Answer :
The differential equation is equivalent to
[tex]\frac{dy}{dx} -\frac{3x(1-y)}{x^{2} +1}[/tex]
If y denotes the fraction of the population who have heard the rumor, then 1 −y represents the fraction of the population who haven’t heard the rumor.
The rate of spread of the rumor (y'(t)) being proportional to the product of the fraction y of the population who have heard the rumor and the fraction who have not heard the rumor can then be rewritten as:
[tex]\frac{dy}{dt} =ky(1-y)[/tex]
for some positive constant k. This is a separable differential equation, so to solve it we separate the variables
[tex]\frac{dy}{y(1-y)} =kdt[/tex]
and integrate
[tex]\int\ \frac{dy}{y(1-y)} =\int k dt < = > ln |\frac{y}{1-y} |=kt+c[/tex]
Exponentiating, we get
[tex]|\frac{y}{1-y}|=e^{kt+c}[/tex]
Denoting ec by A, and taking into account that y ∈ (0, 1) we get
[tex]\frac{y}1-{y} =Ae^{kt} =\frac{Ae^{kt} }{1}[/tex]
Adding the numerators to the denominators in the above equality of fractions we obtain
[tex]y=\frac{y}{(1-y)+y} =\frac{Ae^{kt} }{1+Ae^{kt} }[/tex]
The Initial value:
[tex](x^{2}+1) \frac{dy}{dx} +3y(y-1)=0,y(0)=1[/tex]
The differential equation is equal to
[tex]\frac{dy}{dx} -\frac{3x(1-y)}{x^2+1}[/tex]
the initial condition of our problem is y(0) = 1, the solution must be y ≡ 1
The differential equation is equivalent to
[tex]\frac{dy}{dx} +\frac{3x}{x^2+1} y=\frac{3x}{x^2+1}[/tex]
which is a linear equation, with P(x) = Q(x) =[tex]\frac{3x}{x^2+1}[/tex] we have
[tex]\int P(x)=\frac{3}{2}\int \frac{2x}{x^2+1} dx=\frac{3}{2} ln(x^2+1)[/tex]
It follows that the integrating factor I(x) is then
[tex]I(x)e^{\int P(x)dx} =e^{\frac{3}{2} ln(x^2+1)} =(x^2+1)^{\frac{3}{2} }[/tex]
The general technique for solving linear differential equations yields
[tex]yI(x)=\int Q(x)I(x)dx=\int \frac{3x}{x^2+1} (x^2+1)^{\frac{3}{2} } \\\\=\int 3x(x^2+1)^{\frac{1}{2} } =(x^2+1)^{\frac{3}{2} } +C[/tex]
Dividing by I(x) we get
[tex]y=1+\frac{C}{(x^2+1)^{\frac{3}{2} } }[/tex]
The initial condition y(0) = 1 yields 1 = 1 + c, i.e. c = 0. Therefore y ≡ 1.
The differential equation is separable. We get this by separating the variables
[tex]\frac{dy}{y-1} -\frac{-3x}{x^2+1}[/tex]
Integrating, we obtain
[tex]ln|y-1|=\frac{-3}{2} ln(x^2+1)+c[/tex]
which yields by exponentiation
[tex]|y-1|=\frac{e^c}{(X^2+1)^\frac{3}{2} }[/tex]
substituting ec by a constant A, and allowing A to also be negative or 0, we get
[tex]y-1=\frac{A}{(x^2+1)^\frac{3}{2} }[/tex]
The initial condition y(0) = 1 implies that 1 − 1 = A/1 = A, hence A = 0 and y ≡ 1
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