Reducing to a linear equation

**Deadstar** · January 26th 2010, 03:40 PM

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

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

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

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

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

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

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

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

**Danny** · January 26th 2010, 03:47 PM

Originally Posted by Deadstar

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

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

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

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

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

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

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

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

If $z = y^{1-a}$ then wouldn't $\frac{dz}{dt} = (1-a)y^{-a} \frac{dy}{dt}$ ?

**Deadstar** · January 26th 2010, 04:00 PM

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

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