Key Concepts and Formulas

• Bernoulli's Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ can be transformed into a linear differential equation.
• Substitution for Bernoulli's Equation: Let $z = y^{1-n}$. Then, $\frac{dz}{dx} = (1-n)y^{-n}\frac{dy}{dx}$.
• Integrating Factor for Linear Differential Equation: For a linear differential equation of the form $\frac{dz}{dx} + P'(x)z = Q'(x)$, the integrating factor is given by $I.F. = e^{\int P'(x) dx}$. The solution is $z \cdot (I.F.) = \int Q'(x) \cdot (I.F.) dx + C$.

Step-by-Step Solution

Step 1: Identify and Rearrange the Differential Equation

The given differential equation is $(2xy^2 - y)dx + xdy = 0$. We want to express it in a recognizable form. Divide by $dx$: $2xy^2 - y + x\frac{dy}{dx} = 0$ Rearrange to isolate $\frac{dy}{dx}$: $x\frac{dy}{dx} = y - 2xy^2$ Divide by $x$: $\frac{dy}{dx} = \frac{y}{x} - 2y^2$ Rewrite: $\frac{dy}{dx} - \frac{1}{x}y = -2y^2$ This is a Bernoulli's differential equation.

Step 2: Transform into a Linear Differential Equation

We have the Bernoulli equation $\frac{dy}{dx} - \frac{1}{x}y = -2y^2$. Here, $P(x) = -\frac{1}{x}$, $Q(x) = -2$, and $n = 2$. Let $z = y^{1-n} = y^{-1} = \frac{1}{y}$. Then, $\frac{dz}{dx} = -y^{-2}\frac{dy}{dx} = -\frac{1}{y^2}\frac{dy}{dx}$. Divide the original equation by $y^2$: $\frac{1}{y^2}\frac{dy}{dx} - \frac{1}{x}\frac{1}{y} = -2$ Substitute $z = \frac{1}{y}$ and $\frac{1}{y^2}\frac{dy}{dx} = -\frac{dz}{dx}$: $-\frac{dz}{dx} - \frac{1}{x}z = -2$ Multiply by -1: $\frac{dz}{dx} + \frac{1}{x}z = 2$ This is a linear differential equation.

Step 3: Solve the Linear Differential Equation

We have $\frac{dz}{dx} + \frac{1}{x}z = 2$. Here, $P'(x) = \frac{1}{x}$ and $Q'(x) = 2$. Calculate the integrating factor: $I.F. = e^{\int P'(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$ Multiply the equation by the integrating factor: $x\frac{dz}{dx} + z = 2x$ $zx = \int 2x \, dx + C$ $zx = x^2 + C$

Step 4: Substitute Back and Solve for y

Substitute $z = \frac{1}{y}$: $\frac{x}{y} = x^2 + C$ $y = \frac{x}{x^2 + C}$

Step 5: Find the Intersection Point

The curve passes through the intersection of $2x - 3y = 1$ and $3x + 2y = 8$. Solve this system of equations. Multiply the first equation by 2 and the second by 3: $4x - 6y = 2$ $9x + 6y = 24$ Add the equations: $13x = 26$ $x = 2$ Substitute $x = 2$ into $2x - 3y = 1$: $2(2) - 3y = 1$ $4 - 3y = 1$ $3y = 3$ $y = 1$ The intersection point is $(2, 1)$.

Step 6: Use the Intersection Point to Find C

The curve $y = \frac{x}{x^2 + C}$ passes through $(2, 1)$. $1 = \frac{2}{2^2 + C}$ $1 = \frac{2}{4 + C}$ $4 + C = 2$ $C = -2$

Step 7: Find the Specific Solution

The particular solution is $y = \frac{x}{x^2 - 2}$.

Step 8: Calculate |y(1)|

We want to find $|y(1)|$. $y(1) = \frac{1}{1^2 - 2} = \frac{1}{1 - 2} = \frac{1}{-1} = -1$ $|y(1)| = |-1| = 1$

Common Mistakes & Tips

• Remember to divide the entire Bernoulli equation by $y^n$ before making the substitution.
• Be careful with signs when differentiating the substitution $z = y^{1-n}$.
• Double-check your integration and algebraic manipulations.

Summary

We identified the differential equation as a Bernoulli equation, transformed it into a linear differential equation using an appropriate substitution, solved the linear equation using an integrating factor, found the constant of integration using the intersection point of two lines, and finally calculated the absolute value of $y(1)$. The final answer is 1.

Final Answer The final answer is \boxed{1}.