Find the particular solution of, $y=Ce^{-2x}+De^{-3x}+cos(x)+sin(x)$ [on hold]
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Where, $y = 1$, $dy/dx = 0$ when $x = 0$.
I've tried using simultaneous equations but keep getting $0$ as an answer for both constants, not sure how else to proceed.
differential-equations
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put on hold as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher yesterday
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Where, $y = 1$, $dy/dx = 0$ when $x = 0$.
I've tried using simultaneous equations but keep getting $0$ as an answer for both constants, not sure how else to proceed.
differential-equations
New contributor
RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, max_zorn, José Carlos Santos, Christopher
If this question can be reworded to fit the rules in the help center, please edit the question.
Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
– RocketKangaroo
2 days ago
I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
– projectilemotion
2 days ago
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
2 days ago
add a comment |
up vote
0
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up vote
0
down vote
favorite
Where, $y = 1$, $dy/dx = 0$ when $x = 0$.
I've tried using simultaneous equations but keep getting $0$ as an answer for both constants, not sure how else to proceed.
differential-equations
New contributor
RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Where, $y = 1$, $dy/dx = 0$ when $x = 0$.
I've tried using simultaneous equations but keep getting $0$ as an answer for both constants, not sure how else to proceed.
differential-equations
differential-equations
New contributor
RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
mrtaurho
2,3691726
2,3691726
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asked 2 days ago


RocketKangaroo
163
163
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New contributor
RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, max_zorn, José Carlos Santos, Christopher
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, max_zorn, José Carlos Santos, Christopher
If this question can be reworded to fit the rules in the help center, please edit the question.
Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
– RocketKangaroo
2 days ago
I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
– projectilemotion
2 days ago
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
2 days ago
add a comment |
Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
– RocketKangaroo
2 days ago
I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
– projectilemotion
2 days ago
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
2 days ago
Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
– RocketKangaroo
2 days ago
Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
– RocketKangaroo
2 days ago
I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
– projectilemotion
2 days ago
I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
– projectilemotion
2 days ago
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
2 days ago
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
2 days ago
add a comment |
1 Answer
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3
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From the initial conditions,
$$begin{cases}1=C+D+1,
\0=-2C-3D+1.end{cases}$$
$C=D=0$ is obviously not a solution.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
From the initial conditions,
$$begin{cases}1=C+D+1,
\0=-2C-3D+1.end{cases}$$
$C=D=0$ is obviously not a solution.
add a comment |
up vote
3
down vote
From the initial conditions,
$$begin{cases}1=C+D+1,
\0=-2C-3D+1.end{cases}$$
$C=D=0$ is obviously not a solution.
add a comment |
up vote
3
down vote
up vote
3
down vote
From the initial conditions,
$$begin{cases}1=C+D+1,
\0=-2C-3D+1.end{cases}$$
$C=D=0$ is obviously not a solution.
From the initial conditions,
$$begin{cases}1=C+D+1,
\0=-2C-3D+1.end{cases}$$
$C=D=0$ is obviously not a solution.
answered 2 days ago
Yves Daoust
121k668216
121k668216
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add a comment |
Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
– RocketKangaroo
2 days ago
I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
– projectilemotion
2 days ago
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
2 days ago