Question

1. 4 pts specify the type of the equation and solve for y

\frac{dy}{dx} + \frac{y}{x} - 9x^9 y^2 = 0

Explanation:

Step1: Rewrite the differential equation

The given differential equation is \(\frac{dy}{dx}+\frac{y}{x}-9x^{9}y^{2} = 0\), we can rewrite it as \(\frac{dy}{dx}+\frac{y}{x}=9x^{9}y^{2}\). This is a Bernoulli differential equation of the form \(\frac{dy}{dx}+P(x)y = Q(x)y^{n}\), where \(P(x)=\frac{1}{x}\), \(Q(x) = 9x^{9}\) and \(n = 2\).

Step2: Transform the Bernoulli equation

For a Bernoulli equation \(\frac{dy}{dx}+P(x)y=Q(x)y^{n}\), we use the substitution \(v=y^{1 - n}\). Here \(n = 2\), so \(v=y^{- 1}\), then \(\frac{dv}{dx}=-y^{-2}\frac{dy}{dx}\), or \(\frac{dy}{dx}=-y^{2}\frac{dv}{dx}\)

Substitute \(y=\frac{1}{\sqrt{v}}\) (since \(v = y^{-1}\), \(y=\frac{1}{v}\) actually, my mistake earlier, \(v=y^{1 - n}=y^{-1}\), so \(y=\frac{1}{v}\), \(\frac{dy}{dx}=-\frac{1}{v^{2}}\frac{dv}{dx}\)) into the equation \(\frac{dy}{dx}+\frac{y}{x}=9x^{9}y^{2}\)

We get \(-\frac{1}{v^{2}}\frac{dv}{dx}+\frac{1}{xv}=9x^{9}\cdot\frac{1}{v^{2}}\)

Multiply through by \(-v^{2}\) to get \(\frac{dv}{dx}-\frac{v}{x}=- 9x^{9}\)

Step3: Solve the linear differential equation

The equation \(\frac{dv}{dx}-\frac{v}{x}=-9x^{9}\) is a linear differential equation of the form \(\frac{dv}{dx}+P(x)v=Q(x)\), where \(P(x)=-\frac{1}{x}\) and \(Q(x)=-9x^{9}\)

The integrating factor \(\mu(x)=e^{\int P(x)dx}=e^{\int-\frac{1}{x}dx}=e^{-\ln x}=\frac{1}{x}\)

Multiply the linear differential equation by the integrating factor:

\(\frac{1}{x}\frac{dv}{dx}-\frac{v}{x^{2}}=-9x^{8}\)

The left - hand side is the derivative of \(\frac{v}{x}\) with respect to \(x\), i.e., \(\frac{d}{dx}(\frac{v}{x})=-9x^{8}\)

Step4: Integrate both sides

Integrate both sides with respect to \(x\):

\(\int\frac{d}{dx}(\frac{v}{x})dx=\int - 9x^{8}dx\)

\(\frac{v}{x}=-9\cdot\frac{x^{9}}{9}+C=-x^{9}+C\)

Step5: Substitute back \(v = y^{-1}\)

Since \(v = y^{-1}=\frac{1}{y}\), we have \(\frac{1}{xy}=-x^{9}+C\)

Then \(y=\frac{1}{x(-x^{9}+C)}=\frac{1}{-x^{10}+Cx}\)

Answer:

The equation is a Bernoulli differential equation. The solution is \(y=\frac{1}{Cx - x^{10}}\) (where \(C\) is an arbitrary constant)