- #1

- 5

- 0

(2x^2)yy' = −1

I'm just really not sure how to go about solving this.

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- Thread starter SuperNomad
- Start date

- #1

- 5

- 0

(2x^2)yy' = −1

I'm just really not sure how to go about solving this.

- #2

Mute

Homework Helper

- 1,388

- 10

Why isn't it separable?

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

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

- #3

- 5

- 0

Ah yes, that's actually quite easy, I'm just being dense.

Thanks for the help.

SuperNomad

Thanks for the help.

SuperNomad

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