How to Solve the Special first order differential equation?
A standard form of Bernoulli equation is
$$\frac{dy}{dx} + P(X)y = Q(X) y^n$$
Fact: where n is any real but not 0 or 1. When n=0 then the equation can be solved as a first order linear differential equation
When n=1 then the equation can be solved using separation of variables
For other values of n we can solve it by substituting as follows:
$$u=y^{1-n}$$
and change it into a linear differential equation and then solve.
Example
$$\frac{dy}{dx} + x^4y= x^4y^6$$
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Step#1: It is a Bernoulli equation with P(x)=Q(x)=x^4 and n=6
By substitution
$$u=y^{1-n}$$
$$u=y^5$$
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Step#2: In terms of y which is
$$u=y^{\frac{-1}{5}}$$
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Step#3: Now Differentiate y with respect to x
$$\frac{dy}{dx}= -\frac{1}{5} u^{-\frac{6}{5}} \frac{du}{dx}$$
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Step#4: Substitute dy/dx and y into the original equation
$$\frac{dy}{dx} + x^5y= x^5y^6$$
$$-\frac{1}{5} u^{\frac{-6}{5}}\;\frac{du}{dx }+ x^4\;u^{\frac{-1}{6}}=x^4\; u^{\frac{-6}{5}}$$
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Step#5: Multiply all of the above by
$$-5u^{\frac{1}{5}}$$
$$-\frac{du}{dx}- 5x^4 u=-5x^4 u$$
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Step#6: The substitution worked now we have an equation we can hopefully solve.
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Step#7: Simplify
$$\frac{du}{dx} =5x^4u-5x^4$$
$$\frac{du}{dx} =(u-1)5x^4$$
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Step#8: by Using separation of Variables
$$\frac{du}{u}-1=5x^4$$
$$\frac{dy}{dx}=5x^5dx$$
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Step#9: Integrate on both sides
$$\int \frac{1}{u}-1 du = 5x^4dx$$
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Step#10: Conclusion
$$ln(u-1)=x^5+C$$
$$u-1=e^{x5}+C$$
$$u=(e^{x^5}+c +1)$$
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Step#11: Substitute back
$$y=u(\frac{-1}{5})$$
$$y=(e^{x^5+c}+1)^{(\frac{-1}{5})}$$