Mixing
How to obtain the next step in the differential equation
$begingroup$
Can someone please help me understand step $(*)$ below. How is this step obtained from the previous. The derivative has been removed so do we perform integration? I've tried integration by parts, but do not get anything similar.
For the differential equation $$frac{dx}{dt} =ax,$$
There are only solutions of the form $x(t)=ke^{at}$. To see this, let $u(t)$ be any solution and compute the derivative of $u(t)e^{-at}:$
begin{align}
frac{d}{dt}(u(t)e^{−at}) &=u'(t)e^{−at} +u(t)(−ae^{−at}) \&=au(t)e^{−at} −au(t)e^{−at} quad quad quadquad(*)\&=0
end{align}
ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
Can someone please help me understand step $(*)$ below. How is this step obtained from the previous. The derivative has been removed so do we perform integration? I've tried integration by parts, but do not get anything similar.
For the differential equation $$frac{dx}{dt} =ax,$$
There are only solutions of the form $x(t)=ke^{at}$. To see this, let $u(t)$ be any solution and compute the derivative of $u(t)e^{-at}:$
begin{align}
frac{d}{dt}(u(t)e^{−at}) &=u'(t)e^{−at} +u(t)(−ae^{−at}) \&=au(t)e^{−at} −au(t)e^{−at} quad quad quadquad(*)\&=0
end{align}
ordinary-differential-equations
$endgroup$
2
$begingroup$
I think x should be u instead in the differential equation. Then the step is just substitution
$endgroup$
– Mark
Jan 19 at 16:07
add a comment |
$begingroup$
Can someone please help me understand step $(*)$ below. How is this step obtained from the previous. The derivative has been removed so do we perform integration? I've tried integration by parts, but do not get anything similar.
For the differential equation $$frac{dx}{dt} =ax,$$
There are only solutions of the form $x(t)=ke^{at}$. To see this, let $u(t)$ be any solution and compute the derivative of $u(t)e^{-at}:$
begin{align}
frac{d}{dt}(u(t)e^{−at}) &=u'(t)e^{−at} +u(t)(−ae^{−at}) \&=au(t)e^{−at} −au(t)e^{−at} quad quad quadquad(*)\&=0
end{align}
ordinary-differential-equations
$endgroup$
Can someone please help me understand step $(*)$ below. How is this step obtained from the previous. The derivative has been removed so do we perform integration? I've tried integration by parts, but do not get anything similar.
For the differential equation $$frac{dx}{dt} =ax,$$
There are only solutions of the form $x(t)=ke^{at}$. To see this, let $u(t)$ be any solution and compute the derivative of $u(t)e^{-at}:$
begin{align}
frac{d}{dt}(u(t)e^{−at}) &=u'(t)e^{−at} +u(t)(−ae^{−at}) \&=au(t)e^{−at} −au(t)e^{−at} quad quad quadquad(*)\&=0
end{align}
ordinary-differential-equations
ordinary-differential-equations
asked Jan 19 at 15:59
user606466
2
$begingroup$
I think x should be u instead in the differential equation. Then the step is just substitution
$endgroup$
– Mark
Jan 19 at 16:07
add a comment |
2
$begingroup$
I think x should be u instead in the differential equation. Then the step is just substitution
$endgroup$
– Mark
Jan 19 at 16:07
2
2
$begingroup$
I think x should be u instead in the differential equation. Then the step is just substitution
$endgroup$
– Mark
Jan 19 at 16:07
$begingroup$
I think x should be u instead in the differential equation. Then the step is just substitution
$endgroup$
– Mark
Jan 19 at 16:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $u(t)=ke^{at}$ then $u'(t)=ake^{at}=acdot u(t)$ is what this method is using.
If you struggle to follow this method however, noting this is a Variable Separable Equation and solving is much more intuitive in my opinion.
$$frac{dx}{dt}=ax$$
$$to frac{dx}{x} = a dt$$
$$to int frac{dx}{x} = int a dt$$
$$to ln x =at + C$$
$$to x=e^{at+C}=e^Ce^{at}=ke^{at}$$
Since $e^C$ is just a constant.
answered Jan 19 at 16:10
Rhys HughesRhys Hughes
6,9441530
$endgroup$
add a comment |
$begingroup$
By assumption, $u (t) $ is a solution of differential equation. Hence $u'(t)=acdot u (t) $.
answered Jan 19 at 16:11
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
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$begingroup$
If $u(t)=ke^{at}$ then $u'(t)=ake^{at}=acdot u(t)$ is what this method is using.
If you struggle to follow this method however, noting this is a Variable Separable Equation and solving is much more intuitive in my opinion.
$$frac{dx}{dt}=ax$$
$$to frac{dx}{x} = a dt$$
$$to int frac{dx}{x} = int a dt$$
$$to ln x =at + C$$
$$to x=e^{at+C}=e^Ce^{at}=ke^{at}$$
Since $e^C$ is just a constant.
answered Jan 19 at 16:10
Rhys HughesRhys Hughes
6,9441530
$endgroup$
add a comment |
$begingroup$
If $u(t)=ke^{at}$ then $u'(t)=ake^{at}=acdot u(t)$ is what this method is using.
If you struggle to follow this method however, noting this is a Variable Separable Equation and solving is much more intuitive in my opinion.
$$frac{dx}{dt}=ax$$
$$to frac{dx}{x} = a dt$$
$$to int frac{dx}{x} = int a dt$$
$$to ln x =at + C$$
$$to x=e^{at+C}=e^Ce^{at}=ke^{at}$$
Since $e^C$ is just a constant.
answered Jan 19 at 16:10
Rhys HughesRhys Hughes
6,9441530
$endgroup$
add a comment |
$begingroup$
If $u(t)=ke^{at}$ then $u'(t)=ake^{at}=acdot u(t)$ is what this method is using.
If you struggle to follow this method however, noting this is a Variable Separable Equation and solving is much more intuitive in my opinion.
$$frac{dx}{dt}=ax$$
$$to frac{dx}{x} = a dt$$
$$to int frac{dx}{x} = int a dt$$
$$to ln x =at + C$$
$$to x=e^{at+C}=e^Ce^{at}=ke^{at}$$
Since $e^C$ is just a constant.
answered Jan 19 at 16:10
Rhys HughesRhys Hughes
6,9441530
$endgroup$
If $u(t)=ke^{at}$ then $u'(t)=ake^{at}=acdot u(t)$ is what this method is using.
If you struggle to follow this method however, noting this is a Variable Separable Equation and solving is much more intuitive in my opinion.
$$frac{dx}{dt}=ax$$
$$to frac{dx}{x} = a dt$$
$$to int frac{dx}{x} = int a dt$$
$$to ln x =at + C$$
$$to x=e^{at+C}=e^Ce^{at}=ke^{at}$$
Since $e^C$ is just a constant.
answered Jan 19 at 16:10
Rhys HughesRhys Hughes
6,9441530
edited Jan 19 at 16:27
answered Jan 19 at 16:10
Rhys HughesRhys Hughes
6,9441530
answered Jan 19 at 16:10
Rhys HughesRhys Hughes
6,9441530
answered Jan 19 at 16:10
Rhys HughesRhys Hughes
6,9441530
6,9441530
add a comment |
add a comment |
$begingroup$
By assumption, $u (t) $ is a solution of differential equation. Hence $u'(t)=acdot u (t) $.
answered Jan 19 at 16:11
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
add a comment |
$begingroup$
By assumption, $u (t) $ is a solution of differential equation. Hence $u'(t)=acdot u (t) $.
answered Jan 19 at 16:11
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
add a comment |
$begingroup$
By assumption, $u (t) $ is a solution of differential equation. Hence $u'(t)=acdot u (t) $.
answered Jan 19 at 16:11
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
By assumption, $u (t) $ is a solution of differential equation. Hence $u'(t)=acdot u (t) $.
answered Jan 19 at 16:11
Thomas ShelbyThomas Shelby
3,7192525
answered Jan 19 at 16:11
Thomas ShelbyThomas Shelby
3,7192525
answered Jan 19 at 16:11
Thomas ShelbyThomas Shelby
3,7192525
answered Jan 19 at 16:11
Thomas ShelbyThomas Shelby
3,7192525
3,7192525
add a comment |
add a comment |
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$begingroup$
I think x should be u instead in the differential equation. Then the step is just substitution
$endgroup$
– Mark
Jan 19 at 16:07