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I am once again stuck on one of these questions and I cannot figure it out. Any help is appreciated. Thanks in advance! Inflection Points For many differential equations, the easiest way to find inflection points is to use the differential equation rather than the solution itself. To do this, we can compute y′′ by differentiating y′, remembering to use the chain rule wherever y occurs. Next, we can substitute for y′ by using the differential equation and setting y"=0. Then we can solve for y to find the inflection points. (Keep in mind here that solving for y can also produce some equilibrium solutions, which may not be inflection points!) Use the technique described above to find the inflection point for the solutions of the differential equation y′=cos(y−t). Your answer may contain t. Also, in this problem, use n for any arbitrary integer, 2n for any arbitrary even integer, and 2n+1 for any arbitrary odd integer (you may or may not need this since we are dealing with trig functions).
if you can find y" then you are almost there , can you ?
or would it be +sin(t-x)x'?
your question above says y' =cos(y-t) is the variable on the right y or x ? it actually makes a huge difference.
oh im sorry! i should have replaced x with y.
y is a function of t, so the step-by-step solution would be this \[d^2y/dt^2 = y" = d/dt(\cos(y-t)) \] so by chain rule \[-\sin(y-t) * d/dt (y-t)\] = -sin(y-t)(y'-1)
-sin(y-t) = 0 (y'-1) = 0 --> cos(y-t) - 1 = 0 yeah?
help each question of a 5 multiple exam has 4 possible answers. suppose a student picks an answer at ramdon for each question.find the prob the student slects the correct answer on only the 2nd and 3rd questions
Thank you! i figured it out :)
there are other things that you have to be careful with finding the inflection points, so if you still need help just let me know.