**Problem 01**

$(x + y) \, dx + (x - y) \, dy = 0$

**Solution 01**

Test for exactness

$M = x + y$ ; $\dfrac{\partial M}{\partial y} = 1$

$N = x - y$ ; $\dfrac{\partial N}{\partial x} = 1$

$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$ ; thus, exact!

Step 1: Let

$\dfrac{\partial F}{\partial x} = M$

$\dfrac{\partial F}{\partial x} = x + y$

Step 2: Integrate partially with respect to x, holding y as constant

$\partial F = (x + y) \, \partial x$

$\displaystyle \int \partial F = \int (x + y) \, \partial x$

$F = \frac{1}{2}x^2 + xy + f(y)$ → Equation (1)

Step 3: Differentiate Equation (1) partially with respect to y, holding x as constant

$\dfrac{\partial F}{\partial y} = x + f'(y)$

Step 4: Equate the result of Step 3 to N and collect similar terms. Let

$\dfrac{\partial F}{\partial y} = N$

$x + f'(y) = x - y$

$f'(y) = -y$

Step 5: Integrate partially the result in Step 4 with respect to y, holding x as constant

$\displaystyle \int f'(y) = - \int y \, dy$

$f(y) = -\frac{1}{2}y^2$

Step 6: Substitute f(y) to Equation (1)

$F = \frac{1}{2}x^2 + xy - \frac{1}{2}y^2$

Equate F to ½c

$F = \frac{1}{2}c$

$\frac{1}{2}x^2 + xy - \frac{1}{2}y^2 = \frac{1}{2}c$

$x^2 + 2xy - y^2 = c$ *answer*

## Tags:

- Log in to post comments