What is the Bernoulli Differential Equation Form of 3y^2y' + y^3 = e^-x?

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The Problem:

3y2y' + y3 = e-x

I think maybe I'm supposed to use something to do with the Bernoulli stuff, but I'm not sure. I've tried to figure it out for a while now and I'm stuck.

Thanks for any help you can give.
 
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It's in the form of a Bernoulli DE if you divide everything by 3y^2.
 
Thread 'Direction Fields and Isoclines'

Thread 'Direction Fields and Isoclines'

I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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