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Particular solution to nonhomogenous differential equation $y''-6y'+9y=9t+3$.
Clash Royale CLAN TAG#URR8PPP
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I continue my quest of finishing a problem where I
previously calculated the unique solution
to a second order homogeneous differential equation and got the right answer in the end.
Now though, I have to Determine which of the following function expressions is a (particular) solution of the non-homogeneous differential equation
$$y''-6y'+9y=9t+3 tag1labeloriginal$$
with the following possible answers:
$$ beginarrayl
a) t+1 & d)e^3t+t+1 \
b) t^2+fract2-1 & e)-2te^3t+t+1 \
c) 9t+frac173 & f)e^3t-te^3t \
endarray $$
I understand that the general solution, $y$, equals $y_h+y_p$ where $y_h$ is the general solution to the homogeneous equation $bigl(y_h(t)=2e^3t+te^3t bigr)$ and $y_p$ is any particular solution to the non-homogeneous equation. And if I follow the theory of my text book and Sal Khan from Khan Academy then I have to make a shrewd guess of what $y_p$ might be. So, since the differential equation equals a polynomial my guess of $y_p$ will be the polynomial:
$y_p=At+B tag2labely_p$ which gives us,
$y_p'=A$ and
$y_p''=0$
I then substitute these back into the original equation (reforiginal) and get:
$$1cdot (0) -6 cdot (A) + 9cdot(At+B)=9t+3Rightarrow 9At-6A+9B=9t+3$$
We know that whatever the t-coefficients add up to be on the left-hand side they should be equal to the right-hand side, same goes for the constants. Let's calculate and solve for A and B:
$9A=9 Rightarrow A=1$
$-6 cdot 1 + 9B=3 Rightarrow 9B = 3 + 6 Rightarrow B=1$
We insert A and B into (refy_p)
$y_p=t+1$
$y=y_h+y_p=bigl(2e^3t+te^3tbigr)+(t+1) tag3labely$
Final question: How would $y$ look when added together and why does the results-list say the correct answers are $a), d)$ and $e)$?
Thanks in advance =)
Edit:
I think that the professor wants the particular solution as well as the solution of the non-homogeneous D.E. because of the way he puts parentheses around the word particular, and that's why $a)$ is correct, but then I just need to understand how $d)$ and $e)$ are results as well.
calculus differential-equations
edited Aug 3 at 12:16
user 108128
18.7k41544
asked Aug 3 at 10:29
JayFreddy
244
 |Â
show 6 more comments
up vote
1
down vote
favorite
I continue my quest of finishing a problem where I
previously calculated the unique solution
to a second order homogeneous differential equation and got the right answer in the end.
Now though, I have to Determine which of the following function expressions is a (particular) solution of the non-homogeneous differential equation
$$y''-6y'+9y=9t+3 tag1labeloriginal$$
with the following possible answers:
$$ beginarrayl
a) t+1 & d)e^3t+t+1 \
b) t^2+fract2-1 & e)-2te^3t+t+1 \
c) 9t+frac173 & f)e^3t-te^3t \
endarray $$
I understand that the general solution, $y$, equals $y_h+y_p$ where $y_h$ is the general solution to the homogeneous equation $bigl(y_h(t)=2e^3t+te^3t bigr)$ and $y_p$ is any particular solution to the non-homogeneous equation. And if I follow the theory of my text book and Sal Khan from Khan Academy then I have to make a shrewd guess of what $y_p$ might be. So, since the differential equation equals a polynomial my guess of $y_p$ will be the polynomial:
$y_p=At+B tag2labely_p$ which gives us,
$y_p'=A$ and
$y_p''=0$
I then substitute these back into the original equation (reforiginal) and get:
$$1cdot (0) -6 cdot (A) + 9cdot(At+B)=9t+3Rightarrow 9At-6A+9B=9t+3$$
We know that whatever the t-coefficients add up to be on the left-hand side they should be equal to the right-hand side, same goes for the constants. Let's calculate and solve for A and B:
$9A=9 Rightarrow A=1$
$-6 cdot 1 + 9B=3 Rightarrow 9B = 3 + 6 Rightarrow B=1$
We insert A and B into (refy_p)
$y_p=t+1$
$y=y_h+y_p=bigl(2e^3t+te^3tbigr)+(t+1) tag3labely$
Final question: How would $y$ look when added together and why does the results-list say the correct answers are $a), d)$ and $e)$?
Thanks in advance =)
Edit:
I think that the professor wants the particular solution as well as the solution of the non-homogeneous D.E. because of the way he puts parentheses around the word particular, and that's why $a)$ is correct, but then I just need to understand how $d)$ and $e)$ are results as well.
calculus differential-equations
edited Aug 3 at 12:16
user 108128
18.7k41544
asked Aug 3 at 10:29
JayFreddy
244
I have got $$y_p=t+1$$ check it.
– Dr. Sonnhard Graubner
Aug 3 at 10:37
Yes that's what I got as well, and results-list says that it is correct but d) and e) are correct as well which I don''t quite understand..
– JayFreddy
Aug 3 at 10:40
Let $y=(C_1e^3t+C_2te^3t)$ is general solution and $y=t$ is particulat, then $y_1=2e^3t+5te^3t+t$ is a solution. So if we separate $y_1$ and write $y_1=e^3t+5te^3t+(t+e^3t)$, then $(t+e^3t)$ is a particular solution as well???
– user 108128
Aug 3 at 10:47
How did you get that $y_h=2e^3t + t e^3t$ without any initial data?
– Calvin Khor
Aug 3 at 10:48
While it is true that $y_p = t+1$ is a Particular Solution (in the sense that the phrase presumably has a mathematical definition), its also true that the answers $a,d,e$ are particular solutions (i.e., they are examples of solutions)
– Calvin Khor
Aug 3 at 10:51
 |Â
show 6 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I continue my quest of finishing a problem where I
previously calculated the unique solution
to a second order homogeneous differential equation and got the right answer in the end.
Now though, I have to Determine which of the following function expressions is a (particular) solution of the non-homogeneous differential equation
$$y''-6y'+9y=9t+3 tag1labeloriginal$$
with the following possible answers:
$$ beginarrayl
a) t+1 & d)e^3t+t+1 \
b) t^2+fract2-1 & e)-2te^3t+t+1 \
c) 9t+frac173 & f)e^3t-te^3t \
endarray $$
I understand that the general solution, $y$, equals $y_h+y_p$ where $y_h$ is the general solution to the homogeneous equation $bigl(y_h(t)=2e^3t+te^3t bigr)$ and $y_p$ is any particular solution to the non-homogeneous equation. And if I follow the theory of my text book and Sal Khan from Khan Academy then I have to make a shrewd guess of what $y_p$ might be. So, since the differential equation equals a polynomial my guess of $y_p$ will be the polynomial:
$y_p=At+B tag2labely_p$ which gives us,
$y_p'=A$ and
$y_p''=0$
I then substitute these back into the original equation (reforiginal) and get:
$$1cdot (0) -6 cdot (A) + 9cdot(At+B)=9t+3Rightarrow 9At-6A+9B=9t+3$$
We know that whatever the t-coefficients add up to be on the left-hand side they should be equal to the right-hand side, same goes for the constants. Let's calculate and solve for A and B:
$9A=9 Rightarrow A=1$
$-6 cdot 1 + 9B=3 Rightarrow 9B = 3 + 6 Rightarrow B=1$
We insert A and B into (refy_p)
$y_p=t+1$
$y=y_h+y_p=bigl(2e^3t+te^3tbigr)+(t+1) tag3labely$
Final question: How would $y$ look when added together and why does the results-list say the correct answers are $a), d)$ and $e)$?
Thanks in advance =)
Edit:
I think that the professor wants the particular solution as well as the solution of the non-homogeneous D.E. because of the way he puts parentheses around the word particular, and that's why $a)$ is correct, but then I just need to understand how $d)$ and $e)$ are results as well.
calculus differential-equations
edited Aug 3 at 12:16
user 108128
18.7k41544
asked Aug 3 at 10:29
JayFreddy
244
I continue my quest of finishing a problem where I
previously calculated the unique solution
to a second order homogeneous differential equation and got the right answer in the end.
Now though, I have to Determine which of the following function expressions is a (particular) solution of the non-homogeneous differential equation
$$y''-6y'+9y=9t+3 tag1labeloriginal$$
with the following possible answers:
$$ beginarrayl
a) t+1 & d)e^3t+t+1 \
b) t^2+fract2-1 & e)-2te^3t+t+1 \
c) 9t+frac173 & f)e^3t-te^3t \
endarray $$
I understand that the general solution, $y$, equals $y_h+y_p$ where $y_h$ is the general solution to the homogeneous equation $bigl(y_h(t)=2e^3t+te^3t bigr)$ and $y_p$ is any particular solution to the non-homogeneous equation. And if I follow the theory of my text book and Sal Khan from Khan Academy then I have to make a shrewd guess of what $y_p$ might be. So, since the differential equation equals a polynomial my guess of $y_p$ will be the polynomial:
$y_p=At+B tag2labely_p$ which gives us,
$y_p'=A$ and
$y_p''=0$
I then substitute these back into the original equation (reforiginal) and get:
$$1cdot (0) -6 cdot (A) + 9cdot(At+B)=9t+3Rightarrow 9At-6A+9B=9t+3$$
We know that whatever the t-coefficients add up to be on the left-hand side they should be equal to the right-hand side, same goes for the constants. Let's calculate and solve for A and B:
$9A=9 Rightarrow A=1$
$-6 cdot 1 + 9B=3 Rightarrow 9B = 3 + 6 Rightarrow B=1$
We insert A and B into (refy_p)
$y_p=t+1$
$y=y_h+y_p=bigl(2e^3t+te^3tbigr)+(t+1) tag3labely$
Final question: How would $y$ look when added together and why does the results-list say the correct answers are $a), d)$ and $e)$?
Thanks in advance =)
Edit:
I think that the professor wants the particular solution as well as the solution of the non-homogeneous D.E. because of the way he puts parentheses around the word particular, and that's why $a)$ is correct, but then I just need to understand how $d)$ and $e)$ are results as well.
calculus differential-equations
edited Aug 3 at 12:16
user 108128
18.7k41544
asked Aug 3 at 10:29
JayFreddy
244
edited Aug 3 at 12:16
user 108128
18.7k41544
edited Aug 3 at 12:16
user 108128
18.7k41544
edited Aug 3 at 12:16
user 108128
18.7k41544
18.7k41544
asked Aug 3 at 10:29
JayFreddy
244
asked Aug 3 at 10:29
JayFreddy
244
asked Aug 3 at 10:29
JayFreddy
244
244
I have got $$y_p=t+1$$ check it.
– Dr. Sonnhard Graubner
Aug 3 at 10:37
Yes that's what I got as well, and results-list says that it is correct but d) and e) are correct as well which I don''t quite understand..
– JayFreddy
Aug 3 at 10:40
Let $y=(C_1e^3t+C_2te^3t)$ is general solution and $y=t$ is particulat, then $y_1=2e^3t+5te^3t+t$ is a solution. So if we separate $y_1$ and write $y_1=e^3t+5te^3t+(t+e^3t)$, then $(t+e^3t)$ is a particular solution as well???
– user 108128
Aug 3 at 10:47
How did you get that $y_h=2e^3t + t e^3t$ without any initial data?
– Calvin Khor
Aug 3 at 10:48
While it is true that $y_p = t+1$ is a Particular Solution (in the sense that the phrase presumably has a mathematical definition), its also true that the answers $a,d,e$ are particular solutions (i.e., they are examples of solutions)
– Calvin Khor
Aug 3 at 10:51
 |Â
show 6 more comments
I have got $$y_p=t+1$$ check it.
– Dr. Sonnhard Graubner
Aug 3 at 10:37
Yes that's what I got as well, and results-list says that it is correct but d) and e) are correct as well which I don''t quite understand..
– JayFreddy
Aug 3 at 10:40
Let $y=(C_1e^3t+C_2te^3t)$ is general solution and $y=t$ is particulat, then $y_1=2e^3t+5te^3t+t$ is a solution. So if we separate $y_1$ and write $y_1=e^3t+5te^3t+(t+e^3t)$, then $(t+e^3t)$ is a particular solution as well???
– user 108128
Aug 3 at 10:47
How did you get that $y_h=2e^3t + t e^3t$ without any initial data?
– Calvin Khor
Aug 3 at 10:48
While it is true that $y_p = t+1$ is a Particular Solution (in the sense that the phrase presumably has a mathematical definition), its also true that the answers $a,d,e$ are particular solutions (i.e., they are examples of solutions)
– Calvin Khor
Aug 3 at 10:51
I have got $$y_p=t+1$$ check it.
– Dr. Sonnhard Graubner
Aug 3 at 10:37
I have got $$y_p=t+1$$ check it.
– Dr. Sonnhard Graubner
Aug 3 at 10:37
Yes that's what I got as well, and results-list says that it is correct but d) and e) are correct as well which I don''t quite understand..
– JayFreddy
Aug 3 at 10:40
Yes that's what I got as well, and results-list says that it is correct but d) and e) are correct as well which I don''t quite understand..
– JayFreddy
Aug 3 at 10:40
Let $y=(C_1e^3t+C_2te^3t)$ is general solution and $y=t$ is particulat, then $y_1=2e^3t+5te^3t+t$ is a solution. So if we separate $y_1$ and write $y_1=e^3t+5te^3t+(t+e^3t)$, then $(t+e^3t)$ is a particular solution as well???
– user 108128
Aug 3 at 10:47
Let $y=(C_1e^3t+C_2te^3t)$ is general solution and $y=t$ is particulat, then $y_1=2e^3t+5te^3t+t$ is a solution. So if we separate $y_1$ and write $y_1=e^3t+5te^3t+(t+e^3t)$, then $(t+e^3t)$ is a particular solution as well???
– user 108128
Aug 3 at 10:47
How did you get that $y_h=2e^3t + t e^3t$ without any initial data?
– Calvin Khor
Aug 3 at 10:48
How did you get that $y_h=2e^3t + t e^3t$ without any initial data?
– Calvin Khor
Aug 3 at 10:48
While it is true that $y_p = t+1$ is a Particular Solution (in the sense that the phrase presumably has a mathematical definition), its also true that the answers $a,d,e$ are particular solutions (i.e., they are examples of solutions)
– Calvin Khor
Aug 3 at 10:51
While it is true that $y_p = t+1$ is a Particular Solution (in the sense that the phrase presumably has a mathematical definition), its also true that the answers $a,d,e$ are particular solutions (i.e., they are examples of solutions)
– Calvin Khor
Aug 3 at 10:51
 |Â
show 6 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Let $y=C_1e^3t+C_2te^3t$ is the general solution of the equation and $y=t+1$ is the particular solution, then
$$y_0=C_1e^3t+C_2te^3t+t+1$$
is a particular solution for every specific numbers $C_1$ and $C_2$. For instance
$$y_0=6e^3t+5te^3t+(t+1)=5e^3t+5te^3t+(t+1+e^3t)$$
is a particular solution as well.
answered Aug 3 at 11:01
user 108128
18.7k41544
oh okay, I thought that I had to use the $y(t) = 2e^3t + te^3t$ and not $y(t)=c_1 e^3t+c_2 te^3t$, it makes sence now, thx =)
– JayFreddy
Aug 3 at 11:06
Good luck......
– user 108128
Aug 3 at 11:07
add a comment |Â
up vote
0
down vote
All the solution are in the form
$$y=y_h+y_p=bigl(c_1e^3t+c_2te^3tbigr)+(t+1)$$
and the only compatible answers are $a$,$d$ and $e$.
answered Aug 3 at 10:48
gimusi
63.7k73480
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $y=C_1e^3t+C_2te^3t$ is the general solution of the equation and $y=t+1$ is the particular solution, then
$$y_0=C_1e^3t+C_2te^3t+t+1$$
is a particular solution for every specific numbers $C_1$ and $C_2$. For instance
$$y_0=6e^3t+5te^3t+(t+1)=5e^3t+5te^3t+(t+1+e^3t)$$
is a particular solution as well.
answered Aug 3 at 11:01
user 108128
18.7k41544
oh okay, I thought that I had to use the $y(t) = 2e^3t + te^3t$ and not $y(t)=c_1 e^3t+c_2 te^3t$, it makes sence now, thx =)
– JayFreddy
Aug 3 at 11:06
Good luck......
– user 108128
Aug 3 at 11:07
add a comment |Â
up vote
1
down vote
accepted
Let $y=C_1e^3t+C_2te^3t$ is the general solution of the equation and $y=t+1$ is the particular solution, then
$$y_0=C_1e^3t+C_2te^3t+t+1$$
is a particular solution for every specific numbers $C_1$ and $C_2$. For instance
$$y_0=6e^3t+5te^3t+(t+1)=5e^3t+5te^3t+(t+1+e^3t)$$
is a particular solution as well.
answered Aug 3 at 11:01
user 108128
18.7k41544
oh okay, I thought that I had to use the $y(t) = 2e^3t + te^3t$ and not $y(t)=c_1 e^3t+c_2 te^3t$, it makes sence now, thx =)
– JayFreddy
Aug 3 at 11:06
Good luck......
– user 108128
Aug 3 at 11:07
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $y=C_1e^3t+C_2te^3t$ is the general solution of the equation and $y=t+1$ is the particular solution, then
$$y_0=C_1e^3t+C_2te^3t+t+1$$
is a particular solution for every specific numbers $C_1$ and $C_2$. For instance
$$y_0=6e^3t+5te^3t+(t+1)=5e^3t+5te^3t+(t+1+e^3t)$$
is a particular solution as well.
answered Aug 3 at 11:01
user 108128
18.7k41544
Let $y=C_1e^3t+C_2te^3t$ is the general solution of the equation and $y=t+1$ is the particular solution, then
$$y_0=C_1e^3t+C_2te^3t+t+1$$
is a particular solution for every specific numbers $C_1$ and $C_2$. For instance
$$y_0=6e^3t+5te^3t+(t+1)=5e^3t+5te^3t+(t+1+e^3t)$$
is a particular solution as well.
answered Aug 3 at 11:01
user 108128
18.7k41544
answered Aug 3 at 11:01
user 108128
18.7k41544
answered Aug 3 at 11:01
user 108128
18.7k41544
answered Aug 3 at 11:01
user 108128
18.7k41544
18.7k41544
oh okay, I thought that I had to use the $y(t) = 2e^3t + te^3t$ and not $y(t)=c_1 e^3t+c_2 te^3t$, it makes sence now, thx =)
– JayFreddy
Aug 3 at 11:06
Good luck......
– user 108128
Aug 3 at 11:07
add a comment |Â
oh okay, I thought that I had to use the $y(t) = 2e^3t + te^3t$ and not $y(t)=c_1 e^3t+c_2 te^3t$, it makes sence now, thx =)
– JayFreddy
Aug 3 at 11:06
Good luck......
– user 108128
Aug 3 at 11:07
oh okay, I thought that I had to use the $y(t) = 2e^3t + te^3t$ and not $y(t)=c_1 e^3t+c_2 te^3t$, it makes sence now, thx =)
– JayFreddy
Aug 3 at 11:06
oh okay, I thought that I had to use the $y(t) = 2e^3t + te^3t$ and not $y(t)=c_1 e^3t+c_2 te^3t$, it makes sence now, thx =)
– JayFreddy
Aug 3 at 11:06
Good luck......
– user 108128
Aug 3 at 11:07
Good luck......
– user 108128
Aug 3 at 11:07
add a comment |Â
up vote
0
down vote
All the solution are in the form
$$y=y_h+y_p=bigl(c_1e^3t+c_2te^3tbigr)+(t+1)$$
and the only compatible answers are $a$,$d$ and $e$.
answered Aug 3 at 10:48
gimusi
63.7k73480
add a comment |Â
up vote
0
down vote
All the solution are in the form
$$y=y_h+y_p=bigl(c_1e^3t+c_2te^3tbigr)+(t+1)$$
and the only compatible answers are $a$,$d$ and $e$.
answered Aug 3 at 10:48
gimusi
63.7k73480
add a comment |Â
up vote
0
down vote
up vote
0
down vote
All the solution are in the form
$$y=y_h+y_p=bigl(c_1e^3t+c_2te^3tbigr)+(t+1)$$
and the only compatible answers are $a$,$d$ and $e$.
answered Aug 3 at 10:48
gimusi
63.7k73480
All the solution are in the form
$$y=y_h+y_p=bigl(c_1e^3t+c_2te^3tbigr)+(t+1)$$
and the only compatible answers are $a$,$d$ and $e$.
answered Aug 3 at 10:48
gimusi
63.7k73480
answered Aug 3 at 10:48
gimusi
63.7k73480
answered Aug 3 at 10:48
gimusi
63.7k73480
answered Aug 3 at 10:48
gimusi
63.7k73480
63.7k73480
add a comment |Â
add a comment |Â
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I have got $$y_p=t+1$$ check it.
– Dr. Sonnhard Graubner
Aug 3 at 10:37
Yes that's what I got as well, and results-list says that it is correct but d) and e) are correct as well which I don''t quite understand..
– JayFreddy
Aug 3 at 10:40
Let $y=(C_1e^3t+C_2te^3t)$ is general solution and $y=t$ is particulat, then $y_1=2e^3t+5te^3t+t$ is a solution. So if we separate $y_1$ and write $y_1=e^3t+5te^3t+(t+e^3t)$, then $(t+e^3t)$ is a particular solution as well???
– user 108128
Aug 3 at 10:47
How did you get that $y_h=2e^3t + t e^3t$ without any initial data?
– Calvin Khor
Aug 3 at 10:48
While it is true that $y_p = t+1$ is a Particular Solution (in the sense that the phrase presumably has a mathematical definition), its also true that the answers $a,d,e$ are particular solutions (i.e., they are examples of solutions)
– Calvin Khor
Aug 3 at 10:51