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## Main Question or Discussion Point

This problem is out of Kreyszig's 9E of advanced engineering mathematics.

I don't understand their closed form solution.

Their solution ends up being

y=3tan(3x+c)-9x.

But when they were solving v=y+9x for y,

then taking the derivative, they get:

y'=v'-9, then set that =v^2

--Thats the step I have issues with. They are taking the derivative,

but with respect to what? If it is some dummy variable,

it would be y'=v'-9x' (chain rule?). If it is y with respect to v,

the derivative of y=v-9x would just be y'=v'. Unless x' with respect to

a dummy variable is = 1, of course.

Can someone please enlighten me?

I don't understand their closed form solution.

Their solution ends up being

y=3tan(3x+c)-9x.

But when they were solving v=y+9x for y,

then taking the derivative, they get:

y'=v'-9, then set that =v^2

--Thats the step I have issues with. They are taking the derivative,

but with respect to what? If it is some dummy variable,

it would be y'=v'-9x' (chain rule?). If it is y with respect to v,

the derivative of y=v-9x would just be y'=v'. Unless x' with respect to

a dummy variable is = 1, of course.

Can someone please enlighten me?