Substitution in a differential equation, independent variable

[fiksx]

Homework Statement

$$y'=-\frac{1}{10}y+(cos t)y^2$$
when doing substitute for $z=\frac{1}{y}$
I understand this is $z(t)=\frac{1}{y(t)}$
I know t is independent variable and y is dependent variable
but I want to know what is z role here, is it change the dependent variable?
when $y(t)=\frac{1}{z(t)}$
then
I saw in book it is written $y'=\frac{-z'}{z^2}$, is this means $y(t)'=\frac{dz}{dy}\frac{dy}{dt}$?
is the relation is $z(y(t))$? can someone explain the implicit differentiation in this expression? and what is dependent and independent variable here? thanks!
I understand how to finish this. but I want to know the intuition behind this. I always confuse when doing substitution, what is the dependent and independent variable, and to what variable I need to implicitly differentiate.

Homework Equations

The Attempt at a Solution

 

[epenguin]

I can explain maybe some of the intuition.

The problem is pattern recognition. I will say that almost the only way (as far as ordinary nonmathematician scientific math users are concerned) to solve a differential equation is to know the solution! The so-called 'methods' are just ways of hammering equations until they look like something you recognise.

Here you have a function, (cos t)y, In which the two variables are mixed up. Nasty, maybe you haven't done anything like it before.
But maybe you have gone through what every scientist has to do, linear differentials of the form
Ay' + By = f(t)
With the stuff about general solution of the homogeneous equation, and the particular integral.
Maybe you could get some equation like that, at any rate you could get things involving y and its derivatives on one side and something involving only t on the other?
So you look you see you can get that by dividing the whole equation by y².
Do that and your first term is y'/y².
which you hopefully recognise as - d(1/y)/dt
So it looks like 1/y should be a more suitable variable than y.

 

[tnich]

Homework Statement

$$y'=-\frac{1}{10}y+(cos t)y^2$$
when doing substitute for $z=\frac{1}{y}$
I understand this is $z(t)=\frac{1}{y(t)}$
I know t is independent variable and y is dependent variable
but I want to know what is z role here, is it change the dependent variable?
when $y(t)=\frac{1}{z(t)}$
then
I saw in book it is written $y'=\frac{-z'}{z^2}$, is this means $y(t)'=\frac{dz}{dy}\frac{dy}{dt}$?
is the relation is $z(y(t))$? can someone explain the implicit differentiation in this expression? and what is dependent and independent variable here? thanks!
I understand how to finish this. but I want to know the intuition behind this. I always confuse when doing substitution, what is the dependent and independent variable, and to what variable I need to implicitly differentiate.

Homework Equations

The Attempt at a Solution

$y'=\frac{-z'}{z^2}$ is just the chain rule $\frac{dy}{dt}=\frac{dy}{dz}\frac{dz}{dt}$

 

[fiksx]

I can explain maybe some of the intuition.

The problem is pattern recognition. I will say that almost the only way (as far as ordinary nonmathematician scientific math users are concerned) to solve a differential equation is to know the solution! The so-called 'methods' are just ways of hammering equations until they look like something you recognise.

Here you have a function, (cos t)y, In which the two variables are mixed up. Nasty, maybe you haven't done anything like it before.
But maybe you have gone through what every scientist has to do, linear differentials of the form
Ay' + By = f(t)
With the stuff about general solution of the homogeneous equation, and the particular integral.
Maybe you could get some equation like that, at any rate you could get things involving y and its derivatives on one side and something involving only t on the other?
So you look you see you can get that by dividing the whole equation by y².
Do that and your first term is y'/y².
which you hopefully recognise as - d(1/y)/dt
So it looks like 1/y should be a more suitable variable than y.

thanks but can I say the relation is $y(z(t))$? , both y and z is function of t. so $$1/y $$ is like the integrating factor right?, is there a way to find it in general for all problem? beside the first order linear differential equation, I know it use $\int e$

 

[LCKurtz]

thanks but can I say the relation is $y(z(t))$? , both y and z is function of t. so $1/y$ is like the integrating factor right?, is there a way to find it in general for all problem? beside the first order linear differential equation, I know it use $\int e$

Use double $ or # signs to delimit latex. The double $ gives display mode and the double # gives inline tex.
No, there is no general way for all problems. But this method works for a class of similar problems where you have a linear first order left side and a $y^n$ on the right side. In your case, as others have pointed out, you can divide by the $y^2$ giving$$
y^{-2}y' + \frac 1 {10}y^{-1} =\cos t$$Then if you change the dependent variable to $z=y^{-1}$, you get a linear equation in $z$. The same technique works for $y^n$ on the right side. It is called a Bernoulli substitution, I suppose because he must have first noticed that substitution.

 

[fiksx]

Use double $ or # signs to delimit latex. The double $ gives display mode and the double # gives inline tex.
No, there is no general way for all problems. But this method works for a class of similar problems where you have a linear first order left side and a $y^n$ on the right side. In your case, as others have pointed out, you can divide by the $y^2$ giving$$
y^{-2}y' + \frac 1 {10}y^{-1} =\cos t$$Then if you change the dependent variable to $z=y^{-1}$, you get a linear equation in $z$. The same technique works for $y^n$ on the right side. It is called a Bernoulli substitution, I suppose because he must have first noticed that substitution.

thanks but I want to know when you change dependent variable from z to y $z=y^{-1}$, when I differentiate it with respect x, why I need the chain rule? is it because $$z(y(t))$$ ? the relation is z->y->t?

 

[LCKurtz]

thanks but I want to know when you change dependent variable from z to y $z=y^{-1}$, when I differentiate it with respect x, why I need the chain rule? is it because $$z(y(t))$$ ? the relation is z->y->t?

You mean differentiate with respect to $t$. Yes, $z(t) = y(t)^{-1}$ so $z'(t) = -y(t)^{-2}y'(t)$.