Question:
The general solution of the differential equation
\[
(4xy^2-2xy+2x^2y^2-x^2y)\,dx=dy
\]
is
The general solution of the differential equation
\[
(4xy^2-2xy+2x^2y^2-x^2y)\,dx=dy
\]
is
Show Hint
For separable differential equations, first factor the expression completely. After separating variables, use partial fractions whenever the denominator contains products of linear factors.
Updated On: Jul 29, 2026
- \[ x^2+\frac{x^3}{3} = \log\!\left(\frac{c(2y-1)}{y}\right) \]
- \[ x^2+\frac{x^3}{3} = \log\!\left(\frac{c|2y-1|}{\sqrt y}\right) \]
- \[ x^2+\frac{x^3}{3} = \log\!\big(c(2y^2-y)\big) \]
- \[ x^2+\frac{x^3}{3} = \log\!\left(\frac{c(2y-1)}{y^2}\right) \]
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The Correct Option is A
Solution and Explanation
Concept:
First write the differential equation in separable form and then integrate both sides.
Step 1: Factor the right-hand side. Given, \[ \frac{dy}{dx} = 4xy^2-2xy+2x^2y^2-x^2y. \] Taking common factors, \[ \frac{dy}{dx} = xy(2y-1)(2+x). \] Hence, \[ \frac{dy}{y(2y-1)} = x(x+2)\,dx. \]
Step 2: Resolve into partial fractions. Let \[ \frac{1}{y(2y-1)} = \frac{A}{y} + \frac{B}{2y-1}. \] Then \[ 1=A(2y-1)+By. \] Comparing coefficients, \[ A=-1, \qquad B=2. \] Therefore, \[ \frac{1}{y(2y-1)} = -\frac1y+\frac{2}{2y-1}. \]
Step 3: Integrate both sides. \[ \int \left( -\frac1y+\frac{2}{2y-1} \right)dy = \int x(x+2)\,dx. \] \[ -\log y+\log|2y-1| = \frac{x^3}{3}+x^2+C. \] \[ \log\left|\frac{2y-1}{y}\right| = x^2+\frac{x^3}{3}+C. \]
Step 4: Write the arbitrary constant in logarithmic form. Let \[ C=\log c. \] Then \[ x^2+\frac{x^3}{3} = \log\left(\frac{c(2y-1)}{y}\right). \] Therefore, \[ \boxed{ x^2+\frac{x^3}{3} = \log\left(\frac{c(2y-1)}{y}\right) } \] \[ \boxed{\text{Answer = (A)}} \]
Step 1: Factor the right-hand side. Given, \[ \frac{dy}{dx} = 4xy^2-2xy+2x^2y^2-x^2y. \] Taking common factors, \[ \frac{dy}{dx} = xy(2y-1)(2+x). \] Hence, \[ \frac{dy}{y(2y-1)} = x(x+2)\,dx. \]
Step 2: Resolve into partial fractions. Let \[ \frac{1}{y(2y-1)} = \frac{A}{y} + \frac{B}{2y-1}. \] Then \[ 1=A(2y-1)+By. \] Comparing coefficients, \[ A=-1, \qquad B=2. \] Therefore, \[ \frac{1}{y(2y-1)} = -\frac1y+\frac{2}{2y-1}. \]
Step 3: Integrate both sides. \[ \int \left( -\frac1y+\frac{2}{2y-1} \right)dy = \int x(x+2)\,dx. \] \[ -\log y+\log|2y-1| = \frac{x^3}{3}+x^2+C. \] \[ \log\left|\frac{2y-1}{y}\right| = x^2+\frac{x^3}{3}+C. \]
Step 4: Write the arbitrary constant in logarithmic form. Let \[ C=\log c. \] Then \[ x^2+\frac{x^3}{3} = \log\left(\frac{c(2y-1)}{y}\right). \] Therefore, \[ \boxed{ x^2+\frac{x^3}{3} = \log\left(\frac{c(2y-1)}{y}\right) } \] \[ \boxed{\text{Answer = (A)}} \]
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