- #1
phion
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I'm practicing first-order linear differential equations, and have come across something I find interesting - being able to reduce nonlinear equations to linear equation with appropriate substitutions. I'll start with the well-known Bernoulli equation, and if there are other ways to do this technique please share!
y' + P(x)y = Q(x)y^n
This equation is linear when n=0 and has seperable variables if n=1. So, in the following development, and assuming that n≠0 and n≠1, we can multiply by y^{-n} and (1-n) to obtain
y^{-n}y' + P(x)y^{1-n}=Q(x)
(1-n)y^{-n}y'+(1-n)P(x)y^{1-n}=(1-n)Q(x)
\frac{d}{dx}[y^{1-n}]+(1-n)P(x)y^{1-n}=(1-n)Q(x)
which is a linear equation with the variable y^{1-n}, and if we let z=y^{1-n} we then get
\frac{dz}{dx}+(1-n)P(x)z=(1-n)Q(x).
Now, by multiplying by the integrating factor e^{∫P(x)dx} we can convert the left side of the equation into the derivative of the product ye^{∫P(x)dx}, and we get the general solution of the Bernoulli equation!
y^{(1-n)P(x)dx}=∫(1-n)Q(x)e^{∫(1-n)P(x)dx}dx+C
That's freakin' qute, right?
y' + P(x)y = Q(x)y^n
This equation is linear when n=0 and has seperable variables if n=1. So, in the following development, and assuming that n≠0 and n≠1, we can multiply by y^{-n} and (1-n) to obtain
y^{-n}y' + P(x)y^{1-n}=Q(x)
(1-n)y^{-n}y'+(1-n)P(x)y^{1-n}=(1-n)Q(x)
\frac{d}{dx}[y^{1-n}]+(1-n)P(x)y^{1-n}=(1-n)Q(x)
which is a linear equation with the variable y^{1-n}, and if we let z=y^{1-n} we then get
\frac{dz}{dx}+(1-n)P(x)z=(1-n)Q(x).
Now, by multiplying by the integrating factor e^{∫P(x)dx} we can convert the left side of the equation into the derivative of the product ye^{∫P(x)dx}, and we get the general solution of the Bernoulli equation!
y^{(1-n)P(x)dx}=∫(1-n)Q(x)e^{∫(1-n)P(x)dx}dx+C
That's freakin' qute, right?
