**Substitution Suggested by the Equation**

Example 1

The quantity (2x - y) appears twice in the equation. Let

$z = 2x - y$

$dz = 2~dx - dy$

$dy = 2~dx - dz$

Substitute,

$(z + 1)~dx - 3z(2~dx - dz) = 0$

then continue solving.

Example 2

The quantity (-sin y dy) is the exact derivative of cos y. Let

$z = \cos y$

$dz = -\sin y ~ dy$

Substitute,

$(3 + xz) ~ dx + x^2 ~ dz$

then continue solving.

**Bernoulli's Equation**

Bernoulli's equation is in the form

$dy + P(x)~y~dx = Q(x)~y^n~dx$

If x is the dependent variable, Bernoulli's equation can be recognized in the form $dx + P(y)~x~dy = Q(y)~x^n~dy$.

If n = 1, the variables are separable.

If n = 0, the equation is linear.

If n ≠ 1, Bernoulli's equation.

**Steps in solving Bernoulli's equation**

- Write the equation into the form $dy + Py~dx = Qy^n~dx$.

- Identify $P$, $Q$, and $n$.

- Write the quantity $(1 - n)$ and let $z = y^{(1 - n)}$.

- Determine the integrating factor $u = e^{(1 - n)\int P~dx}$.

- The solution is defined by $\displaystyle zu = (1 - n)\int Qu~dx + C$.

- Bring the result back to the original variable.