(y^2) h(y) = e^(y^2) f(t,y) = sin(ty)+t+e^(y^2) = C f(0,π) = sin(0) + 0 + e^(π^2) = C C = e^(π^2) f(t,y) = sin(ty) + t + e^(y^2) = e^(π^2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK

********************************************* Question: 3.3.10. If N(y, t) * y ' + t^2 + y^2 sin(t) = 0 is exact, what is the most general possible form of N(y, t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the equation is exact: M(y,t) = t^2 + y^2sin(t) M_y = 2ysin(t) f(y,t) = ʃMdt = t^3/3-cos(t)y^2 + g(y) δf/δy = -2ycos(t) + dg/dy δf/δy = N(y,t) N(y,t) = -2ycos(t) + dg/dy confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK

********************************************* Question: 3.3.12. If y = -t - sqrt( 4 - t^2 ) is the solution of the differential equation (y + a t) y ' + (a y + b t) = 0 with initial condition y(0) = y_0, what are the values of a, b and y_0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: M(t,y) = y + at N(t,y) = ay + bt M_y = 1 N_t = b Since M_y must be equal to N_t in order for the DE to be exact, we can determine that b = 1. f(t,y) = ʃMdt = yt + at^2/2 + g(y) δf/δy = t + dg/dy ay + t = N(t,y) ay + t = t + dg/dy dg/dy = ay g(y) = ʃaydy = ay^2/2 f(t,y) = yt + at^2/2 + ay^2/2 f(0,y_0) = ay_0^2/2 f(t,y) = yt + at^2/2 + ay^2/2 = ay_0^2/2 y = [a^2*y_0^2 - ((a^2-1)t^2)^(1/2) - t]/a if t = 0, y = y_0^2 Matching this up to the given solution, a =1, leaving y_0 = (4-t^2)^(1/2), meaning that y_0 = 2. a = 1 b = 1 y_0 = 2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think this is the right way to go at this problem. If I’ve managed to carry all my signs and my calculations are correct, I believe I have the answer. I’m unsure if this is the best way to go about this problem, or if there is a better way. ------------------------------------------------ Self-critique rating: 3"

Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!

`gr31

Query 06

#$&*

course MTH 279

Query 06 Differential Equations*********************************************

Question: 3.3.4. If the equation is exact, solve the equation (6 t + y^3 ) y ' + 3 t^2 y = 0, with y(2) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3t^2*y + (6t+y^3)dy/dt = 0

M = 3t^2*y; N = 6t+y^3

M_y = 3t^2; N_t = 6

The equation is not exact.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

*********************************************

Question: If the equation is exact, solve the equation (6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 = -t, with y(2) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(6t + 3t^3)dy/dt +(6y+9/2t^2y^2+t) = 0

M = (6y+9/2t^2y^2+t) ; N = (6t+3t^3)

M_y = 9yt^2+6 ; N_t = 9t^2+6

The equation is not exact.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: 3.3.6. If the equation is exact, solve the equation y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2) with y(0) = pi.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(t cos(t y) + 2 y e^y^2) dy/dt + ( y cos(t y) + 1) = 0

M = y*cos(y*t)+1 ; N = t*cos(t*y) + 2y*e^(y^2)

M_y = cos(t*y)-y*t*sin(t*y) ; N_t = cos(t*y)-t*y*sin(t*y)

M_y = N_t

The equation is exact.

f(t,y) = ʃMdt = sin(t*y)+t + h(y)

δf/δy = tcos(ty)+dh/dy

tcos(ty) + dh/dy = tcos(ty)+2ye^(y^2)

dh/dy = 2ye^(y^2)

h(y) = e^(y^2)

f(t,y) = sin(ty)+t+e^(y^2) = C

f(0,π) = sin(0) + 0 + e^(π^2) = C

C = e^(π^2)

f(t,y) = sin(ty) + t + e^(y^2) = e^(π^2)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

*********************************************

Question: 3.3.10. If N(y, t) * y ' + t^2 + y^2 sin(t) = 0 is exact, what is the most general possible form of N(y, t)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If the equation is exact:

M(y,t) = t^2 + y^2sin(t)

M_y = 2ysin(t)

f(y,t) = ʃMdt = t^3/3-cos(t)y^2 + g(y)

δf/δy = -2ycos(t) + dg/dy

δf/δy = N(y,t)

N(y,t) = -2ycos(t) + dg/dy

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

*********************************************

Question: 3.3.12. If y = -t - sqrt( 4 - t^2 ) is the solution of the differential equation (y + a t) y ' + (a y + b t) = 0 with initial condition y(0) = y_0, what are the values of a, b and y_0?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

M(t,y) = y + at

N(t,y) = ay + bt

M_y = 1

N_t = b

Since M_y must be equal to N_t in order for the DE to be exact, we can determine that b = 1.

f(t,y) = ʃMdt = yt + at^2/2 + g(y)

δf/δy = t + dg/dy

ay + t = N(t,y)

ay + t = t + dg/dy

dg/dy = ay

g(y) = ʃaydy = ay^2/2

f(t,y) = yt + at^2/2 + ay^2/2

f(0,y_0) = ay_0^2/2

f(t,y) = yt + at^2/2 + ay^2/2 = ay_0^2/2

y = [a^2*y_0^2 - ((a^2-1)t^2)^(1/2) - t]/a

if t = 0, y = y_0^2

Matching this up to the given solution, a =1, leaving y_0 = (4-t^2)^(1/2), meaning that y_0 = 2.

a = 1

b = 1

y_0 = 2

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I think this is the right way to go at this problem. If I’ve managed to carry all my signs and my calculations are correct, I believe I have the answer. I’m unsure if this is the best way to go about this problem, or if there is a better way.

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Self-critique rating: 3"

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Your work looks good. Let me know if you have any questions. &#