Question

The general solution of the differential equation \[ (2xy+y^2)\,dy=(x^2-y^2)\,dx \] is:

• A. \[ x^3-3x^2y-y^3=c \]
• B. \[ x^3-3x^2y+y^3=c \]
• C. \[ x^3-3xy^2+y^3=c \]
• D. \[ x^3-3xy^2-y^3=c \]

Hint:

For homogeneous differential equations, always check whether every term has the same degree. If yes, the substitution \(y=vx\) is usually the fastest and most reliable approach.

Answer 1

The Correct Option is D

Concept: The given differential equation is homogeneous because every term appearing in the numerator and denominator has the same degree. Such equations are solved by using the substitution \[ y=vx. \] This converts the differential equation into a separable equation involving \(v\) and \(x\).

Step 1: Rewrite the equation in derivative form. Given, \[ (2xy+y^2)\,dy = (x^2-y^2)\,dx. \] Dividing by \(dx\), \[ (2xy+y^2)\frac{dy}{dx} = x^2-y^2. \] Hence, \[ \frac{dy}{dx} = \frac{x^2-y^2} {2xy+y^2}. \]

Step 2: Verify homogeneity. Observe that \[ x^2,\quad y^2,\quad xy \] all have degree \(2\). Therefore the differential equation is homogeneous. Use \[ y=vx. \] Then \[ \frac{dy}{dx} = v+x\frac{dv}{dx}. \]

Step 3: Substitute \(y=vx\). Substituting into the equation, \[ v+x\frac{dv}{dx} = \frac{x^2-v^2x^2} {2vx^2+v^2x^2}. \] Canceling \(x^2\), \[ v+x\frac{dv}{dx} = \frac{1-v^2} {2v+v^2}. \]

Step 4: Isolate the derivative term. Subtract \(v\) from both sides: \[ x\frac{dv}{dx} = \frac{1-v^2}{2v+v^2} -v. \] Taking a common denominator, \[ x\frac{dv}{dx} = \frac{1-v^2-2v^2-v^3} {2v+v^2}. \] \[ x\frac{dv}{dx} = \frac{1-3v^2-v^3} {2v+v^2}. \]

Step 5: Separate the variables. \[ \frac{2v+v^2} {1-3v^2-v^3} \,dv = \frac{dx}{x}. \]

Step 6: Integrate both sides. Notice that \[ \frac{d}{dv}(1-3v^2-v^3) = -6v-3v^2 = -3(2v+v^2). \] Let \[ t=1-3v^2-v^3. \] Then \[ dt=-3(2v+v^2)\,dv. \] Thus, \[ (2v+v^2)\,dv = -\frac{dt}{3}. \] Substituting, \[ -\frac13 \int \frac{dt}{t} = \int \frac{dx}{x}. \] Integrating, \[ -\frac13\ln|t| = \ln|x| +C. \] Multiplying by \(-3\), \[ \ln|t| = -3\ln|x| +C_1. \] \[ \ln|t| = \ln\left(\frac{C}{x^3}\right). \] Therefore, \[ t=\frac{C}{x^3}. \]

Step 7: Substitute back. Since \[ t=1-3v^2-v^3, \] we obtain \[ 1-3v^2-v^3 = \frac{C}{x^3}. \] Multiplying by \(x^3\), \[ x^3(1-3v^2-v^3)=C. \] Using \[ v=\frac{y}{x}, \] we get \[ x^3 -3x^3\left(\frac{y^2}{x^2}\right) -x^3\left(\frac{y^3}{x^3}\right) = C. \] Simplifying, \[ x^3-3xy^2-y^3=C. \] Replacing \(C\) by \(c\), \[ x^3-3xy^2-y^3=c. \]

Step 8: Final Answer. \[ \boxed{ x^3-3xy^2-y^3=c } \] Hence the correct option is \[ \boxed{\text{(D)}}. \]