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## Main Question or Discussion Point

it...just...does...not...make...the...slightest...sense...to...me....

Here it goes...

[tex]y' + \frac{y}{x} = 3x^2y^2[/tex]

This is a Bernoulli equation with [tex]P = \frac{1}{2}[/tex], [tex]Q = 3x^2[/tex], and [tex]n = 2[/tex]. We first divide through by [tex]y^2[/tex], obtaining...

[tex]\frac{1}{y^2} \frac{dy}{dx} + \frac{y^-^1}{x} = 3x^2[/tex]

We substitute [tex]z = y^-^1[/tex] and

[tex]\frac{dz}{dx} = -y^-^2\frac{dy}{dx} \Longrightarrow \frac{dy}{dx} = -y^2\frac{dz}{dx}[/tex]

Substituting... [tex]-\frac{dz}{dx} + \frac{z}{x} = 3x^2 \Longrightarrow \frac{dz}{x} - \frac{z}{x} = -3x^2[/tex]

I'll stop there... two things I don't get... how does [tex]P = \frac{1}{2}[/tex] and how does that substitution work...

If you can help, I would really appreciate it.

Here it goes...

[tex]y' + \frac{y}{x} = 3x^2y^2[/tex]

This is a Bernoulli equation with [tex]P = \frac{1}{2}[/tex], [tex]Q = 3x^2[/tex], and [tex]n = 2[/tex]. We first divide through by [tex]y^2[/tex], obtaining...

[tex]\frac{1}{y^2} \frac{dy}{dx} + \frac{y^-^1}{x} = 3x^2[/tex]

We substitute [tex]z = y^-^1[/tex] and

[tex]\frac{dz}{dx} = -y^-^2\frac{dy}{dx} \Longrightarrow \frac{dy}{dx} = -y^2\frac{dz}{dx}[/tex]

Substituting... [tex]-\frac{dz}{dx} + \frac{z}{x} = 3x^2 \Longrightarrow \frac{dz}{x} - \frac{z}{x} = -3x^2[/tex]

I'll stop there... two things I don't get... how does [tex]P = \frac{1}{2}[/tex] and how does that substitution work...

If you can help, I would really appreciate it.