Integrating a diffusion equation
I managed to solve
$$ -frac{1}{2} bigg[f + frac{1}{2}eta frac{df}{deta}bigg] = frac{d}{deta}bigg(ffrac{df}{deta}bigg)$$
without much hassle by using a hint to multiply the whole equation by $eta$ and then to integrate by parts. My question is, why does multiplying by $eta$ make this problem easy to integrate?
integration
asked Jan 4 at 16:09
kroneckerdel69kroneckerdel69
358
add a comment |
I managed to solve
$$ -frac{1}{2} bigg[f + frac{1}{2}eta frac{df}{deta}bigg] = frac{d}{deta}bigg(ffrac{df}{deta}bigg)$$
without much hassle by using a hint to multiply the whole equation by $eta$ and then to integrate by parts. My question is, why does multiplying by $eta$ make this problem easy to integrate?
integration
asked Jan 4 at 16:09
kroneckerdel69kroneckerdel69
358
add a comment |
I managed to solve
$$ -frac{1}{2} bigg[f + frac{1}{2}eta frac{df}{deta}bigg] = frac{d}{deta}bigg(ffrac{df}{deta}bigg)$$
without much hassle by using a hint to multiply the whole equation by $eta$ and then to integrate by parts. My question is, why does multiplying by $eta$ make this problem easy to integrate?
integration
asked Jan 4 at 16:09
kroneckerdel69kroneckerdel69
358
I managed to solve
$$ -frac{1}{2} bigg[f + frac{1}{2}eta frac{df}{deta}bigg] = frac{d}{deta}bigg(ffrac{df}{deta}bigg)$$
without much hassle by using a hint to multiply the whole equation by $eta$ and then to integrate by parts. My question is, why does multiplying by $eta$ make this problem easy to integrate?
integration
integration
asked Jan 4 at 16:09
kroneckerdel69kroneckerdel69
358
asked Jan 4 at 16:09
kroneckerdel69kroneckerdel69
358
asked Jan 4 at 16:09
kroneckerdel69kroneckerdel69
358
asked Jan 4 at 16:09
kroneckerdel69kroneckerdel69
358
asked Jan 4 at 16:09
kroneckerdel69kroneckerdel69
358
358
add a comment |
add a comment |
1 Answer
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Because you have
$$-frac{eta}{2} bigg[f + frac{1}{2}eta frac{df}{deta}bigg] = -frac{1}{4}frac{d}{d eta}(eta^2f)$$
and the expression on the RHS can be integrated in closed form to reduce the order.
answered Jan 4 at 16:54
RRLRRL
49.3k42573
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Because you have
$$-frac{eta}{2} bigg[f + frac{1}{2}eta frac{df}{deta}bigg] = -frac{1}{4}frac{d}{d eta}(eta^2f)$$
and the expression on the RHS can be integrated in closed form to reduce the order.
answered Jan 4 at 16:54
RRLRRL
49.3k42573
add a comment |
Because you have
$$-frac{eta}{2} bigg[f + frac{1}{2}eta frac{df}{deta}bigg] = -frac{1}{4}frac{d}{d eta}(eta^2f)$$
and the expression on the RHS can be integrated in closed form to reduce the order.
answered Jan 4 at 16:54
RRLRRL
49.3k42573
add a comment |
Because you have
$$-frac{eta}{2} bigg[f + frac{1}{2}eta frac{df}{deta}bigg] = -frac{1}{4}frac{d}{d eta}(eta^2f)$$
and the expression on the RHS can be integrated in closed form to reduce the order.
answered Jan 4 at 16:54
RRLRRL
49.3k42573
Because you have
$$-frac{eta}{2} bigg[f + frac{1}{2}eta frac{df}{deta}bigg] = -frac{1}{4}frac{d}{d eta}(eta^2f)$$
and the expression on the RHS can be integrated in closed form to reduce the order.
answered Jan 4 at 16:54
RRLRRL
49.3k42573
answered Jan 4 at 16:54
RRLRRL
49.3k42573
answered Jan 4 at 16:54
RRLRRL
49.3k42573
answered Jan 4 at 16:54
RRLRRL
49.3k42573
49.3k42573
add a comment |
add a comment |
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