Question

If the curve, y=y(x) represented by the solution of the differential equation $(2xy^2-y)dx+xdy=0$, passes through the intersection of the lines, 2x$-$3y=1 and 3x+2y=8, then $|y(1)|$ is equal to ________ .

Hint:

Recognize standard forms of differential equations. An equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ is a Bernoulli equation. The substitution $v = y^{1-n}$ will transform it into a linear differential equation.

Answer 1

Correct Answer: 1

82. Given differential equation: \[ (2xy^2-y)\,dx + x\,dy = 0 \] Rewrite: \[ \frac{dy}{dx} = \frac{y}{x} - 2y^2 \] This is a Bernoulli equation. Divide by \(y^2\): \[ y^{-2}\frac{dy}{dx} - \frac{1}{x}y^{-1} = -2 \] Let \(v=y^{-1}\), then: \[ \frac{dv}{dx} + \frac{1}{x}v = 2 \] Integrating factor: \[ \text{IF}=x \] \[ \frac{d}{dx}(vx)=2x \Rightarrow vx=x^2+C \] \[ \frac{x}{y}=x^2+C \] Intersection of \(2x-3y=1\) and \(3x+2y=8\) gives \((2,1)\). \[ \frac{2}{1}=4+C \Rightarrow C=-2 \] \[ \frac{x}{y}=x^2-2 \] At \(x=1\): \[ y=-1 \Rightarrow |y(1)|=1 \] \[ \boxed{1} \]