Solve Nonlinear First Order Differential Equation | y'(t)=y(t)^3+f(t)

bpcraig
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Maybe I'm just dumb...

<br /> y&#039;(t)=y(t)^3+f(t)<br />

find y(t)

Thanks...
 
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bpcraig said:
Maybe I'm just dumb...

<br /> y&#039;(t)=y(t)^3+f(t)<br />

find y(t)

Thanks...
Welcome bpcrai.
Try substituting z =1/y^2 if permissible. If the magnitude of z is large in the range, we can neglect a constant & separate the variables.

P.S. : I'm afraid everybody's dumb at solving nonlinear differential equations. ;)
 
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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