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Step-by-Step Solution
Step 1: Express the Differential Equation in the Form dy/dx = …
We are given the differential equation
(2x y^2 - y)\,dx + x\,dy = 0.
Rearrange to isolate dy/dx :
x\,dy = -\bigl(2x y^2 - y\bigr)\,dx
\quad\Longrightarrow\quad
\frac{dy}{dx} = -\frac{(2x y^2 - y)}{x}.
Simplify inside the parentheses:
\frac{dy}{dx}
= -\Bigl(2y^2 - \frac{y}{x}\Bigr)
= -2y^2 + \frac{y}{x}.
Hence,
\frac{dy}{dx} + 2y^2 - \frac{y}{x} = 0.
Step 2: Perform the Substitution z = \frac{1}{y}
Let z = \frac{1}{y} . Then y = \frac{1}{z} , and we compute
\frac{dy}{dx} = \frac{d}{dx}\Bigl(\frac{1}{z}\Bigr)
= -\frac{1}{z^2}\,\frac{dz}{dx}.
Hence,
-\frac{1}{y^2}\,\frac{dy}{dx}
= \frac{dz}{dx}.
Step 3: Transform the Original Equation in Terms of z
From
\frac{dy}{dx} + 2y^2 - \frac{y}{x} = 0,
multiply both sides by -\frac{1}{y^2} :
-\frac{1}{y^2}\,\frac{dy}{dx} - 2 + \frac{1}{x}\frac{1}{y} = 0.
Recognize -\frac{1}{y^2}\,\frac{dy}{dx} = \frac{dz}{dx} and \frac{1}{y} = z . Thus,
\frac{dz}{dx} + z\left(\frac{1}{x}\right) = 2.
Step 4: Find the Integrating Factor (IF)
For the linear differential equation
\frac{dz}{dx} + \frac{1}{x} z = 2,
the integrating factor is given by
\text{IF} = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x.
Step 5: Solve for z(x)
Multiply the differential equation by the integrating factor x :
x\,\frac{dz}{dx} + z = 2x.
Observe that the left-hand side is the derivative of x z with respect to x :
\frac{d}{dx}(x\,z) = 2x.
Integrate both sides with respect to x :
x\,z = \int 2x \,dx = x^2 + C.
Therefore,
z = \frac{x^2 + C}{x}.
But z = \frac{1}{y} , so
\frac{1}{y} = \frac{x^2 + C}{x}
\quad\Longrightarrow\quad
\frac{x}{y} = x^2 + C.
Step 6: Use the Initial Condition from Intersection of Given Lines
The curve passes through the intersection of the lines
2x - 3y = 1
\quad\text{and}\quad
3x + 2y = 8.
Solve these simultaneously to find the point of intersection:
From 2x - 3y = 1 , we can, for instance, express x or y ; or use elimination with the other line.
By inspection or systematic solving, the intersection point is (x, y) = (2, 1) .
Substitute x=2 and y=1 into
\frac{x}{y} = x^2 + C.
Thus,
\frac{2}{1} = 2^2 + C
\quad\Longrightarrow\quad
2 = 4 + C
\quad\Longrightarrow\quad
C = -2.
Hence, the general solution becomes
\frac{x}{y} = x^2 - 2.
Step 7: Find y(1) and Its Absolute Value
Substitute x=1 into \frac{x}{y} = x^2 - 2 :
\frac{1}{y} = 1 - 2 = -1.
So y(1) = -1 . The absolute value is
|y(1)| = 1.
Final Answer
|y(1)| = 1 .
