Thread: Reducing to a linear equation

1. Reducing to a linear equation

This is part of a larger question but I'm stick on this part.

I need to change $\displaystyle \frac{dy}{dt} + f(t)y = g(t)y^a$ (*) where $\displaystyle f \textrm{ and } g$ are continuous functions and $\displaystyle a \in \mathbb{R}$ into...

$\displaystyle \frac{dz}{dt} + (1-a)f(t)z = (1-a)g(t)$ using $\displaystyle z = y^{1-a}$.

So what I've done so far is...

Multiply both sides of (*) by $\displaystyle y^{-a}$
$\displaystyle y^{-a}\frac{dy}{dt} + f(t)y^{1-a} = g(t)$

Multiply both sides by $\displaystyle 1-a$
$\displaystyle (1-a)y^{-a}\frac{dy}{dt} + (1-a)f(t)z = (1-a)g(t)$

So we now have $\displaystyle \frac{dz}{dt}\frac{dy}{dt} + (1-a)f(t)z = (1-a)g(t)$ (i think...)

But how do I get rid of the $\displaystyle \frac{dy}{dt}$?

This is part of a larger question but I'm stick on this part.

I need to change $\displaystyle \frac{dy}{dt} + f(t)y = g(t)y^a$ (*) where $\displaystyle f \textrm{ and } g$ are continuous functions and $\displaystyle a \in \mathbb{R}$ into...

$\displaystyle \frac{dz}{dt} + (1-a)f(t)z = (1-a)g(t)$ using $\displaystyle z = y^{1-a}$.

So what I've done so far is...

Multiply both sides of (*) by $\displaystyle y^{-a}$
$\displaystyle y^{-a}\frac{dy}{dt} + f(t)y^{1-a} = g(t)$

Multiply both sides by $\displaystyle 1-a$
$\displaystyle (1-a)y^{-a}\frac{dy}{dt} + (1-a)f(t)z = (1-a)g(t)$

So we now have $\displaystyle \frac{dz}{dt}\frac{dy}{dt} + (1-a)f(t)z = (1-a)g(t)$ (i think...)

But how do I get rid of the $\displaystyle \frac{dy}{dt}$?
If $\displaystyle z = y^{1-a}$ then wouldn't $\displaystyle \frac{dz}{dt} = (1-a)y^{-a} \frac{dy}{dt}$ ?

3. Thanks that's kinda what I thought but wasn't sure.