#$&*
course Mth 279
May 7 around 2:15pm.
Query 19 Differential Equations*********************************************
Question: Find the general solution of the equation
y '' + y = e^t sin(t).
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Linear non-homogeneous equation:
(gen. soln. of non-homo eq) = (gen. soln. of homo eq) + (particular soln. of non-homo)
Y(t) = Yc(t) + Yp(t)
Particular soln: Yp’’-Yp’-2Yp
Y’’ + y = 0
(lambda^2) + 1 = 0
No real roots, need to find complex roots:
Turn to quadratic equation: (-b +-sqrt(b^2-4ac))/(2a)
(+sqrt(4))/2 = 1 (lambda one)
(-sqrt(4))/2 = (lambda two)
Yc(t) = c1e^ lambda one + c2e^ lambda two
@& The general solution is c1 e^(lamda1 t) + c2 e^(lambda2 t) = c1 e^(it) + c2 e^-(it), which gives general real part A cos(t) + B sin(t).*@
Solve for particular now, find a “y=” that is a multiple and set it as Yp and plug into Yp’’-Yp’-2Yp. After you get the particular solution, you can find the gen soln of non-homo, Y(t) = Yc(t) + Yp(t).
@& Right. You will need to find a particular solution; and you need to put your complementary solution into the appropriate form.
The complementary solution is easily found to be y_C = A cos(t) + D sin(t).
Our particular solution can be expected to have the form
y_P = C e^t sin(t) + D e^t cos(t).
Taking the second derivative of this form we obtain
y_P '' = 2 C e^t cos(t) - 2 D e^t sin(t)
so our equation becomes
2 C e^t cos(t) - 2 D e^t sin(t) + C e^t sin(t) + D e^t cos(t) = e^t sin(t)
Dividing through by e^t and simplifying we obtain
(2 C + D) cos(t) + (C - 2 D) sin(t) = sin(t)
The coefficient of cos(t) must be 0, and the coefficient of sin(t) must be 1. We get
2C + D = 0
C - 2 D = 1.
These equations are easily solved, yielding
C = 1/5, D = -2/5
so that
y_P = 1/5 e^t sin(t) - 2/5 e^t cos(t)
Our general solution is therefore
y(t) = y_C (t) + y_P (t) = A cos(t) + B sin(t) + 1/5 e^t sin(t) - 2/5 e^t cos(t).
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question: Find the general solution of the equation
y '' + y ' = 6 t^2
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Linear non-homogeneous equation:
(gen. soln. of non-homo eq) = (gen. soln. of homo eq) + (particular soln. of non-homo)
Y(t) = Yc(t) + Yp(t)
Particular soln: Yp’’-Yp’-2Yp
Y’’ + y’ = 0
(lambda^2) + (lambda) + 0 = 0
Real roots: 0, and -1
(lambda -0)(lambda + 1)
Yc(t) = c1e^(-t) + c2e^(0)
Rewriting: c1e^(-t) + c2
Yp’’-Yp’-2Yp = 6t^2
This is where I get boggled down. How do I find a common y= that I can use in the equation? Would this one be t^2?! I know what to do after I find this though.
@& Close, but you need to use other powers of t as well in your trial solution, and you need arbitrary constants to evaluate.
Complete solution:
The characteristic equation for the complementary solution is
r^2 + r = 0
with solutions
r = 0, r= -1
yielding
y_C = A + B e^-t.
Our particular solution must yield 6 t^2.
There's a rule for this, but let's pretend we forgot it and use a little trial and error.
Trying first y_P = a t^2 + b t + c, we don't get too far. Our left-hand side includes y '' and y ' terms, neither of which would give us a multiple of t^2.
So we need at least one higher power of t. We try y_P = a t^3 + b t^2 + c t + d.
y_P ' = 3 a t^2 + 2 b t + c
y_P '' = 6 a t + 2 b
and our equation becomes
6 a t + 2 b + 3 a t^2 + 2 b t + c = 6 t^2.
Rearranging by powers of t we get
3 a t^2 + (6 a + 2 b) t + (2 b + c) = 6 t^2,
so that
3 a = 6
6 a + 2 b = 0
2 b + c = 0
with solutions
a = 2
b = -6
c = 12
Thus
y_P = 2 t^2 - 6 t + 12 + d,
where d can be any constant (since only y ' and y '' appear in the equation, any constant d will disappear).
We might as well let d = 0 so that
y_P = 2 t^2 - 6 t + 12
and our general solution is
y(t) = y_C + y_P = A + B e^-t + 2 t^2 - 6 t + 12
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question: Find the general solution of the equation
y '' + y ' = cos(t).
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Linear non-homogeneous equation:
(gen. soln. of non-homo eq) = (gen. soln. of homo eq) + (particular soln. of non-homo)
Y(t) = Yc(t) + Yp(t)
Particular soln: Yp’’-Yp’-2Yp
Y’’ + y’ = 0
(lambda^2) + (lambda) + 0 = 0
Real roots: 0, and -1
(lambda -0)(lambda + 1)
Yc(t) = c1e^(-t) + c2e^(0)
Rewriting: c1e^(-t) + c2
Yp’’-Yp’-2Yp = cos(t)
Same question as before...How do I find a common y= that I can use in the equation? Would this one be cos(t)?! I know what to do after I find this though, plug the derivatives into the particular solution equation and then you can solve for the gen soln.
@&
Our complementary solution is the same as in the preceding question
y_C = A + B e^-t
Our particular solution is expected to be of the form
y_P = C cos(t) + D sin(t)
so that
y_P ' = -C sin(t) + D cos(t)
and
y_P '' = -C cos(t) - D sin(t).
Substituting y_P into the equation we therefore obtain
-C cos(t) - D sin(t) - C sin(t) + D cos(t) = cos(t).
Rearranging we have
(-C + D) cos(t) + (-C - D ) sin(t) = cos(t)
so
-C + D = 1
-C - D = 0
with solutions
C = -1/2, D = 1/2.
Our particular solution is therefore
y_P = -1/2 cos(t) + 1/2 sin(t)
and our general solution is
y = y_C + y_P = A + B e^-t -1/2 cos(t) + 1/2 sin(t)
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question: Give the expected form of the particular solution to the given equation, but do not actually solve for the constants:
y '' - 2 y ' + 3 y = 2 e^-t cos(t) + t^2 + t e^(3 t)
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
There is no real roots, need to find complex roots (lambda1, lambda2)
Y(t) = c1e^(lambda1 t) + c2e^(lambda2 t)
Yp’’ + Yp’ + 2Yp = 2 e^-t cos(t) + t^2 + t e^(3 t)
Assume y= t? (same question as previous 2 questions)
Find y’’, y’ and plug y into Yp’’ + Yp’ + 2Yp and solve.
@&
The complementary solution is A e^(3 t) + B e^(-t)
The term 2 e^-t cos(t) is expected to result from a combination of multiples of e^-t cos(t) and e^(-t) sin (t).
The term t^2 is expected to result from a second-degree polynomial of the form a t^2 + b t + c, with a = 1. The constants b and c can be adjusted so that the various derivatives of the other terms cancel.
The term t e^(3 t) would be expected to result from a combination of multiples of t e^(3 t) and e^(3 t), except that e^(3 t) is a solution of the homogeneous equation. So we will try a combination of t^2 e^(3 t) and t e^(3 t).
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
@& You need to go through much more detail in your solutions, but you do have the correct overview. Check my notes.*@
#$&*
course Mth 279
May 7 around 2:15pm.
Query 19 Differential Equations*********************************************
Question: Find the general solution of the equation
y '' + y = e^t sin(t).
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Linear non-homogeneous equation:
(gen. soln. of non-homo eq) = (gen. soln. of homo eq) + (particular soln. of non-homo)
Y(t) = Yc(t) + Yp(t)
Particular soln: Yp’’-Yp’-2Yp
Y’’ + y = 0
(lambda^2) + 1 = 0
No real roots, need to find complex roots:
Turn to quadratic equation: (-b +-sqrt(b^2-4ac))/(2a)
(+sqrt(4))/2 = 1 (lambda one)
(-sqrt(4))/2 = (lambda two)
Yc(t) = c1e^ lambda one + c2e^ lambda two
@& The general solution is c1 e^(lamda1 t) + c2 e^(lambda2 t) = c1 e^(it) + c2 e^-(it), which gives general real part A cos(t) + B sin(t).*@
Solve for particular now, find a “y=” that is a multiple and set it as Yp and plug into Yp’’-Yp’-2Yp. After you get the particular solution, you can find the gen soln of non-homo, Y(t) = Yc(t) + Yp(t).
@& Right. You will need to find a particular solution; and you need to put your complementary solution into the appropriate form.
The complementary solution is easily found to be y_C = A cos(t) + D sin(t).
Our particular solution can be expected to have the form
y_P = C e^t sin(t) + D e^t cos(t).
Taking the second derivative of this form we obtain
y_P '' = 2 C e^t cos(t) - 2 D e^t sin(t)
so our equation becomes
2 C e^t cos(t) - 2 D e^t sin(t) + C e^t sin(t) + D e^t cos(t) = e^t sin(t)
Dividing through by e^t and simplifying we obtain
(2 C + D) cos(t) + (C - 2 D) sin(t) = sin(t)
The coefficient of cos(t) must be 0, and the coefficient of sin(t) must be 1. We get
2C + D = 0
C - 2 D = 1.
These equations are easily solved, yielding
C = 1/5, D = -2/5
so that
y_P = 1/5 e^t sin(t) - 2/5 e^t cos(t)
Our general solution is therefore
y(t) = y_C (t) + y_P (t) = A cos(t) + B sin(t) + 1/5 e^t sin(t) - 2/5 e^t cos(t).
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question: Find the general solution of the equation
y '' + y ' = 6 t^2
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Linear non-homogeneous equation:
(gen. soln. of non-homo eq) = (gen. soln. of homo eq) + (particular soln. of non-homo)
Y(t) = Yc(t) + Yp(t)
Particular soln: Yp’’-Yp’-2Yp
Y’’ + y’ = 0
(lambda^2) + (lambda) + 0 = 0
Real roots: 0, and -1
(lambda -0)(lambda + 1)
Yc(t) = c1e^(-t) + c2e^(0)
Rewriting: c1e^(-t) + c2
Yp’’-Yp’-2Yp = 6t^2
This is where I get boggled down. How do I find a common y= that I can use in the equation? Would this one be t^2?! I know what to do after I find this though.
@& Close, but you need to use other powers of t as well in your trial solution, and you need arbitrary constants to evaluate.
Complete solution:
The characteristic equation for the complementary solution is
r^2 + r = 0
with solutions
r = 0, r= -1
yielding
y_C = A + B e^-t.
Our particular solution must yield 6 t^2.
There's a rule for this, but let's pretend we forgot it and use a little trial and error.
Trying first y_P = a t^2 + b t + c, we don't get too far. Our left-hand side includes y '' and y ' terms, neither of which would give us a multiple of t^2.
So we need at least one higher power of t. We try y_P = a t^3 + b t^2 + c t + d.
y_P ' = 3 a t^2 + 2 b t + c
y_P '' = 6 a t + 2 b
and our equation becomes
6 a t + 2 b + 3 a t^2 + 2 b t + c = 6 t^2.
Rearranging by powers of t we get
3 a t^2 + (6 a + 2 b) t + (2 b + c) = 6 t^2,
so that
3 a = 6
6 a + 2 b = 0
2 b + c = 0
with solutions
a = 2
b = -6
c = 12
Thus
y_P = 2 t^2 - 6 t + 12 + d,
where d can be any constant (since only y ' and y '' appear in the equation, any constant d will disappear).
We might as well let d = 0 so that
y_P = 2 t^2 - 6 t + 12
and our general solution is
y(t) = y_C + y_P = A + B e^-t + 2 t^2 - 6 t + 12
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question: Find the general solution of the equation
y '' + y ' = cos(t).
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Linear non-homogeneous equation:
(gen. soln. of non-homo eq) = (gen. soln. of homo eq) + (particular soln. of non-homo)
Y(t) = Yc(t) + Yp(t)
Particular soln: Yp’’-Yp’-2Yp
Y’’ + y’ = 0
(lambda^2) + (lambda) + 0 = 0
Real roots: 0, and -1
(lambda -0)(lambda + 1)
Yc(t) = c1e^(-t) + c2e^(0)
Rewriting: c1e^(-t) + c2
Yp’’-Yp’-2Yp = cos(t)
Same question as before...How do I find a common y= that I can use in the equation? Would this one be cos(t)?! I know what to do after I find this though, plug the derivatives into the particular solution equation and then you can solve for the gen soln.
@&
Our complementary solution is the same as in the preceding question
y_C = A + B e^-t
Our particular solution is expected to be of the form
y_P = C cos(t) + D sin(t)
so that
y_P ' = -C sin(t) + D cos(t)
and
y_P '' = -C cos(t) - D sin(t).
Substituting y_P into the equation we therefore obtain
-C cos(t) - D sin(t) - C sin(t) + D cos(t) = cos(t).
Rearranging we have
(-C + D) cos(t) + (-C - D ) sin(t) = cos(t)
so
-C + D = 1
-C - D = 0
with solutions
C = -1/2, D = 1/2.
Our particular solution is therefore
y_P = -1/2 cos(t) + 1/2 sin(t)
and our general solution is
y = y_C + y_P = A + B e^-t -1/2 cos(t) + 1/2 sin(t)
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
Question: Give the expected form of the particular solution to the given equation, but do not actually solve for the constants:
y '' - 2 y ' + 3 y = 2 e^-t cos(t) + t^2 + t e^(3 t)
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
There is no real roots, need to find complex roots (lambda1, lambda2)
Y(t) = c1e^(lambda1 t) + c2e^(lambda2 t)
Yp’’ + Yp’ + 2Yp = 2 e^-t cos(t) + t^2 + t e^(3 t)
Assume y= t? (same question as previous 2 questions)
Find y’’, y’ and plug y into Yp’’ + Yp’ + 2Yp and solve.
@&
The complementary solution is A e^(3 t) + B e^(-t)
The term 2 e^-t cos(t) is expected to result from a combination of multiples of e^-t cos(t) and e^(-t) sin (t).
The term t^2 is expected to result from a second-degree polynomial of the form a t^2 + b t + c, with a = 1. The constants b and c can be adjusted so that the various derivatives of the other terms cancel.
The term t e^(3 t) would be expected to result from a combination of multiples of t e^(3 t) and e^(3 t), except that e^(3 t) is a solution of the homogeneous equation. So we will try a combination of t^2 e^(3 t) and t e^(3 t).
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
@& You need to go through much more detail in your solutions, but you do have the correct overview. Check my notes.*@