Solve the Bernoulli differential equation.

Question

Given the following Bernoulli Differential Equations $$\dfrac{\mathrm dy}{\mathrm dx}+2xy=2x^3y^3$$ Can someone please help? Thanks in advance

Answer 1

To solve Bernoulli equation of the form

$\dfrac{\mathrm dy}{\mathrm dx}+yP(x)=y^nQ(x)$ we divide both sides by $y^n$ and then put $y^{1−n}=v$ to reduce it to linear differential equation.

In this case the differential equation is

$\dfrac{\mathrm dy}{\mathrm dx}+2xy=2x^3y^3......(1)$

(1) Divide both sides by $y^3$

$\therefore\;$ equation(1) because $\dfrac{1}{y^3}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{2x}{y^2}=2x^3$

$y^{-3}\;\dfrac{\mathrm dy}{\mathrm dx}+2xy^{-2}=2x^3......(2)$

Put $y^{-2}=v$

$\Rightarrow-2y^{-3}\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dv}{\mathrm dx}$

$\therefore\;$ equation(2) becomes

$=\dfrac{-1}{2}\dfrac{\mathrm dv}{\mathrm dx}+2vx=2x^3$

$\Rightarrow \dfrac{\mathrm dv}{\mathrm dx}-4vx=-4x^3$ (which is Linear)

$I.F.=e^{\int P(x)\mathrm dx}$

$=e^{\int-4x}$

$=e^{-2x^2}$

$\therefore\;$ General solution is

$ve^{-2x^2}=\int-4x^3e^{-2x^2}\mathrm dx$

on R.H.S. put $-2x^2=t$

$-4x \;\mathrm dx=\mathrm dt$

$\displaystyle ve^{-2x^2}=\int-\dfrac{t}{2}e^t\;\mathrm dt$

$\Rightarrow v=x^2+\dfrac{1}{2}+ce^{2x^2}$

Put back $v=\dfrac{1}{y^2}$

$\Rightarrow \dfrac{1}{y^2}=x^2+\dfrac{1}{2}+ce^{2x^2}$(required general solution)

Answer 2

We have Bernoulli Differential Equation: $$y'+P(x)y=Q(x)y^n \tag{1}$$

We divide both sides by $y^3$ to obtain: $$\frac{y'}{y^3}+2\frac{x}{y^2}=2x^3$$ Then, let $v(x)=\frac{1}{y^2}$ since the substitution $v(x)=y^{1-n}$ reduces any Bernoulli differential equation to a linear one. This leads to $\frac{dv(x)}{dx}=-2\frac{y'}{y^3}$ by Implicit differentiation. Substituting leads to the following linear differential equation: $$-\frac{1}{2}\frac{dv}{dx}+2xv=2x^3$$ $$\frac{dv}{dx}-4xv=-4x^3 \tag{2}$$ Now, you can solve this easily using an Integrating Factor or by using Variation of Parameters.

Answer 3

Hint : Divide throughout by (y^3).. Put y^(-2) = v. So that

(-2/y^3)dy/dx=dv/dx

Your original equation then becomes a linear one in and can be solved for v.