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Unable to Find My Error for First Order Differential Problem
Clash Royale CLAN TAG#URR8PPP
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The Question: Given $f'(x) = 5f(x)$ and $f(3) = 2$ find the equation for $f(x)$
My Problem: My attempt to solve it produced the wrong answer. I got $f(x) = frac285x - 1$ but the correct answer is $f(x) = 2e^5(x-3)$ (you can follow this link to see how it is solved).
Note: I am not asking how to get the correct answer, as I have already found and understood such an explanation. I am instead asking why my process is incorrect.
My Method:
- $fracdf(x)dx = 5f(x)$
- $df(x) = 5f(x)dx$
- $intdf(x) = int5f(x)dx$
- $f(x) = 5f(x)cdot x + C$
- $f(3) = 5f(3)cdot 3 + C$
- $2 = 30 + C$
- $C = -28$
- $f(x) = 5f(x)cdot x - 28$
- $28 = f(x)(5x - 1)$
- $f(x) = frac285x - 1$
Explanation: I know my answer is wrong, as it does not satisfy the condition $f'(x) = 5f(x)$, yet I am intrigued by this problem as I cannot seem to figure out what I did wrong. My answer does satisfy the condition $f(3) = 2$ and its derivative - $f'(x) = -frac140(5x - 1)^2$ - is suggestive of some similarity to the original differential (eg numerator is multiplied by 5). Perhaps someone can enlighten me regarding where I went astray. Thank you.
calculus differential-equations
asked Jul 20 at 0:56
GodOrGovern
133
add a comment |Â
up vote
2
down vote
favorite
The Question: Given $f'(x) = 5f(x)$ and $f(3) = 2$ find the equation for $f(x)$
My Problem: My attempt to solve it produced the wrong answer. I got $f(x) = frac285x - 1$ but the correct answer is $f(x) = 2e^5(x-3)$ (you can follow this link to see how it is solved).
Note: I am not asking how to get the correct answer, as I have already found and understood such an explanation. I am instead asking why my process is incorrect.
My Method:
- $fracdf(x)dx = 5f(x)$
- $df(x) = 5f(x)dx$
- $intdf(x) = int5f(x)dx$
- $f(x) = 5f(x)cdot x + C$
- $f(3) = 5f(3)cdot 3 + C$
- $2 = 30 + C$
- $C = -28$
- $f(x) = 5f(x)cdot x - 28$
- $28 = f(x)(5x - 1)$
- $f(x) = frac285x - 1$
Explanation: I know my answer is wrong, as it does not satisfy the condition $f'(x) = 5f(x)$, yet I am intrigued by this problem as I cannot seem to figure out what I did wrong. My answer does satisfy the condition $f(3) = 2$ and its derivative - $f'(x) = -frac140(5x - 1)^2$ - is suggestive of some similarity to the original differential (eg numerator is multiplied by 5). Perhaps someone can enlighten me regarding where I went astray. Thank you.
calculus differential-equations
asked Jul 20 at 0:56
GodOrGovern
133
1
For what it's worth, this looks like a high-quality question to me. You made an earnest attempt and showed what you did in detail. I would just delete the disclaimer.
– David K
Jul 20 at 1:07
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The Question: Given $f'(x) = 5f(x)$ and $f(3) = 2$ find the equation for $f(x)$
My Problem: My attempt to solve it produced the wrong answer. I got $f(x) = frac285x - 1$ but the correct answer is $f(x) = 2e^5(x-3)$ (you can follow this link to see how it is solved).
Note: I am not asking how to get the correct answer, as I have already found and understood such an explanation. I am instead asking why my process is incorrect.
My Method:
- $fracdf(x)dx = 5f(x)$
- $df(x) = 5f(x)dx$
- $intdf(x) = int5f(x)dx$
- $f(x) = 5f(x)cdot x + C$
- $f(3) = 5f(3)cdot 3 + C$
- $2 = 30 + C$
- $C = -28$
- $f(x) = 5f(x)cdot x - 28$
- $28 = f(x)(5x - 1)$
- $f(x) = frac285x - 1$
Explanation: I know my answer is wrong, as it does not satisfy the condition $f'(x) = 5f(x)$, yet I am intrigued by this problem as I cannot seem to figure out what I did wrong. My answer does satisfy the condition $f(3) = 2$ and its derivative - $f'(x) = -frac140(5x - 1)^2$ - is suggestive of some similarity to the original differential (eg numerator is multiplied by 5). Perhaps someone can enlighten me regarding where I went astray. Thank you.
calculus differential-equations
asked Jul 20 at 0:56
GodOrGovern
133
The Question: Given $f'(x) = 5f(x)$ and $f(3) = 2$ find the equation for $f(x)$
My Problem: My attempt to solve it produced the wrong answer. I got $f(x) = frac285x - 1$ but the correct answer is $f(x) = 2e^5(x-3)$ (you can follow this link to see how it is solved).
Note: I am not asking how to get the correct answer, as I have already found and understood such an explanation. I am instead asking why my process is incorrect.
My Method:
- $fracdf(x)dx = 5f(x)$
- $df(x) = 5f(x)dx$
- $intdf(x) = int5f(x)dx$
- $f(x) = 5f(x)cdot x + C$
- $f(3) = 5f(3)cdot 3 + C$
- $2 = 30 + C$
- $C = -28$
- $f(x) = 5f(x)cdot x - 28$
- $28 = f(x)(5x - 1)$
- $f(x) = frac285x - 1$
Explanation: I know my answer is wrong, as it does not satisfy the condition $f'(x) = 5f(x)$, yet I am intrigued by this problem as I cannot seem to figure out what I did wrong. My answer does satisfy the condition $f(3) = 2$ and its derivative - $f'(x) = -frac140(5x - 1)^2$ - is suggestive of some similarity to the original differential (eg numerator is multiplied by 5). Perhaps someone can enlighten me regarding where I went astray. Thank you.
calculus differential-equations
asked Jul 20 at 0:56
GodOrGovern
133
edited Jul 20 at 1:13
asked Jul 20 at 0:56
GodOrGovern
133
asked Jul 20 at 0:56
GodOrGovern
133
asked Jul 20 at 0:56
GodOrGovern
133
133
1
For what it's worth, this looks like a high-quality question to me. You made an earnest attempt and showed what you did in detail. I would just delete the disclaimer.
– David K
Jul 20 at 1:07
add a comment |Â
1
For what it's worth, this looks like a high-quality question to me. You made an earnest attempt and showed what you did in detail. I would just delete the disclaimer.
– David K
Jul 20 at 1:07
1
1
For what it's worth, this looks like a high-quality question to me. You made an earnest attempt and showed what you did in detail. I would just delete the disclaimer.
– David K
Jul 20 at 1:07
For what it's worth, this looks like a high-quality question to me. You made an earnest attempt and showed what you did in detail. I would just delete the disclaimer.
– David K
Jul 20 at 1:07
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Your mistake is in going from (3) to (4).
Since $f(x)$ is a function of $x$, you cannot treat it as a constant in the integral. Basically, you should not say
$$int 5 f(x) dx = 5 f(x) cdot x + C$$
edited Jul 21 at 14:38
Lolita
52318
answered Jul 20 at 1:02
Michael Biro
10.7k21731
add a comment |Â
up vote
1
down vote
as @Michael has pointed out, Line 3 is a bit complicated. Because you don't know how to integrate $int f(x)dx$
$$f'=5f$$
$$implies frac f'f=5$$
$$(ln f)'=5$$
Integrate
$$ln f=5int dx=5x+K$$
Therefore
$$f(x)=Ke^5x$$
$$f(3)=2 implies K=frac 2 e^15$$
$$f(x)=frac 2e^5x e^15=2e^5(x-3)$$
answered Jul 20 at 9:13
Isham
10.6k3829
1
Did you use the condition $f(3) = 2$ in the question?
– john fowles
Jul 20 at 9:44
No sorry I will add some lines thanks @johnfowles
– Isham
Jul 20 at 9:46
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
Your mistake is in going from (3) to (4).
Since $f(x)$ is a function of $x$, you cannot treat it as a constant in the integral. Basically, you should not say
$$int 5 f(x) dx = 5 f(x) cdot x + C$$
edited Jul 21 at 14:38
Lolita
52318
answered Jul 20 at 1:02
Michael Biro
10.7k21731
add a comment |Â
up vote
1
down vote
accepted
Your mistake is in going from (3) to (4).
Since $f(x)$ is a function of $x$, you cannot treat it as a constant in the integral. Basically, you should not say
$$int 5 f(x) dx = 5 f(x) cdot x + C$$
edited Jul 21 at 14:38
Lolita
52318
answered Jul 20 at 1:02
Michael Biro
10.7k21731
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your mistake is in going from (3) to (4).
Since $f(x)$ is a function of $x$, you cannot treat it as a constant in the integral. Basically, you should not say
$$int 5 f(x) dx = 5 f(x) cdot x + C$$
edited Jul 21 at 14:38
Lolita
52318
answered Jul 20 at 1:02
Michael Biro
10.7k21731
Your mistake is in going from (3) to (4).
Since $f(x)$ is a function of $x$, you cannot treat it as a constant in the integral. Basically, you should not say
$$int 5 f(x) dx = 5 f(x) cdot x + C$$
edited Jul 21 at 14:38
Lolita
52318
answered Jul 20 at 1:02
Michael Biro
10.7k21731
edited Jul 21 at 14:38
Lolita
52318
edited Jul 21 at 14:38
Lolita
52318
edited Jul 21 at 14:38
Lolita
52318
52318
answered Jul 20 at 1:02
Michael Biro
10.7k21731
answered Jul 20 at 1:02
Michael Biro
10.7k21731
answered Jul 20 at 1:02
Michael Biro
10.7k21731
10.7k21731
add a comment |Â
add a comment |Â
up vote
1
down vote
as @Michael has pointed out, Line 3 is a bit complicated. Because you don't know how to integrate $int f(x)dx$
$$f'=5f$$
$$implies frac f'f=5$$
$$(ln f)'=5$$
Integrate
$$ln f=5int dx=5x+K$$
Therefore
$$f(x)=Ke^5x$$
$$f(3)=2 implies K=frac 2 e^15$$
$$f(x)=frac 2e^5x e^15=2e^5(x-3)$$
answered Jul 20 at 9:13
Isham
10.6k3829
1
Did you use the condition $f(3) = 2$ in the question?
– john fowles
Jul 20 at 9:44
No sorry I will add some lines thanks @johnfowles
– Isham
Jul 20 at 9:46
add a comment |Â
up vote
1
down vote
as @Michael has pointed out, Line 3 is a bit complicated. Because you don't know how to integrate $int f(x)dx$
$$f'=5f$$
$$implies frac f'f=5$$
$$(ln f)'=5$$
Integrate
$$ln f=5int dx=5x+K$$
Therefore
$$f(x)=Ke^5x$$
$$f(3)=2 implies K=frac 2 e^15$$
$$f(x)=frac 2e^5x e^15=2e^5(x-3)$$
answered Jul 20 at 9:13
Isham
10.6k3829
1
Did you use the condition $f(3) = 2$ in the question?
– john fowles
Jul 20 at 9:44
No sorry I will add some lines thanks @johnfowles
– Isham
Jul 20 at 9:46
add a comment |Â
up vote
1
down vote
up vote
1
down vote
as @Michael has pointed out, Line 3 is a bit complicated. Because you don't know how to integrate $int f(x)dx$
$$f'=5f$$
$$implies frac f'f=5$$
$$(ln f)'=5$$
Integrate
$$ln f=5int dx=5x+K$$
Therefore
$$f(x)=Ke^5x$$
$$f(3)=2 implies K=frac 2 e^15$$
$$f(x)=frac 2e^5x e^15=2e^5(x-3)$$
answered Jul 20 at 9:13
Isham
10.6k3829
as @Michael has pointed out, Line 3 is a bit complicated. Because you don't know how to integrate $int f(x)dx$
$$f'=5f$$
$$implies frac f'f=5$$
$$(ln f)'=5$$
Integrate
$$ln f=5int dx=5x+K$$
Therefore
$$f(x)=Ke^5x$$
$$f(3)=2 implies K=frac 2 e^15$$
$$f(x)=frac 2e^5x e^15=2e^5(x-3)$$
answered Jul 20 at 9:13
Isham
10.6k3829
edited Jul 20 at 9:47
answered Jul 20 at 9:13
Isham
10.6k3829
answered Jul 20 at 9:13
Isham
10.6k3829
answered Jul 20 at 9:13
Isham
10.6k3829
10.6k3829
1
Did you use the condition $f(3) = 2$ in the question?
– john fowles
Jul 20 at 9:44
No sorry I will add some lines thanks @johnfowles
– Isham
Jul 20 at 9:46
add a comment |Â
1
Did you use the condition $f(3) = 2$ in the question?
– john fowles
Jul 20 at 9:44
No sorry I will add some lines thanks @johnfowles
– Isham
Jul 20 at 9:46
1
1
Did you use the condition $f(3) = 2$ in the question?
– john fowles
Jul 20 at 9:44
Did you use the condition $f(3) = 2$ in the question?
– john fowles
Jul 20 at 9:44
No sorry I will add some lines thanks @johnfowles
– Isham
Jul 20 at 9:46
No sorry I will add some lines thanks @johnfowles
– Isham
Jul 20 at 9:46
add a comment |Â
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1
For what it's worth, this looks like a high-quality question to me. You made an earnest attempt and showed what you did in detail. I would just delete the disclaimer.
– David K
Jul 20 at 1:07