Query 06

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course Mth 279

Sorry this took a while, I was stuck on 3 for a little while, but I think I got it.4/10 7

Query 06 Differential Equations

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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.

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Your solution:

the equation is not exact.

using the differential equation partial M / partial y = partial N / partial t

it is exact if and only if that statement is true.

so in this equation, M(t,y) = 3t^2y ; N(t,y) = 6t+y^3

taking partial integration of M / y and N/t

we get that 1 does not equal 6. therefore, this equation is not exact

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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. *

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Your solution:

This equation is indeed again, not exact. therefore I can not solve it as an exact equation.

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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.

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Your solution:

rearanging to get to the m + n dy = 0 form . so the partial of m wrt y is -tysin(ty) + cos(ty) and the partial of N wrt t is also -tysin(ty) + cos(ty)

these equal eachother, meaning they are equal, so we solve for psi. set psi_t = to our origional M equation of ycos(ty) + 1.

notice, we can set d/dy psi = psi_y + psi_t*y',

then integrate that function wrt t

to get psi_t = sin(ty) + t + h(y), we are calling h(y) our arbitrary constant that is a function of y, so that when we take the derivative wrt y, we have h(y)'

now integrate this wrt y, we have -cos(ty)/t + ty + h(y)'

knowing from our origional re-written equation d/dy psi = psi_y + psi_t*y', we can set h(y)' = psi_y from before.

so rearange to solve for h(y)' = -cos(ty) + 2ye^(y^2) -ty + cos(ty)/t

now integrate wrt y to solve for h(y) = -sin(ty)/t + e^(y^2) -ty^/2 + sin(ty)/t^2 + c

now this is where I'm confused, I know that psi = c, so the point of finding c is to find out what psi equals, but i'm unsure of how to use my initial conditions of y(0) = pi

because in this h(y) equation, I have t values, and if I only sub pi into the h(y) equation, i'm left with a variable in there, therefore, h(y) is not a constant.

help? maybe an intermediate step was messed up?

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You're on the right track. A little off with the integration.

Detailed solution:

We test the equation to see if it is exact, i.e., of the form

M dy + N dt = 0

with M_t = N_x.

The equation can be rearranged to

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

so it is of the form M dy + N dt = 0 with

M = t cos(t y) + 2y e^(y^2)

and

N = y cos(t y) + 1

M_t = cos(t y) - y t sin(t y)

and

N_y = cos(t y) - y t sin(t y)

These are equal, so our equation is of the form

dF = 0

with

F_y = t cos(t y) + 2y e^(y^2)

and

F_t = y cos(t y) + 1.

F_y = t cos(t y) + 2y e^(y^2) implies that

F = integral( (t cos(t y) + 2 y e^(y^2) ) dy = -sin(t y) + e^(y^2) + g(t),

where g(t) is our integration constant (y being the variable of integration, any function of t has derivative zero with respect to y and so is constant in this integral).

F_t = y cos(t y) + 1 implies that

F = integral( (y cos(t y) + 1) dt) = -sin(t y) + h(y),

where h(y) is constant with respect to t, the variable of integration.

If h(y) = e^(y^2) and g(t) = 0, our result is

F = -sin(t y) + e^(y^2) .

dF = 0 means that

F = c,

where c is constant, so

-sin(t y) + e^(y^2) = c.

The initial condition is that y(0) = pi, so we have

-sin(0) + e^(pi^2) = c

and c = e^(pi^2).

Our implicit solution is therefore given by the equation

-sin(t y) + e^(y^2) = e^(pi^2).

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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)?

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Your solution:

just looking at the value of M(yt) = t^2 +y^2sin(t), the partial of m wrt y is 2ysin(t)

so if it is exact, then partial of N wrt t should equal 2ysin(t), so integrate this wrt t to get the origional N(ty) equation

integrated wrt 2y int(sin(t) dt ) = -2ycos(t) + c

so the most general possible form of N(ty) is -2ycos(t) + c.....not sure if the c should be in there. but thats basically it.

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When you integrate with respect to t, since any function of y alone is constant with respect to t, your integration constant is a general function of y.

In detail:

The equation is of the form

N(y, t) dy + (t^2 + y^2) sin(t) dt = 0

It is exact if N_t = M_y, where

M_y = 2 y sin(t).

N_t = M_y so that

N = integral(M_y dt) = integral ( 2 y sin(t) dt ) = -2 y cos(t) + g(y),

where g(y) is the most general integration constant for integration with respect to t.

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Given Solution:

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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?

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Your solution:

just looking at the equation; -b +- sqrt(b^2-4ac ) / 2a

then a has to be 1

and b has to be 2

as for y_0, it should be -1 - sqrt(3)

@&

It's a little more complicated than that:

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?

This equation is exact, with M = (y + a t) so that M_y = a, and N = (a y + b t) so that N_t = a.

Thus the equation is of the form dF = 0 with

F = integral( (y + at) dy) = y^2 / 2 + a t y + g(t)

and

F = integral( (a y + b t) dt) = a y t + b t^2 / 2 + h(y).

We reconcile the two forms of F by letting g(t) = b t^2 / 2 and h(y) = y^2 / 2, so that

F = y^2 / 2 + b t^2 / 2 + a t y.

dF = 0 has solution F = c, where c is constant, so the implicit solution is

y^2 / 2 + b t^2 / 2 + a t y = c.

If y = -t - sqrt( 4 - t^2 ) then y^2 = t^2 + (4 - t^2) + 2 t sqrt(4 - t^2) = 4 + 2 t sqr(4 - t^2). Substituting for y in the equation we therefore obtain

4 + 2 t sqr(4 - t^2) + b t^2 / 2 + a t ( -t - sqrt( 4 - t^2) ) = c

(2 t - a t) sqrt( 4 - t^2) + (b/2 - a) t^2 + 4 = c

Thus

2 t - a t = 0

b/2 - a = 0

c = 4

and it follows that

a = 2, b = 4

and when t = 0 we have

y_0^2 / 2 = 4

so that y_0 = 2 sqrt(2).

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Given Solution:

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You're on the right track. Check my notes.

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