First order bernoulli differential equations

Form

$$y^{\prime }+P(x)y=Q(x)y^n$$

Note: A Bernoulli equation is just a linear where $n$ = 0 or 1, its also solved in a similar way with a very nice "bare bones" formula to solve them quickly. But first lets go through how the formula is derived like before.

In order to get this into a solvable form, were going to first divide the whole equation by $y^n$

$$[y^{\prime }+P(x)y=Q(x)y^n]\frac{1}{y^n}$$

$$\Rightarrow$$

$$y^{\prime }{y}^{-n}+P(x)y^{1-n}=Q(x)$$

now were going to use the substitution

$$v=y^{1-n}$$

Next were going differentiate this with implicit differentiation to give us

$$v^{\prime }=(1-n)y^{-n}{y}^{\prime }$$

rearranging to solve for $y^{\prime }$

$$y^{\prime }=\frac{v^{\prime }}{(1-n)y^{-n}}$$

we now have

$$y^{1-n}=v\text{and}{y}^{\prime }=\frac{v^{\prime }}{(1-n)y^{-n}}$$

Directly substituting this into the original equation gives us

$$y^{\prime }{y}^{-n}+P(x)y^{1-n}=Q(x)$$

$$\Rightarrow$$

$$\frac{v^{\prime }{y}^{-n}}{(1-n)y^{-n}}+P(x)v=Q(x)$$

$$\Rightarrow$$

$$\frac{v^{\prime }}{(1-n)}+P(x)v=Q(x)$$

$$\Rightarrow$$

$$v^{\prime }+P(x)(1-n)v=Q(x)(1-n)$$

This is now a first order linear ODE that we can solve, and then substitute the $y$ back in for the full answer.

Going through all the same steps, the integrating factor will be

$$I(x)=e^{\int (1-n)P\left(x\right)dx}$$

and the final general solution will be

$$y^{1-n}=\frac{1}{I(x)}(\int (1-n)Q(x)I(x)dx+c)$$

again putting everything together in one place the formula that can be followed to solve a Bernoulli Equation is...

Form

$$y^{\prime }+P(x)y=Q(x)y^n$$

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General Solution

$$I(x)=e^{\int (1-n)P(x)dx}$$

$$y^{1-n}=\frac{1}{I(x)}(\int (1-n)Q(x)I(x)dx+c)$$

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Examples:

$$\frac{dy}{dx}+\frac{y}{x}=8x^9{y}^2$$

First we find our $P(x)$, $Q(x)$, and $n$

$$P(x)=\frac{1}{x}Q(x)=8x^{9}n=2$$

Using this we find our integrating factor

$$I(x)=e^{(\int [1-2]\left[\frac{1}{x}\right]dx)}$$

$$e^{(\int -\frac{1}{x}dx)}=e^{-\ln |x|}$$

simplifying using the properties of logarithms

$$-\ln |x|=\ln |x^{-1}|=\ln \left|\frac{1}{x}\right|$$

$$e^{-\ln |x|}=e^{\ln \left|\frac{1}{x}\right|}=\frac{1}{x}$$

Our integrating factor is therefore

$$I(x)=\frac{1}{x}$$

Plugging this into the second part of our general solution formula we get

$$y^{1-2}=x(\int (1-2)(8x^9)\frac{1}{x}dx)$$

$$y^{-1}=x(-\int 8x^{8}dx)$$

$$y^{-1}=x(-\frac{8x^9}{9}+c)$$

distributing the $x$ and finding the implicit solution in terms of $y$

$$y=\frac{1}{-\frac{8x^{10}}{9}+cx}$$

For a final bit of simplification, we can multiply the $cx$ by $\frac{9}{9}$, have the ''$c$'' eat the 9 in its numerator which would give us

$$y=\frac{1}{\frac{-8x^{10}+cx}{9}}$$

factoring out a -1 and simplifying the denominator yields us a final answer of

$$y=\frac{-9}{8x^{10}-cx}$$

$$\frac{dx}{dt}+3t^3{x}^5+\frac{x}{t}=0$$

subtracting the $3t^3{x}^5$ to get it into the correct form

$$\frac{dx}{dt}+\frac{x}{t}=-3t^3{x}^5$$

Finding our $P(t)$, $Q(t)$, and $n$

$$P(t)=\frac{1}{t}Q(t)=-3t^{3}n=5$$

using this to get our integrating factor

$$I(t)=e^{(\int (1-5)\frac{1}{t}dt)}$$

$$e^{-4\ln |t|}=e^{\ln |t^{-4}|}=e^{\ln \left|\frac{1}{t^4}\right|}=\frac{1}{t^4}$$

$$I(t)=\frac{1}{t^4}$$

Plugging this into our general solution formula

$$x^{-4}=t^4[\int (1-5)(-3t^3)\left(\frac{1}{t^4}\right)]$$

$$x^{-4}=t^4[\int \frac{12}{t}dt]$$

$$x^{-4}=t^4(12\ln |t|+c)$$

Here we have our solution given in an explicit form.

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