Question

The general solution of the differential equation $(x+y)dx + xdy = 0$ is

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

Answer 1

Answer: (C)

C

Answer 2

The given differential equation is $(x+y)dx + xdy = 0$. This can be solved using two methods: as a linear differential equation or as an exact differential equation.

Method 1: Linear Differential Equation

1. Rearrange the equation: $\frac{dy}{dx} + \frac{1}{x}y = -1$

2. Identify $P(x)$ and $Q(x)$: $P(x) = \frac{1}{x}$ and $Q(x) = -1$.

3. Calculate the Integrating Factor (IF): $IF = e^{\int P(x)dx} = e^{\int \frac{1}{x}dx} = x$

4. Apply the general solution formula: $y \cdot IF = \int Q(x) \cdot IF \ dx + C$ $xy = \int -x \ dx + C$ $xy = -\frac{x^2}{2} + C$

5. Rearrange the solution: $x^2 + 2xy = 2C$ $x^2 + 2xy = c$

Method 2: Exact Differential Equation

1. Identify $M(x,y)$ and $N(x,y)$: $M(x,y) = x+y$ and $N(x,y) = x$.

2. Check for exactness: $\frac{\partial M}{\partial y} = 1$ and $\frac{\partial N}{\partial x} = 1$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact.

3. Find the potential function $f(x,y)$: $f(x,y) = \int (x+y) dx + g(y) = \frac{x^2}{2} + xy + g(y)$

4. Differentiate $f(x,y)$ with respect to $y$: $\frac{\partial f}{\partial y} = x + g'(y) = x$, so $g'(y) = 0$.

5. Integrate $g'(y)$: $g(y) = C_1$

6. Substitute back into $f(x,y)$: $f(x,y) = \frac{x^2}{2} + xy = c$ $x^2 + 2xy = c$

Both methods yield the same solution: $x^2 + 2xy = c$.