First order linear differential equation by substitution.Watch

Thread starter 4 years ago
#1
Solve (x+y)dy/dx = y

Use the substitution y=xu

I seem to not get the answer. So far this is my working:

(x+xu)dy/dx=xu
dy/dx=[xu]/[x+xu]
dy/dx=u/(1+u)

dy/dx = u+xdu/dx by implicit and product rule, so I substituted this in for dy/dx:

u+xdu/dx = u/(1+u)

xdu/dx = [u/(1+u)]-u

1/x dx = [(1+u)/u]-1/u du

int(1/x dx) = int[(1+u)/u] - 1/u du

lnmod(x) = lnmod(u) +u - lnmod(u) +k

lnmod(x) = u+k
mod(x) = (e^u)(e^k)
x = ke^u = ke^(y/x)

I can't rearrange this to get the answer which is y=ke^(x/y)
0
4 years ago
#2
(Original post by Protoxylic)
Solve (x+y)dy/dx = y

Use the substitution y=xu

I seem to not get the answer. So far this is my working:

(x+xu)dy/dx=xu
dy/dx=[xu]/[x+xu]
dy/dx=u/(1+u)

dy/dx = u+xdu/dx by implicit and product rule, so I substituted this in for dy/dx:

u+xdu/dx = u/(1+u)

xdu/dx = [u/(1+u)]-u

1/x dx = [(1+u)/u]-1/u du

int(1/x dx) = int[(1+u)/u] - 1/u du

lnmod(x) = lnmod(u) +u - lnmod(u) +k

lnmod(x) = u+k
mod(x) = (e^u)(e^k)
x = ke^u = ke^(y/x)

I can't rearrange this to get the answer which is y=ke^(x/y)
You have used

but this is not correct.

I would make into one fraction before finding the reciprocal.
0
Thread starter 4 years ago
#3
(Original post by notnek)
You have used

but this is not correct.

I would make into one fraction before finding the reciprocal.
Ah, damn. So reciprocal of [(1+u)/u] - u = -[(1+u)/u^2]. And integrating that gets 1/u - lnmod(u). So the full expanded equation is lnmod(x)=1/u - lnmod(u) + k. It follows that ux = y so lnmod(y) = 1/u + k. mod(y) = [e^(1/u)][e^k] and therefore y=ke^(x/y)?
0
4 years ago
#4
(Original post by Protoxylic)
Ah, damn. So reciprocal of [(1+u)/u] - u = -[(1+u)/u^2]. And integrating that gets 1/u - lnmod(u). So the full expanded equation is lnmod(x)=1/u - lnmod(u) + k. It follows that ux = y so lnmod(y) = 1/u + k. mod(y) = [e^(1/u)][e^k] and therefore y=ke^(x/y)?
That's correct.

You shouldn't really use the same letter for your constant of integration and your value in front of e^(x/y) since they're not the same.

You've said e^k = k which isn't true. It would be better to use a different letter e.g. e^k = A and so you end up with y=Ae^(x/y)
0
Thread starter 4 years ago
#5
(Original post by notnek)
That's correct.

You shouldn't really use the same letter for your constant of integration and your value in front of e^(x/y) since they're not the same.

You've said e^k = k which isn't true. It would be better to use a different letter e.g. e^k = A and so you end up with y=Ae^(x/y)
Oh of course, I think I just got a bit slack and said to myself: "They're both constant, so k will do"
0
4 years ago
#6
(Original post by Protoxylic)
Oh of course, I think I just got a bit slack and said to myself: "They're both constant, so k will do"
It's not the biggest crime since the final answer is still correct but I think it's best to always keep working clear and correct.
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