A Bernoulli differential equation is one of the form

dy/dx+P(x)y=Q(x)y^n.

Observe that, if n=0 or 1, the Bernoulli equation is linear. For other values of n, the substitution u=y1−n transforms the Bernoulli equation into the linear equation

du/dx+(1−n)P(x)u=(1−n)Q(x).

Use an appropriate substitution to solve the equation

y′−5/x*y=y^5/x^9,

and find the solution that satisfies y(1)=1.

The solution to the differential equation y′−5/x*y=y^5/x^9, by using the appropriate substitution is [tex]\mathbf{y^{-4} = \dfrac{1}{3}x^{-8} + \dfrac{4}{3}x^{-20}}[/tex] . The general solution that

satisfies y(1) = 1 is [tex]\mathbf{y(x) = \Big (\dfrac{3x^{20}}{4-x^{12}} \Big) ^{1/4}}[/tex]

The given equation is well expressed as:

[tex]\mathbf{y' - \dfrac{5}{x}y =\dfrac{y^5}{y^9} \ , \ y(1) = 1}[/tex]

We are to apply the Bernoulli equation approach to solve the given differential equation.

So, the first step is to divide the differential equation by [tex]\mathbf{y^5}[/tex], by doing so, we have

[tex]\mathbf{y^{-5}y' - \dfrac{5}{x}y^{-4} =\dfrac{1}{y^9} }[/tex]

Now, let make:

[tex]\mathbf{y^{-4} = u}[/tex] and;

[tex]\mathbf{-4y^{-5} \dfrac{dy}{dx}= \dfrac{du}{dx}}[/tex]

∴

From the equation [tex]\mathbf{y^{-5}y' - \dfrac{5}{x}y^{-4} =\dfrac{1}{y^9} }[/tex] ;

[tex]\mathbf{y^{-5} \dfrac{dy}{dx}= -\dfrac{1}{4} \dfrac{du}{dx}}[/tex]

We can now have the equation to be differentiated as:

[tex]\mathbf{-\dfrac{1}{4}\dfrac{du}{dx} - \dfrac{5}{x}u = \dfrac{1}{x^9}}[/tex]

Multiplying both sides by 4, we have:

[tex]\mathbf{\dfrac{du}{dx} + \dfrac{20}{x}u = -\dfrac{4}{x^9}}[/tex]

Hence, we can infer that the linear equation in terms of u is:

[tex]\mathbf{\dfrac{du}{dx} + \dfrac{20}{x}u = -\dfrac{4}{x^9}}[/tex]

Now, taking the integration factor: [tex]\mathbf{\mu(x) = e^ {^{\Big \int \frac{20}{x}\ dx }} = e^{20 \ In \ x} = x^{20} }[/tex]

We need to multiply the integration factor [tex]\mathbf{ \mu(x)}[/tex] with the equation

[tex]\mathbf{\dfrac{du}{dx} + \dfrac{20}{x}u = -\dfrac{4}{x^9}}[/tex] , so we have:

[tex]\mathbf{x^{20} \Big(\dfrac{du}{dx} + \dfrac{20}{x}u = -\dfrac{4}{x^9}(x^{20} )\Big) }[/tex]

[tex]\mathbf{x^{20} \dfrac{du}{dx} + 20x^{19}u = -4x^{11}} }[/tex]

Taking the integral of both sides;

[tex]\mathbf{\int (x^{20} u)' dx = \int -4x^{11} \ dx }[/tex]

[tex]\mathbf{ x^{20} u= -4\Big ( \dfrac{x^{12} }{12} \Big ) +c }[/tex]

[tex]\mathbf{ x^{20} u= -\dfrac{1}{3}x^{12} +c }[/tex]

[tex]\mathbf{u = -\dfrac{1}{3} x^{-8} +cx ^{-20}}[/tex]

Recall that: [tex]\mathbf{u = y^{-4}}[/tex], if we replace the value of u with [tex]\mathbf{ y^{-4}}[/tex] in the above equation, we will have:

[tex]\mathbf{ y^{-4} = -\dfrac{1}{3}x^{-8} +cx ^{-20}}[/tex]

Using the given initial condition y(1) = 1 to the equation above to determine the value of c, we have:

[tex]\mathbf{\dfrac{1}{y^4(1)} = -\dfrac{1}{3}(1)^{-8} + c(1) ^{-20} }[/tex]

[tex]\mathbf{\dfrac{1}{1} = -\dfrac{1}{3}+c}[/tex]

[tex]\mathbf{c = \dfrac{4}{3}}[/tex]

∴

[tex]\mathbf{y^{-4} = -\dfrac{1}{3} x^{-8} + \dfrac{4}{3}x^{-20}}[/tex]

Now, to solve for y(x); we have:

[tex]\mathbf{\dfrac{1}{y^{4} }= -\dfrac{1}{3x^8} + \dfrac{4/3}{x^{20}}}}[/tex]

[tex]\mathbf{\dfrac{1}{y^4}= \dfrac{1}{3x^8}\Big(-1 + \dfrac{4}{x^{12}} \Big)} }[/tex]

[tex]\mathbf{\dfrac{1}{y^4} = \dfrac{4-x^{12}}{3x^{20}}}[/tex]

[tex]\mathbf{y^4 = \dfrac{3x^{20}}{4-x^{12}}}[/tex]

[tex]\mathbf{y =\Big( \dfrac{3x^{20}}{4-x^{12} }\Big) ^{1/4}}[/tex]

Therefore, the general solution that satisfies y(1) = 1 is [tex]\mathbf{y(x) = \Big (\dfrac{3x^{20}}{4-x^{12}} \Big) ^{1/4}}[/tex]

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