How to solve the following differential equation $B'(t)=1100exp({frac{ln{0.5}}{5}t})+frac{ln{0.5}}{6}B(t)$?
$begingroup$
I have constructed this differential equation $B'(t)=1100expleft(frac{ln{0.5}}{5}tright)+frac{ln{0.5}}{6}B(t)$
I have constructed this equation when i asked this question here,
but i'm unable to solve this equation from methods that i know.
I'm unable to separate it, and it's not of the form $y'+p(t)y=q(t)$.
What should be done here?
ordinary-differential-equations derivatives
edited Dec 17 '18 at 19:15
amWhy
192k28225439
asked Dec 17 '18 at 19:02
user3133165user3133165
1838
$endgroup$
|
show 3 more comments
$begingroup$
I have constructed this differential equation $B'(t)=1100expleft(frac{ln{0.5}}{5}tright)+frac{ln{0.5}}{6}B(t)$
I have constructed this equation when i asked this question here,
but i'm unable to solve this equation from methods that i know.
I'm unable to separate it, and it's not of the form $y'+p(t)y=q(t)$.
What should be done here?
ordinary-differential-equations derivatives
edited Dec 17 '18 at 19:15
amWhy
192k28225439
asked Dec 17 '18 at 19:02
user3133165user3133165
1838
$endgroup$
2
$begingroup$
why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
$endgroup$
– amWhy
Dec 17 '18 at 19:04
4
$begingroup$
"it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
$endgroup$
– Did
Dec 17 '18 at 19:09
2
$begingroup$
$q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
$endgroup$
– amWhy
Dec 17 '18 at 19:13
1
$begingroup$
Yes I've got it.
$endgroup$
– user3133165
Dec 17 '18 at 19:14
2
$begingroup$
$ln(0.5)/6$ is a constant function of $t$
$endgroup$
– Martín Vacas Vignolo
Dec 17 '18 at 19:18
|
show 3 more comments
$begingroup$
I have constructed this differential equation $B'(t)=1100expleft(frac{ln{0.5}}{5}tright)+frac{ln{0.5}}{6}B(t)$
I have constructed this equation when i asked this question here,
but i'm unable to solve this equation from methods that i know.
I'm unable to separate it, and it's not of the form $y'+p(t)y=q(t)$.
What should be done here?
ordinary-differential-equations derivatives
edited Dec 17 '18 at 19:15
amWhy
192k28225439
asked Dec 17 '18 at 19:02
user3133165user3133165
1838
$endgroup$
I have constructed this differential equation $B'(t)=1100expleft(frac{ln{0.5}}{5}tright)+frac{ln{0.5}}{6}B(t)$
I have constructed this equation when i asked this question here,
but i'm unable to solve this equation from methods that i know.
I'm unable to separate it, and it's not of the form $y'+p(t)y=q(t)$.
What should be done here?
ordinary-differential-equations derivatives
ordinary-differential-equations derivatives
edited Dec 17 '18 at 19:15
amWhy
192k28225439
asked Dec 17 '18 at 19:02
user3133165user3133165
1838
edited Dec 17 '18 at 19:15
amWhy
192k28225439
asked Dec 17 '18 at 19:02
user3133165user3133165
1838
edited Dec 17 '18 at 19:15
amWhy
192k28225439
edited Dec 17 '18 at 19:15
amWhy
192k28225439
edited Dec 17 '18 at 19:15
amWhy
192k28225439
192k28225439
asked Dec 17 '18 at 19:02
user3133165user3133165
1838
asked Dec 17 '18 at 19:02
user3133165user3133165
1838
asked Dec 17 '18 at 19:02
user3133165user3133165
1838
1838
2
$begingroup$
why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
$endgroup$
– amWhy
Dec 17 '18 at 19:04
4
$begingroup$
"it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
$endgroup$
– Did
Dec 17 '18 at 19:09
2
$begingroup$
$q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
$endgroup$
– amWhy
Dec 17 '18 at 19:13
1
$begingroup$
Yes I've got it.
$endgroup$
– user3133165
Dec 17 '18 at 19:14
2
$begingroup$
$ln(0.5)/6$ is a constant function of $t$
$endgroup$
– Martín Vacas Vignolo
Dec 17 '18 at 19:18
|
show 3 more comments
2
$begingroup$
why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
$endgroup$
– amWhy
Dec 17 '18 at 19:04
4
$begingroup$
"it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
$endgroup$
– Did
Dec 17 '18 at 19:09
2
$begingroup$
$q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
$endgroup$
– amWhy
Dec 17 '18 at 19:13
1
$begingroup$
Yes I've got it.
$endgroup$
– user3133165
Dec 17 '18 at 19:14
2
$begingroup$
$ln(0.5)/6$ is a constant function of $t$
$endgroup$
– Martín Vacas Vignolo
Dec 17 '18 at 19:18
2
2
$begingroup$
why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
$endgroup$
– amWhy
Dec 17 '18 at 19:04
$begingroup$
why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
$endgroup$
– amWhy
Dec 17 '18 at 19:04
4
4
$begingroup$
"it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
$endgroup$
– Did
Dec 17 '18 at 19:09
$begingroup$
"it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
$endgroup$
– Did
Dec 17 '18 at 19:09
2
2
$begingroup$
$q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
$endgroup$
– amWhy
Dec 17 '18 at 19:13
$begingroup$
$q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
$endgroup$
– amWhy
Dec 17 '18 at 19:13
1
1
$begingroup$
Yes I've got it.
$endgroup$
– user3133165
Dec 17 '18 at 19:14
$begingroup$
Yes I've got it.
$endgroup$
– user3133165
Dec 17 '18 at 19:14
2
2
$begingroup$
$ln(0.5)/6$ is a constant function of $t$
$endgroup$
– Martín Vacas Vignolo
Dec 17 '18 at 19:18
$begingroup$
$ln(0.5)/6$ is a constant function of $t$
$endgroup$
– Martín Vacas Vignolo
Dec 17 '18 at 19:18
|
show 3 more comments
1 Answer
1
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oldest
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$begingroup$
The equation is of type $y'+p(t)y=q(t)$.
The solution for $B(t)$ will be $B(t)=frac{int{mu(t)q(t)dt +c}}{mu(t)}$,$mu(t)=e^{int{}p(t)dt}$.
answered Dec 18 '18 at 9:45
user3133165user3133165
1838
$endgroup$
add a comment |
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$begingroup$
The equation is of type $y'+p(t)y=q(t)$.
The solution for $B(t)$ will be $B(t)=frac{int{mu(t)q(t)dt +c}}{mu(t)}$,$mu(t)=e^{int{}p(t)dt}$.
answered Dec 18 '18 at 9:45
user3133165user3133165
1838
$endgroup$
add a comment |
$begingroup$
The equation is of type $y'+p(t)y=q(t)$.
The solution for $B(t)$ will be $B(t)=frac{int{mu(t)q(t)dt +c}}{mu(t)}$,$mu(t)=e^{int{}p(t)dt}$.
answered Dec 18 '18 at 9:45
user3133165user3133165
1838
$endgroup$
add a comment |
$begingroup$
The equation is of type $y'+p(t)y=q(t)$.
The solution for $B(t)$ will be $B(t)=frac{int{mu(t)q(t)dt +c}}{mu(t)}$,$mu(t)=e^{int{}p(t)dt}$.
answered Dec 18 '18 at 9:45
user3133165user3133165
1838
$endgroup$
The equation is of type $y'+p(t)y=q(t)$.
The solution for $B(t)$ will be $B(t)=frac{int{mu(t)q(t)dt +c}}{mu(t)}$,$mu(t)=e^{int{}p(t)dt}$.
answered Dec 18 '18 at 9:45
user3133165user3133165
1838
answered Dec 18 '18 at 9:45
user3133165user3133165
1838
answered Dec 18 '18 at 9:45
user3133165user3133165
1838
answered Dec 18 '18 at 9:45
user3133165user3133165
1838
1838
add a comment |
add a comment |
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2
$begingroup$
why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
$endgroup$
– amWhy
Dec 17 '18 at 19:04
4
$begingroup$
"it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
$endgroup$
– Did
Dec 17 '18 at 19:09
2
$begingroup$
$q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
$endgroup$
– amWhy
Dec 17 '18 at 19:13
1
$begingroup$
Yes I've got it.
$endgroup$
– user3133165
Dec 17 '18 at 19:14
2
$begingroup$
$ln(0.5)/6$ is a constant function of $t$
$endgroup$
– Martín Vacas Vignolo
Dec 17 '18 at 19:18