First order differential eqn dy/dx + Py = Qy^n

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The discussion focuses on transforming the first-order differential equation dy/dx + Py = Qy^n into a linear equation using the substitution z = y^-(n-1). A participant initially struggled with the transformation, encountering complications with fractions instead of the expected result. However, they later resolved the issue independently, identifying a mistake in their differentiation involving negative indices. The conversation highlights the importance of careful differentiation when applying substitutions in differential equations. Ultimately, the transformation aligns with Bernoulli's equation principles.
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show that the substitution z = y^-(n-1) transforms the general equation dy/dx + Py = Qy^n, where P and Q are functions of x, into the linear equation dz/dx - P(n-1)z = -Q(n-1). (Bernoulli's equation)


Well, I looked up Bernoulli's stuff on internet, found the usual air flow equation but not this one. In fact I followed the standard procedure to transform the equation, but what I got is a whole bunch of fraction rather than just the -(n-1).

Please help.
 
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never mind

i managed to solved it myself, please ignore this post. i actually made mistake in my differentiation with the negative indices.
 

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Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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