Solving a first-order nonlinear ordinary differential equation (for a physics problem)
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The following DE describes a physical problem involving electronic circuits. How can I solve it?
$$x(t)cdot a+x'(t)cdot b+ccdotlnleft(1+dcdot x(t)right)=0spaceLongleftrightarrowspace x(t)=dots$$
All the variables are real and positive and $x(0)=x_0$
I've no idea where to start what so ever. Thanks for any help or ideas
ordinary-differential-equations logarithms physics
asked Dec 28 '18 at 15:44
KlopjasKlopjas
714
$endgroup$
|
show 2 more comments
$begingroup$
The following DE describes a physical problem involving electronic circuits. How can I solve it?
$$x(t)cdot a+x'(t)cdot b+ccdotlnleft(1+dcdot x(t)right)=0spaceLongleftrightarrowspace x(t)=dots$$
All the variables are real and positive and $x(0)=x_0$
I've no idea where to start what so ever. Thanks for any help or ideas
ordinary-differential-equations logarithms physics
asked Dec 28 '18 at 15:44
KlopjasKlopjas
714
$endgroup$
$begingroup$
It is seperable. Having said that, it looks quite hard to integrate the resulting equation...$$t=-intfrac{b,dx}{cln(1+dx)+ax}$$Are you sure this should be integrable? Are you given any more information?
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– John Doe
Dec 28 '18 at 15:48
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The final result must look like a downgoing exponential function, that is the thing I expect.
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– Klopjas
Dec 28 '18 at 15:52
1
$begingroup$
@JohnDoe: You notice that in your integral above, you've used "dx" in two utterly different ways. That seems OK to me, but it might confuse a beginner...
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– John Hughes
Dec 28 '18 at 15:56
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@JohnHughes Yes it does confuse me :(
$endgroup$
– Klopjas
Dec 28 '18 at 15:58
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Ah yes, that is true. I suppose thats the issue with using $d$ as a constant! We can let $d=D$, giving $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$ if that helps.
$endgroup$
– John Doe
Dec 28 '18 at 16:00
|
show 2 more comments
$begingroup$
The following DE describes a physical problem involving electronic circuits. How can I solve it?
$$x(t)cdot a+x'(t)cdot b+ccdotlnleft(1+dcdot x(t)right)=0spaceLongleftrightarrowspace x(t)=dots$$
All the variables are real and positive and $x(0)=x_0$
I've no idea where to start what so ever. Thanks for any help or ideas
ordinary-differential-equations logarithms physics
asked Dec 28 '18 at 15:44
KlopjasKlopjas
714
$endgroup$
The following DE describes a physical problem involving electronic circuits. How can I solve it?
$$x(t)cdot a+x'(t)cdot b+ccdotlnleft(1+dcdot x(t)right)=0spaceLongleftrightarrowspace x(t)=dots$$
All the variables are real and positive and $x(0)=x_0$
I've no idea where to start what so ever. Thanks for any help or ideas
ordinary-differential-equations logarithms physics
ordinary-differential-equations logarithms physics
asked Dec 28 '18 at 15:44
KlopjasKlopjas
714
asked Dec 28 '18 at 15:44
KlopjasKlopjas
714
edited Dec 28 '18 at 15:51
Klopjas
asked Dec 28 '18 at 15:44
KlopjasKlopjas
714
asked Dec 28 '18 at 15:44
KlopjasKlopjas
714
asked Dec 28 '18 at 15:44
KlopjasKlopjas
714
714
$begingroup$
It is seperable. Having said that, it looks quite hard to integrate the resulting equation...$$t=-intfrac{b,dx}{cln(1+dx)+ax}$$Are you sure this should be integrable? Are you given any more information?
$endgroup$
– John Doe
Dec 28 '18 at 15:48
$begingroup$
The final result must look like a downgoing exponential function, that is the thing I expect.
$endgroup$
– Klopjas
Dec 28 '18 at 15:52
1
$begingroup$
@JohnDoe: You notice that in your integral above, you've used "dx" in two utterly different ways. That seems OK to me, but it might confuse a beginner...
$endgroup$
– John Hughes
Dec 28 '18 at 15:56
$begingroup$
@JohnHughes Yes it does confuse me :(
$endgroup$
– Klopjas
Dec 28 '18 at 15:58
$begingroup$
Ah yes, that is true. I suppose thats the issue with using $d$ as a constant! We can let $d=D$, giving $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$ if that helps.
$endgroup$
– John Doe
Dec 28 '18 at 16:00
|
show 2 more comments
$begingroup$
It is seperable. Having said that, it looks quite hard to integrate the resulting equation...$$t=-intfrac{b,dx}{cln(1+dx)+ax}$$Are you sure this should be integrable? Are you given any more information?
$endgroup$
– John Doe
Dec 28 '18 at 15:48
$begingroup$
The final result must look like a downgoing exponential function, that is the thing I expect.
$endgroup$
– Klopjas
Dec 28 '18 at 15:52
1
$begingroup$
@JohnDoe: You notice that in your integral above, you've used "dx" in two utterly different ways. That seems OK to me, but it might confuse a beginner...
$endgroup$
– John Hughes
Dec 28 '18 at 15:56
$begingroup$
@JohnHughes Yes it does confuse me :(
$endgroup$
– Klopjas
Dec 28 '18 at 15:58
$begingroup$
Ah yes, that is true. I suppose thats the issue with using $d$ as a constant! We can let $d=D$, giving $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$ if that helps.
$endgroup$
– John Doe
Dec 28 '18 at 16:00
$begingroup$
It is seperable. Having said that, it looks quite hard to integrate the resulting equation...$$t=-intfrac{b,dx}{cln(1+dx)+ax}$$Are you sure this should be integrable? Are you given any more information?
$endgroup$
– John Doe
Dec 28 '18 at 15:48
$begingroup$
It is seperable. Having said that, it looks quite hard to integrate the resulting equation...$$t=-intfrac{b,dx}{cln(1+dx)+ax}$$Are you sure this should be integrable? Are you given any more information?
$endgroup$
– John Doe
Dec 28 '18 at 15:48
$begingroup$
The final result must look like a downgoing exponential function, that is the thing I expect.
$endgroup$
– Klopjas
Dec 28 '18 at 15:52
$begingroup$
The final result must look like a downgoing exponential function, that is the thing I expect.
$endgroup$
– Klopjas
Dec 28 '18 at 15:52
1
1
$begingroup$
@JohnDoe: You notice that in your integral above, you've used "dx" in two utterly different ways. That seems OK to me, but it might confuse a beginner...
$endgroup$
– John Hughes
Dec 28 '18 at 15:56
$begingroup$
@JohnDoe: You notice that in your integral above, you've used "dx" in two utterly different ways. That seems OK to me, but it might confuse a beginner...
$endgroup$
– John Hughes
Dec 28 '18 at 15:56
$begingroup$
@JohnHughes Yes it does confuse me :(
$endgroup$
– Klopjas
Dec 28 '18 at 15:58
$begingroup$
@JohnHughes Yes it does confuse me :(
$endgroup$
– Klopjas
Dec 28 '18 at 15:58
$begingroup$
Ah yes, that is true. I suppose thats the issue with using $d$ as a constant! We can let $d=D$, giving $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$ if that helps.
$endgroup$
– John Doe
Dec 28 '18 at 16:00
$begingroup$
Ah yes, that is true. I suppose thats the issue with using $d$ as a constant! We can let $d=D$, giving $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$ if that helps.
$endgroup$
– John Doe
Dec 28 '18 at 16:00
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Note that this solution is only valid for $|Dx_0|<1$. The integral cannot be evaluated in closed form generally, but it can be approximated using a Taylor series.
Given $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$one can expand the logarithm as a Taylor series, and ignore all terms excluding the lowest two order terms $Dx-frac{D^2x^2}2$. $$begin{align}t-t_0&=-int^xfrac{b,du}{(a+cD)u-frac{cD^2}2u^2}=frac2{cD^2}int^xfrac{b,du}{u^2-frac{2(a+cD)}{cD^2}u}end{align}$$Let $A=frac{(a+cD)}{cD^2}$.
$$begin{align}t-t_0&=frac2{cD^2}int^xfrac{b,du}{(u-2A)u}\&=-frac{2b}{cD^2}frac1{2A}int^xduleft(frac{1}u-frac{1}{u-2A}right)\&=frac{b}{cAD^2}ln left(frac{x-2A}{x}right)end{align}$$
Then inverting, $$x-2A=xexpleft(frac{cAD^2}b(t-t_0)right)\x=frac{2A}{1-expleft(frac{cAD^2}b(t-t_0)right)}$$
Finally, applying the initial condition, $$x=frac{x_0}{expleft(frac{cAD^2}btright)+frac{x_0}{2A}left(1-expleft(frac{cAD^2}btright)right)}$$
We initially assumed $x_0$ was small, and we can see that with this assumption, the corresponding solution ensures that $x$ stays small as time passes. So for small $x_0$, this is a good approximation to the true solution.
answered Dec 28 '18 at 16:30
John DoeJohn Doe
11.2k11239
$endgroup$
1
$begingroup$
You either meant to write $|Dx_0|<1$ or $|x_0|ll 1$. As it is, the usefulness of your first condition strongly depends on the size of $D$. // One could present an easier calculation for this quadratic approximation of the logarithm by solving $0=ax+bx'+c(d,x-frac12 d^2,x^2)$ as a Bernoulli equation; setting $u=x^{-1}$, finding $0=au-bu'+c(d,u-frac12d^2)$ which now is linear, etc.
$endgroup$
– LutzL
Dec 28 '18 at 16:43
add a comment |
$begingroup$
Going by @LutzL's suggestion we can linearize the equation for small $x$
$$ bx' + (a+cd)x = 0 $$
since $ln(1+dx) approx dx$
Then an approximation given by
$$ x(t) = x_0expleft(-frac{(a+cd)}{b}tright) $$
which resembles a decaying exponential as you've mentioned. Note that this approximation is only accurate if $|dx_0| ll 1$ and is only valid for $t ge 0$
answered Dec 29 '18 at 5:25
DylanDylan
14k31127
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that this solution is only valid for $|Dx_0|<1$. The integral cannot be evaluated in closed form generally, but it can be approximated using a Taylor series.
Given $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$one can expand the logarithm as a Taylor series, and ignore all terms excluding the lowest two order terms $Dx-frac{D^2x^2}2$. $$begin{align}t-t_0&=-int^xfrac{b,du}{(a+cD)u-frac{cD^2}2u^2}=frac2{cD^2}int^xfrac{b,du}{u^2-frac{2(a+cD)}{cD^2}u}end{align}$$Let $A=frac{(a+cD)}{cD^2}$.
$$begin{align}t-t_0&=frac2{cD^2}int^xfrac{b,du}{(u-2A)u}\&=-frac{2b}{cD^2}frac1{2A}int^xduleft(frac{1}u-frac{1}{u-2A}right)\&=frac{b}{cAD^2}ln left(frac{x-2A}{x}right)end{align}$$
Then inverting, $$x-2A=xexpleft(frac{cAD^2}b(t-t_0)right)\x=frac{2A}{1-expleft(frac{cAD^2}b(t-t_0)right)}$$
Finally, applying the initial condition, $$x=frac{x_0}{expleft(frac{cAD^2}btright)+frac{x_0}{2A}left(1-expleft(frac{cAD^2}btright)right)}$$
We initially assumed $x_0$ was small, and we can see that with this assumption, the corresponding solution ensures that $x$ stays small as time passes. So for small $x_0$, this is a good approximation to the true solution.
answered Dec 28 '18 at 16:30
John DoeJohn Doe
11.2k11239
$endgroup$
1
$begingroup$
You either meant to write $|Dx_0|<1$ or $|x_0|ll 1$. As it is, the usefulness of your first condition strongly depends on the size of $D$. // One could present an easier calculation for this quadratic approximation of the logarithm by solving $0=ax+bx'+c(d,x-frac12 d^2,x^2)$ as a Bernoulli equation; setting $u=x^{-1}$, finding $0=au-bu'+c(d,u-frac12d^2)$ which now is linear, etc.
$endgroup$
– LutzL
Dec 28 '18 at 16:43
add a comment |
$begingroup$
Note that this solution is only valid for $|Dx_0|<1$. The integral cannot be evaluated in closed form generally, but it can be approximated using a Taylor series.
Given $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$one can expand the logarithm as a Taylor series, and ignore all terms excluding the lowest two order terms $Dx-frac{D^2x^2}2$. $$begin{align}t-t_0&=-int^xfrac{b,du}{(a+cD)u-frac{cD^2}2u^2}=frac2{cD^2}int^xfrac{b,du}{u^2-frac{2(a+cD)}{cD^2}u}end{align}$$Let $A=frac{(a+cD)}{cD^2}$.
$$begin{align}t-t_0&=frac2{cD^2}int^xfrac{b,du}{(u-2A)u}\&=-frac{2b}{cD^2}frac1{2A}int^xduleft(frac{1}u-frac{1}{u-2A}right)\&=frac{b}{cAD^2}ln left(frac{x-2A}{x}right)end{align}$$
Then inverting, $$x-2A=xexpleft(frac{cAD^2}b(t-t_0)right)\x=frac{2A}{1-expleft(frac{cAD^2}b(t-t_0)right)}$$
Finally, applying the initial condition, $$x=frac{x_0}{expleft(frac{cAD^2}btright)+frac{x_0}{2A}left(1-expleft(frac{cAD^2}btright)right)}$$
We initially assumed $x_0$ was small, and we can see that with this assumption, the corresponding solution ensures that $x$ stays small as time passes. So for small $x_0$, this is a good approximation to the true solution.
answered Dec 28 '18 at 16:30
John DoeJohn Doe
11.2k11239
$endgroup$
1
$begingroup$
You either meant to write $|Dx_0|<1$ or $|x_0|ll 1$. As it is, the usefulness of your first condition strongly depends on the size of $D$. // One could present an easier calculation for this quadratic approximation of the logarithm by solving $0=ax+bx'+c(d,x-frac12 d^2,x^2)$ as a Bernoulli equation; setting $u=x^{-1}$, finding $0=au-bu'+c(d,u-frac12d^2)$ which now is linear, etc.
$endgroup$
– LutzL
Dec 28 '18 at 16:43
add a comment |
$begingroup$
Note that this solution is only valid for $|Dx_0|<1$. The integral cannot be evaluated in closed form generally, but it can be approximated using a Taylor series.
Given $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$one can expand the logarithm as a Taylor series, and ignore all terms excluding the lowest two order terms $Dx-frac{D^2x^2}2$. $$begin{align}t-t_0&=-int^xfrac{b,du}{(a+cD)u-frac{cD^2}2u^2}=frac2{cD^2}int^xfrac{b,du}{u^2-frac{2(a+cD)}{cD^2}u}end{align}$$Let $A=frac{(a+cD)}{cD^2}$.
$$begin{align}t-t_0&=frac2{cD^2}int^xfrac{b,du}{(u-2A)u}\&=-frac{2b}{cD^2}frac1{2A}int^xduleft(frac{1}u-frac{1}{u-2A}right)\&=frac{b}{cAD^2}ln left(frac{x-2A}{x}right)end{align}$$
Then inverting, $$x-2A=xexpleft(frac{cAD^2}b(t-t_0)right)\x=frac{2A}{1-expleft(frac{cAD^2}b(t-t_0)right)}$$
Finally, applying the initial condition, $$x=frac{x_0}{expleft(frac{cAD^2}btright)+frac{x_0}{2A}left(1-expleft(frac{cAD^2}btright)right)}$$
We initially assumed $x_0$ was small, and we can see that with this assumption, the corresponding solution ensures that $x$ stays small as time passes. So for small $x_0$, this is a good approximation to the true solution.
answered Dec 28 '18 at 16:30
John DoeJohn Doe
11.2k11239
$endgroup$
Note that this solution is only valid for $|Dx_0|<1$. The integral cannot be evaluated in closed form generally, but it can be approximated using a Taylor series.
Given $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$one can expand the logarithm as a Taylor series, and ignore all terms excluding the lowest two order terms $Dx-frac{D^2x^2}2$. $$begin{align}t-t_0&=-int^xfrac{b,du}{(a+cD)u-frac{cD^2}2u^2}=frac2{cD^2}int^xfrac{b,du}{u^2-frac{2(a+cD)}{cD^2}u}end{align}$$Let $A=frac{(a+cD)}{cD^2}$.
$$begin{align}t-t_0&=frac2{cD^2}int^xfrac{b,du}{(u-2A)u}\&=-frac{2b}{cD^2}frac1{2A}int^xduleft(frac{1}u-frac{1}{u-2A}right)\&=frac{b}{cAD^2}ln left(frac{x-2A}{x}right)end{align}$$
Then inverting, $$x-2A=xexpleft(frac{cAD^2}b(t-t_0)right)\x=frac{2A}{1-expleft(frac{cAD^2}b(t-t_0)right)}$$
Finally, applying the initial condition, $$x=frac{x_0}{expleft(frac{cAD^2}btright)+frac{x_0}{2A}left(1-expleft(frac{cAD^2}btright)right)}$$
We initially assumed $x_0$ was small, and we can see that with this assumption, the corresponding solution ensures that $x$ stays small as time passes. So for small $x_0$, this is a good approximation to the true solution.
answered Dec 28 '18 at 16:30
John DoeJohn Doe
11.2k11239
edited Dec 31 '18 at 22:37
answered Dec 28 '18 at 16:30
John DoeJohn Doe
11.2k11239
answered Dec 28 '18 at 16:30
John DoeJohn Doe
11.2k11239
answered Dec 28 '18 at 16:30
John DoeJohn Doe
11.2k11239
11.2k11239
1
$begingroup$
You either meant to write $|Dx_0|<1$ or $|x_0|ll 1$. As it is, the usefulness of your first condition strongly depends on the size of $D$. // One could present an easier calculation for this quadratic approximation of the logarithm by solving $0=ax+bx'+c(d,x-frac12 d^2,x^2)$ as a Bernoulli equation; setting $u=x^{-1}$, finding $0=au-bu'+c(d,u-frac12d^2)$ which now is linear, etc.
$endgroup$
– LutzL
Dec 28 '18 at 16:43
add a comment |
1
$begingroup$
You either meant to write $|Dx_0|<1$ or $|x_0|ll 1$. As it is, the usefulness of your first condition strongly depends on the size of $D$. // One could present an easier calculation for this quadratic approximation of the logarithm by solving $0=ax+bx'+c(d,x-frac12 d^2,x^2)$ as a Bernoulli equation; setting $u=x^{-1}$, finding $0=au-bu'+c(d,u-frac12d^2)$ which now is linear, etc.
$endgroup$
– LutzL
Dec 28 '18 at 16:43
1
1
$begingroup$
You either meant to write $|Dx_0|<1$ or $|x_0|ll 1$. As it is, the usefulness of your first condition strongly depends on the size of $D$. // One could present an easier calculation for this quadratic approximation of the logarithm by solving $0=ax+bx'+c(d,x-frac12 d^2,x^2)$ as a Bernoulli equation; setting $u=x^{-1}$, finding $0=au-bu'+c(d,u-frac12d^2)$ which now is linear, etc.
$endgroup$
– LutzL
Dec 28 '18 at 16:43
$begingroup$
You either meant to write $|Dx_0|<1$ or $|x_0|ll 1$. As it is, the usefulness of your first condition strongly depends on the size of $D$. // One could present an easier calculation for this quadratic approximation of the logarithm by solving $0=ax+bx'+c(d,x-frac12 d^2,x^2)$ as a Bernoulli equation; setting $u=x^{-1}$, finding $0=au-bu'+c(d,u-frac12d^2)$ which now is linear, etc.
$endgroup$
– LutzL
Dec 28 '18 at 16:43
add a comment |
$begingroup$
Going by @LutzL's suggestion we can linearize the equation for small $x$
$$ bx' + (a+cd)x = 0 $$
since $ln(1+dx) approx dx$
Then an approximation given by
$$ x(t) = x_0expleft(-frac{(a+cd)}{b}tright) $$
which resembles a decaying exponential as you've mentioned. Note that this approximation is only accurate if $|dx_0| ll 1$ and is only valid for $t ge 0$
answered Dec 29 '18 at 5:25
DylanDylan
14k31127
$endgroup$
add a comment |
$begingroup$
Going by @LutzL's suggestion we can linearize the equation for small $x$
$$ bx' + (a+cd)x = 0 $$
since $ln(1+dx) approx dx$
Then an approximation given by
$$ x(t) = x_0expleft(-frac{(a+cd)}{b}tright) $$
which resembles a decaying exponential as you've mentioned. Note that this approximation is only accurate if $|dx_0| ll 1$ and is only valid for $t ge 0$
answered Dec 29 '18 at 5:25
DylanDylan
14k31127
$endgroup$
add a comment |
$begingroup$
Going by @LutzL's suggestion we can linearize the equation for small $x$
$$ bx' + (a+cd)x = 0 $$
since $ln(1+dx) approx dx$
Then an approximation given by
$$ x(t) = x_0expleft(-frac{(a+cd)}{b}tright) $$
which resembles a decaying exponential as you've mentioned. Note that this approximation is only accurate if $|dx_0| ll 1$ and is only valid for $t ge 0$
answered Dec 29 '18 at 5:25
DylanDylan
14k31127
$endgroup$
Going by @LutzL's suggestion we can linearize the equation for small $x$
$$ bx' + (a+cd)x = 0 $$
since $ln(1+dx) approx dx$
Then an approximation given by
$$ x(t) = x_0expleft(-frac{(a+cd)}{b}tright) $$
which resembles a decaying exponential as you've mentioned. Note that this approximation is only accurate if $|dx_0| ll 1$ and is only valid for $t ge 0$
answered Dec 29 '18 at 5:25
DylanDylan
14k31127
edited Dec 31 '18 at 12:45
answered Dec 29 '18 at 5:25
DylanDylan
14k31127
answered Dec 29 '18 at 5:25
DylanDylan
14k31127
answered Dec 29 '18 at 5:25
DylanDylan
14k31127
14k31127
add a comment |
add a comment |
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It is seperable. Having said that, it looks quite hard to integrate the resulting equation...$$t=-intfrac{b,dx}{cln(1+dx)+ax}$$Are you sure this should be integrable? Are you given any more information?
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– John Doe
Dec 28 '18 at 15:48
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The final result must look like a downgoing exponential function, that is the thing I expect.
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– Klopjas
Dec 28 '18 at 15:52
1
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@JohnDoe: You notice that in your integral above, you've used "dx" in two utterly different ways. That seems OK to me, but it might confuse a beginner...
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– John Hughes
Dec 28 '18 at 15:56
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@JohnHughes Yes it does confuse me :(
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– Klopjas
Dec 28 '18 at 15:58
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Ah yes, that is true. I suppose thats the issue with using $d$ as a constant! We can let $d=D$, giving $$t=-intfrac{b,dx}{cln(1+Dx)+ax}$$ if that helps.
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– John Doe
Dec 28 '18 at 16:00