Query_07

#$&*

course Mth 279

3/20

Query 07 Differential Equations*********************************************

Question: 3.4.2. Solve the equation y ' = 2 t y ( 1 - y), with y(0) = -1, as a Bernoulli equation.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

->Bernoulli equation form: y' + p(t)y = q(t)y^n

y ' = 2 t y ( 1 - y)

y ' = 2 t y - 2ty^2

y ' - 2 t y = - 2ty^2

This equation is now in the correct form.

p(t) = -2t

q(t) = -2t

n = 2

Using v = y^m, which means y = v^(1/m), and m = (1 - n) from the Bernoulli equation.

The equation becomes:

dv/dt + (1-n)p(t)v = (1 - n)q(t)

dv/dt + (1-2)(-2t)v = (1 - 2)(-2t)

dv/dt + (-1)(-2t)v = (- 1)(-2t)

dv/dt + (2t)v = (2t)

This Equation is now First Order linear, and can be more easily solved

dv/dt + (2t)v = (2t)

dv/dt = (2t) - (2t)v

dv/dt = (2t) (1 - v)

dv/(1 - v) = 2t dt

Integration of both sides yeilds:

ln(1 - v) = t^2 + c

1 - v = Ce^(t^2)

@&

The derivative of ln | 1 - v | is - 1 / (1 - v)

The correct antiderivative would be the n negative of this one.

That negative could as well go on the other side, changing the sign of t^2.

*@

v = 1 - Ce^(t^2)

Remember that y = v^(1/m) = v^(1/(1 - n))

y = v^(1/(1 - 2)) = v^(-1)

y = (1 - Ce^(t^2))^(-1) = 1/(1 - Ce^(t^2))

-----------------------------------------------------

Now that we have this Equation lets solve for C with y(0) = -1

y(0) = 1/(1 - Ce^(0^2)) = 1/(1 - Ce^(0)) = 1/(1 - C))

1/(1 - C) = -1

-(1 - C) = 1

C - 1 = 1

C = 2

The Equation becomes:

y = 1/(1 - 2e^(t^2))

@&

I don't believe this will check out, due to the sign error I pointed out above.

However that error doesn't affect the value of the constant, and you get the very similar solution

y = 1/(1 - 2e^(-t^2))

*@

confidence rating #$&*:

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

3

.............................................

Given Solution:

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

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

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

Question: 3.4.6. Solve the equation y ' - y = t y^(1/3), y(0) = -9 as a Bernoulli equation.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

->Bernoulli equation form: y' + p(t)y = q(t)y^n

y ' - y = t y^(1/3)

Since the Equation is already in the form:

p(t) = -1

q(t) = t

n = 1/3

Using v = y^m, which means y = v^(1/m), and m = (1 - n) from the Bernoulli equation.

The equation becomes:

dv/dt + (1-n)p(t)v = (1 - n)q(t)

dv/dt + (1-(1/3))(-1)v = (1 - (1/3))(t)

dv/dt + (2/3)(-1)v = (2/3)(t)

dv/dt + (- 2/3)v = (2/3)(t)

This Equation can now be solved using a previous method.

I chose to use u(t) = e^(int(p(t))dt) for the form y' + p(t)y = g(t)

For this equation:

p(t) = (-2/3)

int(p(t)) dt = (-2/3)t = -2t/3

u(t) = e^(-2t/3)

v' + (- 2/3)v = (2/3)(t)

v' e^(-2t/3) + (- 2/3)ve^(-2t/3) = (2/3)(t)e^(-2t/3)

v' e^(-2t/3) - (2/3)ve^(-2t/3) = (2/3)(t)e^(-2t/3)

(ve^(-2t/3))' = (2/3)te^(-2t/3)

Integration of both sides yeilds:

ve^(-2t/3) = Int ((2/3)te^(-2t/3)) dt

Integration of the Right hand side required Integration by parts

udv = uv - Int(vdu)

u = t

dv = e^(-2t/3) dt

du = dt

v = Int(dv) = (-3/2)e^(-2t/3)

The Equation becomes:

ve^(-2t/3) = (2/3)((-3/2)te^(-2t/3) + (3/2) Int(e^(-2t/3)) dt)

ve^(-2t/3) = -te^(-2t/3) + (-3/2)e^(-2t/3)

ve^(-2t/3) = -te^(-2t/3) - (3/2)e^(-2t/3)

Dividing Out u(t) = e^(-2t/3), the Equation becomes:

v = -t - (3/2) + c

Remember that y = v^(1/m) = v^(1/(1 - n))

y = v^(1/(1 - 1/3)) = v^(3/2)

y = (-t - (3/2) + c)^(3/2)

-----------------------------------------------------

Now that we have this Equation lets solve for C with y(0) = -9

y(0) = (-0 - (3/2) + c)^(3/2)

(- (3/2) + c)^(3/2) = -9

(- (3/2) + c)^(3) = 81

(- (3/2) + c) = cube root (81)

C = cube root (81) + (3/2) = 5.83 approx.

Equation becomes:

y = (-t - (3/2) + cube root (81) + (3/2))^(3/2)

y = (-t + cube root (81))^(3/2)

y = (cube root (81) - t)^(3/2)

confidence rating #$&*:

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

3

.............................................

Given Solution:

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

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

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

Question: 3.4.8. Solve the equation y ' = - (y + 1) + t ( y + 1)^(-2) as a Bernoulli equation.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

This one in particular I could not figure out how to get it into the form:

->Bernoulli equation form: y' + p(t)y = q(t)y^n

If its possible to get it into this form, could you show me how.

As you can see in the other 2 problems, I can work this one just as easily as the other 2, once in that form.

I can rework this one after I see it in that form bc, I have tried a few things but I cannot seem to get that form.

What I was getting was mult. 'y' s and a y^(-1) and y^(-2) in the problem if it helps.

@&

You can use the substitution u = y + 1.

u ' = y ' so the new equation is

u ' = - u + t u^(-2),

which is easily seem to be a Bernoulli equation.

*@

confidence rating #$&*:

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

0

.............................................

Given Solution:

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

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:"

@&

Not bad. Check my notes.

*@