Relation between independent/dependent variable for substitution in ODE

467 Views Asked by At

I have the following ODE: $$ y'=y^2\cos t-\frac{1}{10}y $$ substituting $z=1/y$, I understand this is $z(t)=1/y(t)$. I know $t$ is the independent variable and $y$ is dependent variable.

What is the role of $z$ here? Does it change the dependent variable when $y(t)=1/z(t)$?

I saw in the book that it is written: $$ y'=\frac{-z'}{z^2} $$ does this mean that: $$ y'(t)=\frac{dz}{dy}\frac{dy}{dt} $$ is the relation $z(y(t))$?

Can someone explain the implicit differentiation in this expression? What are the dependent and independent variables here?

I understand how to finish this, but I want to know the intuition behind. I always confuse what are the dependent and independent variables when doing substitution, and to what variable I need to implicitly differentiate.

2

There are 2 best solutions below

6
On BEST ANSWER

$y'$ in your equation means $ \frac {dy}{dt}$ since z is a func tion of the variable t

$$y'=\frac {dy}{dt}$$ $$y'=\frac {dy}{dz}\frac {dz}{dt}=\frac {dy}{dz}z'$$ Since you have $y=\frac 1 z$ then $$\frac {dy}{dz}=\frac {d}{dz}\frac 1z=-\frac {1}{z^2}$$ Therefore, $$y'=\frac {dy}{dz}z'=-\frac {z'}{z^2}$$ The simpliest way is to differentiate with respect to the variable t

$$y=\frac 1 z$$ Where z is a function of t $$\frac {dy}{dt}=\frac d {dt}\left(\frac 1 z\right)$$ $$y'=\frac {-z'}{z^2}$$

2
On

write your equation in the form $$-\frac{y'(t)}{y(t)^2}-\frac{1}{10y(t)}=-\cos(t)$$ substituting $$u(t)=\frac{1}{y(t)}$$ then you will get $$u'(t)-\frac{u(t)}{10}=-\cos(t)$$ can you finish?