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: Understanding the Question:
We are given a first-order, first-degree differential equation. By observing the degree of each term (all are degree 2), we can identify this as a homogeneous differential equation.

Step 2: Key Formula or Approach:
For a homogeneous differential equation of the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$, we use the standard substitution $y = vx$.
Differentiating both sides with respect to $x$ gives: $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
This substitution will convert the homogeneous equation into a variable-separable form.

Step 3: Detailed Explanation:
First, rewrite the given equation to isolate $\frac{dy}{dx}$:
$$(x^2 + xy)dy = -(3xy + y^2)dx$$ $$\frac{dy}{dx} = -\frac{3xy + y^2}{x^2 + xy}$$ Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$:
$$v + x\frac{dv}{dx} = -\frac{3x(vx) + (vx)^2}{x^2 + x(vx)}$$ Factor out $x^2$ from the numerator and denominator:
$$v + x\frac{dv}{dx} = -\frac{x^2(3v + v^2)}{x^2(1 + v)}$$ $$v + x\frac{dv}{dx} = -\frac{3v + v^2}{1 + v}$$ Move $v$ to the right side:
$$x\frac{dv}{dx} = -\frac{3v + v^2}{1 + v} - v$$ $$x\frac{dv}{dx} = \frac{-3v - v^2 - v(1 + v)}{1 + v}$$ $$x\frac{dv}{dx} = \frac{-3v - v^2 - v - v^2}{1 + v} = \frac{-2v^2 - 4v}{1 + v}$$ Separate the variables $v$ and $x$:
$$\frac{1 + v}{2v^2 + 4v} dv = -\frac{1}{x} dx$$ Multiply by 2 to make the numerator the exact derivative of the denominator:
$$\frac{2 + 2v}{v^2 + 2v} dv = -2\frac{1}{x} dx$$ Integrate both sides:
$$\int \frac{2v + 2}{v^2 + 2v} dv = -2 \int \frac{1}{x} dx$$ $$\log|v^2 + 2v| = -2\log|x| + \log C$$ Using logarithm properties ($n\log a = \log a^n$ and $\log a + \log b = \log(ab)$):
$$\log|v^2 + 2v| + \log x^2 = \log C$$ $$\log|(v^2 + 2v)x^2| = \log C$$ $$(v^2 + 2v)x^2 = C$$ Finally, substitute back $v = \frac{y}{x}$:
$$\left(\frac{y^2}{x^2} + \frac{2y}{x}\right)x^2 = C$$ $$y^2 + 2xy = C$$ Wait, let's look at the options. We need an $x^2$ outside. Let's trace back.
If we integrate $\frac{v+1}{2(v^2+2v)} dv = -\frac{dx}{x} \implies \frac{1}{4} \log(v^2+2v) = -\log x + \log c_1$.
$\log(v^2+2v) = -4\log x + \log C$.
$(v^2+2v)x^4 = C$.
$(y^2/x^2 + 2y/x)x^4 = C \implies x^2y^2 + 2x^3y = C \implies x^2(y^2 + 2xy) = C$.

Step 4: Final Answer:
The general solution is $x^2(2xy + y^2) = c$, corresponding to option (D).