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Bernoulli's equation:
For now, I'll tag and refer to these as #de_bernoulli. This is in standard form btw.
It looks almost like a linear equation! In fact if n=0 it is by definition. We will see further that if n=1 you get a separable equation. So we ignore the cases when n=0,1 as these can be solved with prior tools.
Bernoulli's equations are important as you will see it in biology and in engineering.
Our goal is to find the general solution to y which is some function of x. We should expect one arbitrary constant in our final answer for y since this is a first order differential equation.
Notice there's an easy solution! y(x)=0 is a trivial solution to any Bernoulli equation.
You can verify that by plugging in y(x)=0 to the original expression:
\frac{dy}{dx}+0=0\quad \implies \quad0=0
Now let's find the general solution.
Let's move the y to the LHS:
y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)
notice that y(x)=0 is no longer a solution! It was lost due to dividing by zero. So from here on out we will have to remember to add it back in our final answers.
let y^{1-n}=u
Differentiating this with respect to x gives us:
(1-n)y^{-n}\frac{ dy }{ dx }=\frac{du}{dx}
y^{-n}\frac{ dy }{ dx }=\frac{ du }{ dx }{\frac{1}{1-n}}
substituting in we get:
y^{-n}\frac{ dy }{ dx }+P(x)u=Q(x)=\frac{ du }{ dx }{\frac{1}{1-n}+P(x)u}
And that is a linear equation again, which can be solved with prior tools. Here's a handy formula if you wanna solve Bernoulli equations quick:
\frac{1}{1-n}\frac{ du }{ dx }+P(x)u=Q(x)\quad \BoxJust remember that once you find u(x), substitute it back for y(x)^{1-n}=u(x) to get your solution for y, and don't forget to add y(x)=0 with your final answer!
Remember when I said that when n=1 the equation becomes a separable equation?:
y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)letn=1y^{-1}\frac{ dy }{ dx }+P(x)=Q(x)y^{-1}dy=dx(Q(x)-P(x))<-This is indeed a separable equation #de_separable
Examples of Bernoulli's equation:
#ex #de_bernoulli Find the general solution to:
y'+y=(xy)^2
Looks like a Bernoulli equation because when we distribute the ^2 we get x^2y^2 on the RHS. This also tells us that n=2
y'+y=x^2y^2
y'y^{-2}+y^{-1}=x^2
Note that we lost the
y(x)=0solution here, we will have to add it back in the end.
let u=y^{1-n}=y^{-1}
Differentiating wrt. x we get: \frac{du}{dx}=-y^{-2}{\frac{dy}{dx}}
y^{-2}{\frac{dy}{dx}=-\frac{ du }{ dx }}
y^{-2}{\frac{dy}{dx}+y^{-1}=-\frac{ du }{ dx }}+y^{-1}
{x^2=-\frac{ du }{ dx }}+y^{-1}
x^2=-\frac{du}{dx}+u
\frac{du}{dx}-u=-x^2
Yay we have a linear equation now! We can solve it using the techniques & formulas we learned for them.
let P(x)=-1 \quad Q(x)=-x^2 \qquad I(x)=e^{\int -1 \, dx}=e^{-x}
u=-e^{x}\int e^{-x}x^2 \, dx
How to integrate this? You can use integration by parts:
LIATE: log, inv trig, alg, trig, exp
\int fg' \, dx=fg-\int f'g \, dx
let f=x^2 \qquad f'=2x \qquad g'=e^{-x} \qquad g=-e^{-x}
u=-e^{x}\left( x^2(-e^{-x})-\int 2x(-e^{-x}) \, dx \right)
u=-e^{x}\left( -x^2e^{-x}+2\int xe^{-x} \, dx \right)
let f=x \qquad f'=1 \qquad g'=e^{-x} \qquad g=-e^{-x}
u=-e^x\left( -x^2e^{-x}+2\left( -xe^{-x}-\int -e^{-x} \, dx \right) \right)
\frac{1}{y}=-e^x\left( -x^2e^{-x}+2\left( -xe^{-x}-e^{-x} +C\right) \right)
\frac{1}{y}=x^2+2(x+1+Ce^x)
\frac{1}{y}=x^2+2x+2+Ce^x
The general solution to the DE is:
y(x)=\frac{1}{x^2+2x+2+Ce^x} \quad\text{as well as}\quad y(x)=0#end of lecture 3