Mathematics, 06.11.2019 04:31 vant

Abernoulli 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.

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Answers

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

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

The given equation is well expressed as:

$\mathbf{y' - \dfrac{5}{x}y =\dfrac{y^5}{y^9} \ , \ y(1) = 1}$

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 $\mathbf{y^5}$ , by doing so, we have

$\mathbf{y^{-5}y' - \dfrac{5}{x}y^{-4} =\dfrac{1}{y^9} }$

Now, let make:

$\mathbf{y^{-4} = u}$ and; $\mathbf{-4y^{-5} \dfrac{dy}{dx}= \dfrac{du}{dx}}$

∴

From the equation $\mathbf{y^{-5}y' - \dfrac{5}{x}y^{-4} =\dfrac{1}{y^9} }$ ;

$\mathbf{y^{-5} \dfrac{dy}{dx}= -\dfrac{1}{4} \dfrac{du}{dx}}$

We can now have the equation to be differentiated as:

$\mathbf{-\dfrac{1}{4}\dfrac{du}{dx} - \dfrac{5}{x}u = \dfrac{1}{x^9}}$ Multiplying both sides by 4, we have: $\mathbf{\dfrac{du}{dx} + \dfrac{20}{x}u = -\dfrac{4}{x^9}}$

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

$\mathbf{\dfrac{du}{dx} + \dfrac{20}{x}u = -\dfrac{4}{x^9}}$

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

We need to multiply the integration factor $\mathbf{ \mu(x)}$ with the equation

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

$\mathbf{x^{20} \Big(\dfrac{du}{dx} + \dfrac{20}{x}u = -\dfrac{4}{x^9}(x^{20} )\Big) }$ $\mathbf{x^{20} \dfrac{du}{dx} + 20x^{19}u = -4x^{11}} }$

Taking the integral of both sides;

$\mathbf{\int (x^{20} u)' dx = \int -4x^{11} \ dx }$ $\mathbf{ x^{20} u= -4\Big ( \dfrac{x^{12} }{12} \Big ) +c }$ $\mathbf{ x^{20} u= -\dfrac{1}{3}x^{12} +c }$ $\mathbf{u = -\dfrac{1}{3} x^{-8} +cx ^{-20}}$

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

$\mathbf{ y^{-4} = -\dfrac{1}{3}x^{-8} +cx ^{-20}}$

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

$\mathbf{\dfrac{1}{y^4(1)} = -\dfrac{1}{3}(1)^{-8} + c(1) ^{-20} }$ $\mathbf{\dfrac{1}{1} = -\dfrac{1}{3}+c}$ $\mathbf{c = \dfrac{4}{3}}$

∴

$\mathbf{y^{-4} = -\dfrac{1}{3} x^{-8} + \dfrac{4}{3}x^{-20}}$

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

$\mathbf{\dfrac{1}{y^{4} }= -\dfrac{1}{3x^8} + \dfrac{4/3}{x^{20}}}}$ $\mathbf{\dfrac{1}{y^4}= \dfrac{1}{3x^8}\Big(-1 + \dfrac{4}{x^{12}} \Big)} }$ $\mathbf{\dfrac{1}{y^4} = \dfrac{4-x^{12}}{3x^{20}}}$ $\mathbf{y^4 = \dfrac{3x^{20}}{4-x^{12}}}$ $\mathbf{y =\Big( \dfrac{3x^{20}}{4-x^{12} }\Big) ^{1/4}}$

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

Learn more about Bernoulli differential equation here:

link

Answer from: gracynamos

$y'-\dfrac5xy=\dfrac{y^5}{x^9}$

Divide through both sides by y^5 :

$y^{-5}y'-\dfrac5xy^{-4}=x^{-9}$

Now let $z=y^{-4}$ , so that $z'=-4y^{-5}y'$ . Then

$-\dfrac{z'}4-\dfrac5xz=x^{-9}$

$z'+\dfrac{20}xz=-4x^{-9}$

Multiply both sides by $x^{20}$ :

$x^{20}z'+20x^{19}z=\left(x^{20}z\right)'=-4x^{11}$

Integrate both sides to get

$x^{20}z=-\dfrac{x^{12}}3+C\implies z=-\dfrac1{3x^8}+\dfrac C{x^{20}}$

Solve for :

$y=\left(\dfrac C{x^{20}}-\dfrac1{3x^8}\right)^{-1/4}$

Given that y(1)=1 , we have

$1=\left(C-\dfrac13\right)^{-1/4}\implies C=\dfrac43$

so the particular solution is

$y=\left(\dfrac4{3x^{20}}-\dfrac1{3x^8}\right)^{-1/4}$

which we can rewrite as

$\boxed{y=\dfrac{3^{1/4}}{x^5\left(4-x^{12}\right)^{1/4}}}$

Answer from: Quest

The first one is 2/3the second is 1/160the third is 4/125

Answer from: Quest

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (

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Abernoulli differential equation is one of the form

dy/dx+p(x)y=q(x)y^n.

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