Question

The general solution of the differential equation \[ \frac{dy}{dx}=\frac{2xy-3y^2}{2x^2+3xy} \] is

• A. \(\displaystyle 3\log\left|\frac{y}{x}\right|=-\frac{x}{y}+c\)
• B. \(\displaystyle \log|xy|=2xy+c\)
• C. \(\displaystyle 3\log|xy|=\frac{2x}{y}+c\)
• D. \(\displaystyle \log\left|\frac{y}{x}\right|=xy+c\)

Hint:

Whenever numerator and denominator have the same degree, the differential equation is usually homogeneous.

Answer 1

The Correct Option is C

Step 1: Spot a homogeneous equation.
Both top and bottom of $\dfrac{dy}{dx}=\dfrac{2xy-3y^2}{2x^2+3xy}$ are degree two, so substitute $y=vx$, with $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.
Step 2: Substitute and simplify the right side.
\[ v+x\frac{dv}{dx}=\frac{2vx^2-3v^2x^2}{2x^2+3vx^2}=\frac{2v-3v^2}{2+3v}. \]
Step 3: Isolate $x\dfrac{dv}{dx}$.
\[ x\frac{dv}{dx}=\frac{2v-3v^2}{2+3v}-v=\frac{2v-3v^2-2v-3v^2}{2+3v}=\frac{-6v^2}{2+3v}. \]
Step 4: Separate the variables.
\[ \frac{2+3v}{v^2}\,dv=-6\frac{dx}{x}\Rightarrow\left(\frac{2}{v^2}+\frac{3}{v}\right)dv=-6\frac{dx}{x}. \]
Step 5: Integrate both sides.
\[ -\frac{2}{v}+3\log|v|=-6\log|x|+c. \]
Step 6: Put $v=\dfrac{y}{x}$ and tidy.
$-\dfrac{2x}{y}+3\log\left|\dfrac{y}{x}\right|=-6\log|x|+c$. Combining the logs gives \[ 3\log|xy|=\frac{2x}{y}+c. \] \[ \boxed{3\log|xy|=\dfrac{2x}{y}+c} \]