- #1

- 5

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solve by x.

y^3+x^2y^2-y*sinx=1

derivate:

3y^2*y' + 2xy^2 + 2yx^2*y' - (y'sinx + ycosx) = 0

but to derivate this once again Im not sure how to proceed.

My solution:

[3y^2*y'+6y*y''] + [2y^2 + 4xyy'] + .....

solutoin manual:

6y*y'' + 2x2yy' + [2x^2*1*y'+2x^2*y*y'']

I've been searching for websites where I can read about this, but no luck.

What I think is done in the solution manual is that when x is x^1 it cannot be further derivated, because we kinda "loose" information?

I need some guidance or explination on how to handle equations that got:

y*y'

x*y^2

x*y*y'

Thanks in advance!