The Student Room Group

differential equation with substitution

i have done the question but i cant show the equation it wants so please if anybody could look my work and tell me where i went wrong
Reply 1
my work the end equation is written on top
Reply 2
dy/dz isnt right as
y = z^-1/2
dy/dz = -1/2 z^-3/2

Looks like you put in dz/dy rather than dy/dz.
(edited 10 months ago)
Reply 3
Original post by mqb2766
dy/dz isnt right as
y = z^-1/2
dy/dz = -1/2 z^-3/2

Looks like you put in dz/dy rather than dy/dz.


what? where is dy/dz?
Reply 4
Original post by mikeft
what? where is dy/dz?


Im presuming you did the chain rule for dy/dx
dy/dx = dy/dz dz/dx = ...
You should be clear about what you tried to do there anyways.
Reply 5
Original post by mqb2766
Im presuming you did the chain rule for dy/dx
dy/dx = dy/dz dz/dx = ...
You should be clear about what you tried to do there anyways.

oh yeah with chain rule it makes sense but do i have to do chain rule in these kind of questions? i thought chain rule was not needed, i always differentiated y with respect to x and then just put the d?/d? something of the equation i need to get to in this case i have to put dz/dx next to the derevative is that wrong?
Reply 6
Original post by mikeft
oh yeah with chain rule it makes sense but do i have to do chain rule in these kind of questions? i thought chain rule was not needed, i always differentiated y with respect to x and then just put the d?/d? something of the equation i need to get to in this case i have to put dz/dx next to the derevative is that wrong?

Not sure I fully understand what youre saying, but yes you need to explicitly use the chain rule as youre introducing and transforming the equation to another variable z so you think like y(z(x)) so
dy/dx = dy/dz dz/dx
and you end up with dy/dx replaced by dz/dx (and similar for higher derivatives) multiplied by an expression which in this case cancels with the y^3 on the right.

You wrote
dy/dx = dz/dy dz/dx
which isnt correct
(edited 10 months ago)
Reply 7
Original post by mqb2766
Not sure I fully understand what youre saying, but yes you need to explicitly use the chain rule as youre introducing and transforming the equation to another variable z so you think like y(z(x)) so
dy/dx = dy/dz dz/dx
and you end up with dy/dx replaced by dz/dx (and similar for higher derivatives) multiplied by an expression which in this case cancels with the y^3 on the right.

You wrote
dy/dx = dz/dy dz/dx
which isnt correct

in this question how do i usse chain rule?
for part a
Original post by mikeft
in this question how do i usse chain rule?
for part a

Same again. dy/dx = (dy/dz)(dz/dx)
Reply 9
Original post by old_engineer
Same again. dy/dx = (dy/dz)(dz/dx)


when do i not do chain rule then, i think there is a case where it is not needed
Reply 10
Original post by mikeft
when do i not do chain rule then, i think there is a case where it is not needed


In the context of this thread, you use the chain rule as above. When do you think otherwise?
Reply 11
Original post by mqb2766
In the context of this thread, you use the chain rule as above. When do you think otherwise?


i asked my teacher and she told me i need chain rule only when i dont have the verible i need for the question
Reply 12
Original post by mikeft
i asked my teacher and she told me i need chain rule only when i dont have the verible i need for the question


Not sure what you mean. This thread is about defining a new variable and transforming the ode to be in terms of the new variable. This needs the chain rule as the variables/functions are chained. If youre suggesting a question (could you post an example/question?) where the variables/functions are not chained, then obviously the chain rule does not apply.

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