Question

The general solution of the differential equation $(3xy + y^2)dx + (x^2 + xy)dy = 0$ is

• A. $x^2(2xy - y^2) = c$
• B. $x^2(y^2 - 2xy) = c$
• C. $x(2xy + y^2) = c$
• D. $x^2(2xy + y^2) = c$

Hint:

To avoid tricky log constant manipulations, always attach the constant of integration as $\log C$ whenever all other integrated terms are in $\log$ form. This allows you to combine everything into a single polynomial equation seamlessly!

Answer 1

The Correct Option is D

Step 1: See it is homogeneous.
Every term in $(3xy + y^2)dx + (x^2 + xy)dy = 0$ has degree two, so put $y = vx$, which gives $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.

Step 2: Substitute and separate.
After substituting and tidying, $x\dfrac{dv}{dx} = \dfrac{-2v^2 - 4v}{1 + v}$, which separates to $\dfrac{(1+v)\,dv}{2v^2 + 4v} = -\dfrac{dx}{x}$.

Step 3: Integrate.
This gives $\tfrac14\log(v^2 + 2v) = -\log x + \log c$, so $(v^2 + 2v)x^4 = c$.

Step 4: Put $v = y/x$ back.
$$\left(\frac{y^2}{x^2} + \frac{2y}{x}\right)x^4 = c \Rightarrow x^2(y^2 + 2xy) = c$$
\[ \boxed{x^2(2xy + y^2) = c} \]