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 \(y^{n}\)
\[\left[y^{\prime}+P(x)y = Q(x)y^{n}\right] \frac{1}{y^{n}}\]\[\Rightarrow\]\[y^{\prime}y^{-n}+P(x)y^{1-n} = Q(x)\]
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
\[y^{1-n} = v \quad \text{and} \quad y^{\prime}=\frac{v^{\prime}}{(1-n)y^{-n}}\]
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
\[I(x) = e^{\int \left(1-n\right)P\left(x\right)dx}\]
and the final general solution will be
\[y^{1-n}= \frac{1}{I(x)}\left(\int (1-n) Q(x)I(x)dx +c\right)\]
again putting everything together in one place the formula that can be followed to solve a Bernoulli Equation is...
Form
\[y^{\prime}+P(x)y = Q(x)y^n\]
General Solution
\[I(x) = e^{\int (1-n)P(x)dx}\]\[y^{1-n}= \frac{1}{I(x)}\left(\int (1-n) Q(x)I(x)dx +c\right)\]
Examples:
\[\frac{dy}{dx} + \frac{y}{x} = 8x^{9}y^{2}\]First we find our \(P(x)\), \(Q(x)\), and \(n\)
\[P(x) = \frac{1}{x} \qquad\qquad Q(x) = 8x^{9} \qquad\qquad n = 2\]Using this we find our integrating factor
\[I(x) = e^\left(\int [1-2]\left[ \frac{1}{x}\right]\,dx\right)\]\[e^{\left(\int - \frac{1}{x}\,dx\right)} = e^{-\ln|x|}\]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
\[y^{1-2} = x\left( \int(1-2) \, (8x^{9}) \, \frac{1}{x} \,dx\right)\]\[y^{-1} = x \left(-\int8x^{8}\,dx\right)\]\[y^{-1} = x\left(- \frac{8x^{9}}{9}+c \right)\]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 \(n\)
\[P(t) = \frac{1}{t} \qquad\qquad Q(t) = -3t^{3} \qquad\qquad n = 5\]using this to get our integrating factor
\[I(t) = e^{\left( \int(1-5) \frac{1}{t} \,dt \right)}\]\[e^{-4\ln|t|} = e^{\ln |t^{-4}|} = e^{\ln\left| \frac{1}{t^{4}}\right|} = \frac{1}{t^{4}}\]\[I(t) = \frac{1}{t^{4}}\]Plugging this into our general solution formula
\[x^{-4} = t^{4}\left[\int(1-5)(-3t^{3})\left( \frac{1}{t^{4}}\right)\right]\]\[x^{-4} = t^{4}\left[\int \frac{12}{t}\,dt\right]\]\[x^{-4} = t^{4}(12\ln|t| + c)\]Here we have our solution given in an explicit form.