Find the particular solution of, $y=Ce^{-2x}+De^{-3x}+cos(x)+sin(x)$ [closed]
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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
edited Nov 19 at 19:54
mrtaurho
2,7691927
asked Nov 19 at 19:31
RocketKangaroo
163
closed as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher Nov 20 at 13:24
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.
add a comment |
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
edited Nov 19 at 19:54
mrtaurho
2,7691927
asked Nov 19 at 19:31
RocketKangaroo
163
closed as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher Nov 20 at 13:24
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
Nov 19 at 19:35
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
Nov 19 at 19:39
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
Nov 19 at 19:42
add a comment |
up vote
0
down vote
favorite
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
edited Nov 19 at 19:54
mrtaurho
2,7691927
asked Nov 19 at 19:31
RocketKangaroo
163
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
edited Nov 19 at 19:54
mrtaurho
2,7691927
asked Nov 19 at 19:31
RocketKangaroo
163
edited Nov 19 at 19:54
mrtaurho
2,7691927
asked Nov 19 at 19:31
RocketKangaroo
163
edited Nov 19 at 19:54
mrtaurho
2,7691927
edited Nov 19 at 19:54
mrtaurho
2,7691927
edited Nov 19 at 19:54
mrtaurho
2,7691927
2,7691927
asked Nov 19 at 19:31
RocketKangaroo
163
asked Nov 19 at 19:31
RocketKangaroo
163
asked Nov 19 at 19:31
RocketKangaroo
163
163
closed as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher Nov 20 at 13:24
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.
closed as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher Nov 20 at 13:24
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
Nov 19 at 19:35
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
Nov 19 at 19:39
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
Nov 19 at 19:42
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
Nov 19 at 19:35
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
Nov 19 at 19:39
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
Nov 19 at 19:42
Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
– RocketKangaroo
Nov 19 at 19:35
Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
– RocketKangaroo
Nov 19 at 19:35
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
Nov 19 at 19:39
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
Nov 19 at 19:39
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
Nov 19 at 19:42
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
Nov 19 at 19:42
add a comment |
1 Answer
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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.
answered Nov 19 at 19:44
Yves Daoust
123k668219
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.
answered Nov 19 at 19:44
Yves Daoust
123k668219
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.
answered Nov 19 at 19:44
Yves Daoust
123k668219
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.
answered Nov 19 at 19:44
Yves Daoust
123k668219
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 Nov 19 at 19:44
Yves Daoust
123k668219
answered Nov 19 at 19:44
Yves Daoust
123k668219
answered Nov 19 at 19:44
Yves Daoust
123k668219
answered Nov 19 at 19:44
Yves Daoust
123k668219
123k668219
add a comment |
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
Nov 19 at 19:35
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
Nov 19 at 19:39
@projectilemotion: yep, sorry, my bad.
– Yves Daoust
Nov 19 at 19:42