Question:
The general solution of the differential equation
\[
\frac{dy}{dx}=\frac{2xy-3y^2}{2x^2+3xy}
\]
is
The general solution of the differential equation
\[
\frac{dy}{dx}=\frac{2xy-3y^2}{2x^2+3xy}
\]
is
Show Hint
Whenever numerator and denominator have the same degree, the differential equation is usually homogeneous.
Updated On: Jun 17, 2026
- \(\displaystyle 3\log\left|\frac{y}{x}\right|=-\frac{x}{y}+c\)
- \(\displaystyle \log|xy|=2xy+c\)
- \(\displaystyle 3\log|xy|=\frac{2x}{y}+c\)
- \(\displaystyle \log\left|\frac{y}{x}\right|=xy+c\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
This is a homogeneous differential equation because numerator and denominator are homogeneous functions of the same degree.
Use substitution:
\[
y=vx
\]
Then,
\[
\frac{dy}{dx}=v+x\frac{dv}{dx}
\]
Step 1: Substitute \(y=vx\). Given, \[ \frac{dy}{dx} = \frac{2xy-3y^2}{2x^2+3xy} \] Substituting \(y=vx\), \[ v+x\frac{dv}{dx} = \frac{2x(vx)-3(vx)^2}{2x^2+3x(vx)} \] \[ = \frac{2vx^2-3v^2x^2}{2x^2+3vx^2} \] \[ = \frac{2v-3v^2}{2+3v} \] Thus, \[ x\frac{dv}{dx} = \frac{2v-3v^2}{2+3v}-v \] Taking LCM, \[ = \frac{2v-3v^2-2v-3v^2}{2+3v} \] \[ = \frac{-6v^2}{2+3v} \] Hence, \[ \frac{2+3v}{v^2}dv = -6\frac{dx}{x} \]
Step 2: Integrate both sides. \[ \int \left(\frac2{v^2}+\frac3v\right)dv = -6\int \frac{dx}{x} \] \[ -\,\frac2v+3\log|v| = -6\log|x|+c \] Substituting \(v=\frac{y}{x}\), \[ -\frac{2x}{y} + 3\log\left|\frac{y}{x}\right| = -6\log|x|+c \] Simplifying logarithms, \[ 3\log|xy| = \frac{2x}{y}+c \] Hence, \[ \boxed{ 3\log|xy|=\frac{2x}{y}+c } \]
Step 1: Substitute \(y=vx\). Given, \[ \frac{dy}{dx} = \frac{2xy-3y^2}{2x^2+3xy} \] Substituting \(y=vx\), \[ v+x\frac{dv}{dx} = \frac{2x(vx)-3(vx)^2}{2x^2+3x(vx)} \] \[ = \frac{2vx^2-3v^2x^2}{2x^2+3vx^2} \] \[ = \frac{2v-3v^2}{2+3v} \] Thus, \[ x\frac{dv}{dx} = \frac{2v-3v^2}{2+3v}-v \] Taking LCM, \[ = \frac{2v-3v^2-2v-3v^2}{2+3v} \] \[ = \frac{-6v^2}{2+3v} \] Hence, \[ \frac{2+3v}{v^2}dv = -6\frac{dx}{x} \]
Step 2: Integrate both sides. \[ \int \left(\frac2{v^2}+\frac3v\right)dv = -6\int \frac{dx}{x} \] \[ -\,\frac2v+3\log|v| = -6\log|x|+c \] Substituting \(v=\frac{y}{x}\), \[ -\frac{2x}{y} + 3\log\left|\frac{y}{x}\right| = -6\log|x|+c \] Simplifying logarithms, \[ 3\log|xy| = \frac{2x}{y}+c \] Hence, \[ \boxed{ 3\log|xy|=\frac{2x}{y}+c } \]
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