Query 05

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course mth 279

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))

b)interval of existence: -infinity sign

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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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

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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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 #$&*:

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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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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:

seeing that the problem is a seperable de

in this form:n(y)*y'+m(t)=0

now by rewriting the equation

y'=y^2+2y+2

y'/y^2+2y+2=1

y'/(y^2+2y+2) -1=0

n(y)=1/(y^2+2y+2) m(t)=-1

by taking anti derivatives:

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

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

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

INTEGRAL sign 1/y^2+2y+2 dy

=integral sign 1/(y^2+2y+1)1 dy

=integral sign 1/(y+1)^2+1 dy

let u=y+1 du=dy

integral sign 1/(y+1)^2+1 dy

=tan^-1 (u)

=tan^-1 (y+1)

now plugging back in the soln

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

tan^-1 (y+1)-t=C

tan^-1 (y+1)-t=C

by apllyin the intial conditions y(0)=0 can solve for C

tan^-1 (0+1)-0=C

tan^-1 (1)=C

pie/4=C

by plugging in C,and getting the impct soln

tan^-1 (y+1)-t=pie/4

now solving for exp soln

tan^-1 (y+1)-t=pie/4

tan^-1 (y+1)=pie/4+t

y+1=tan(pie/4+t)

y=tan(pie/4+t)-1

y(t)=tan(pie/4+t)-1

interval of existence:pie-4/4

@&

Good, but the right-hand side of the equation included sin(t) as a factor.

*@

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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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

There is only one graph that has a negative slope which is graph c

y'=y^3

as y increases the value of y' gets larger:graph a

y'=y(4-y)

graph b

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