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Numerade Educator



Problem 23 Hard Difficulty

A Bernoulli differential equation (named after James Bernoulli) is of the form
$ \frac {dy}{dx} + P(x)y = Q(x)y^n $
Observe that, if $ n = 0 $ or 1, the Bernoulli equation is linear. For other values of $ n, $ show that the substitution $ u = y^{1-n} $ transforms the Bernoulli equation into the linear equation
$ \frac {du}{dx} + ( 1 - n)P(x)u = (1 - n)Q(x) $


Then the Bernoulli differerntial equation becomes $\rightarrow \frac{u^{n /(1-n)}}{1-n} \frac{d u}{d x}+P(x) u^{1 /(1-n)}=$
$Q(x) u^{n /(1-n)}$ or $\frac{d u}{d x}+(1-n) P(x) u=Q(x)(1-n)$


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Video Transcript

okay, We know that we have to essentially prove this, which means the first stop we can do is the change will. Which means we can write our d'you Jax as one minus and times wide to negative on D y over D backs. In other words, writing this simply in terms of d Y over Jax. We have wide toothy on over one month's on D'You over DX and again, we use the chain rule on the power rule power room into increased, exported by one divide by the new exponents, which means we now have well, the fact that we're multiplying both sides of the equation by one minus and divided by wives theon. So again, just recall the fact that we're still looking at the same standard form of the equation with our pee vacs and our cue of axe. Given this, we have now proven we have now shown how the substitution transforms the equation into the given linear equation.