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Differential Equations I - Homework Set 10 | MATH 231, Assignments of Differential Equations

Material Type: Assignment; Class: Differential Equations I; Subject: Mathematics; University: University of Tennessee - Knoxville; Term: Unknown 2008;

Typology: Assignments

Pre 2010

Uploaded on 08/27/2009

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Math 231, Spring 2008- HW set 10 (due 4/24) (some problems
are from [Tenenbaum-Pollard])
1. The following first-order equations for y=y(t) are of ‘homogeneous
type’. Find the ‘general solution’, or find the solution to the initial-value
problem (in implicit form if necessary).
(i)ty0ytsin(y/t) = 0 Ans.y = 2tarctan(Ct)
(ii)ydt +tlog( y
t)dy 2tdy = 0 Ans.y =C(1 + log t/y)
(iii)y0=ty y2
t2, y(1) = 1 Ans.y =t/(1 + log t), t > 0
(iv)(t2+y2)dt = 2tydy, y(1) = 0 Ans.y =pt2+t, t (−∞,1]
2. For the following equations, state for which y0the given initial-value
problem is solvable, and what the interval of definition of the solution would
be. (It is possible the answer can’t be more precise than ‘some open interval
including t0’).
(i)ty0ytsin(y/t) = 0, y(1) = y0
(ii)(2t2+ 1)y00 4ty0+ 4y= 0, y(0) = y0
(iii)t2y00 +ty0y=t2ety(1) = y0
(iv)ydt +tlog( y
t)dt 2tdy = 0 y(1) = y0
3. For the following second-order homogeneous equations (y=y(t)),
a solution is given. Use the ‘reduction of order’ method to find a second,
linearly independent solution.
(i)2t2y00 + 3ty0y= 0, y1=t
(ii)(2t2+ 1)y00 4ty0+ 4y= 0, y1=t
4. For equation (i) in problem 3, reduce to ‘standard form’ (where the
coefficient of y00 is 1), then apply the Liouville transformation:
y(t) = z(t)e1
2P(t), P (t) = Zp(t)dt,
1
pf2

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Math 231, Spring 2008- HW set 10 (due 4/24) (some problems are from [Tenenbaum-Pollard])

  1. The following first-order equations for y = y(t) are of ‘homogeneous type’. Find the ‘general solution’, or find the solution to the initial-value problem (in implicit form if necessary).

(i)ty′^ − y − t sin(y/t) = 0 Ans.y = 2t arctan(Ct)

(ii)ydt + t log( y t

)dy − 2 tdy = 0 Ans.y = C(1 + log t/y)

(iii)y′^ =

ty − y^2 t^2 , y(1) = 1 Ans.y = t/(1 + log t), t > 0

(iv)(t^2 + y^2 )dt = 2tydy, y(−1) = 0 Ans.y =

t^2 + t, t ∈ (−∞, −1]

  1. For the following equations, state for which y 0 the given initial-value problem is solvable, and what the interval of definition of the solution would be. (It is possible the answer can’t be more precise than ‘some open interval including t 0 ’). (i)ty′^ − y − t sin(y/t) = 0, y(1) = y 0 (ii)(2t^2 + 1)y′′^ − 4 ty′^ + 4y = 0, y(0) = y 0 (iii)t^2 y′′^ + ty′^ − y = t^2 e−t^ y(1) = y 0

(iv)ydt + t log( y t )dt − 2 tdy = 0 y(1) = y 0

  1. For the following second-order homogeneous equations (y = y(t)), a solution is given. Use the ‘reduction of order’ method to find a second, linearly independent solution.

(i)2t^2 y′′^ + 3ty′^ − y = 0, y 1 =

t

(ii)(2t^2 + 1)y′′^ − 4 ty′^ + 4y = 0, y 1 = t

  1. For equation (i) in problem 3, reduce to ‘standard form’ (where the coefficient of y′′^ is 1), then apply the Liouville transformation:

y(t) = z(t)e−^

1 2 P^ (t), P (t) =

p(t)dt,

to find an equivalent equation for z(t) of the form:

z′′^ + v(t)z = 0

(that is, find v(t) explicitly. Note p(t) = 3/ 2 t here.)

  1. Use variation of parameters to find a particular solution to the non- homogeneous equations below (for y = y(t)), then write down the general so- lution (for equations with non-constant coefficients, two solutions y 1 (t), y 2 (t) of the homogeneous problem are given):

(i) y′′^ − 3 y′^ + 2y = cos(e−t) Ans.yp(t) = −e^2 t^ cos(e−t).

(ii)t^2 y′′^ − ty′^ + y = t y 1 = t, y 2 = t log t, Ans.yp(t) = t 2

(log t)^2

(iii)t^2 y′′^ + ty′^ − y = t^2 e−t, y 1 = t, y 2 = t−^1 Ans.yp(t) = e−t(1 + t−^1 )