Form:

$$x' = a(t)x + f(t)$$

Homogeneous if $f(t)=0$, form:

$$x' = a(t)x$$

$a(t), f(t)$ are called coefficients of the equation.

$$frac{dx}{x} = a(t)dt qquad Longrightarrow qquad ln|x| = int a(t)dt + C \ |x| = e^{int a(t)dt + C} = e^C e^{int a(t)dt}$$

Let $A$ be a constant, which can be either >0, <0 or =0

$$x(t) = Ae^{int a(t)dt}$$

The equation

$$x' = ax + f$$

1. Rewrite the equation as

   $$x' - ax = f$$

2. Multiply by the integrating factor

   $$u(t) = e^{-int a(t)dt}$$

   So that the equation becomes

   $$(ux)' = u(x' - ax) =uf$$

3. Integrate this equation

   $$u(t)x(t) = int u(t)f(t)dt + C$$

4. Solve for x(t)

The equation

$$y' = ay + f$$

1. Get a partialicular solution to the homogeneous equation $y_h’ = ay_h$

   $$y_h(t) = e^{int a(t)dt}$$

2. Substitute $y = vy_h$ into the inhomogeneous equation to find v, or remember that

   $$v' = frac{f}{y_h}$$

   And solve $v$

3. Write the v into the general solution $y = vy_h$

Every solution to the inhomogeneous equation is of the form

$$y(t) = y_p(t) + Ay_h(t)$$

Where $A$ is an arbitrary constant,

$y_p$ is a partialicular solution to the inhomogeneous equation $y’ = a(t)y + f(t)$,

and $y_h$ is a partialicular solution to the associated homogeneous equation $y’ = a(t)y$.

Let $omega = P(x,y)dx + Q(x,y)dy$ be a differential form where both P and Q are continuous and differentiable

(a) if $omega$ is exact, then

$$frac{partial P}{partial y} = frac{partial Q}{partial x}$$

(b) if (a) is true in a rectangle R, then $omega$ is exact in R

If the equation $P(x, y)dx + Q(x, y)dy = 0$ is exact, the solution is given by $F(x, y) = C$, where F is found by

1. Solve $frac{partial F}{partial x} = P$ by integration:

   $$F(x, y) = int P(x, y)dx + phi (y)$$

2. Solve $frac{partial F}{partial y} = Q$

   $$frac{partial F}{partial y} = frac{partial}{partial y} int P(x, y)dx + phi'(y) = Q(x, y)$$

The form $Pdx + Qdy$ has an integrating factor depending on one of the variables under the following conditions.

• If

  $$h = frac{1}{Q} left(frac{partial P}{partial y} - frac{partial Q}{partial x} right)$$

  is a function of x only, then $mu (x) = e^{int h(x)dx}$ is an integrating factor.

• If

  $$g = frac{1}{P} (frac{partial P}{partial y} - frac{partial Q}{partial x})$$

  is a function of y only, then $mu(y) = e^{-int g(y)dy}$ is an integrating factor.

If the equation has the form

$$P(x)dx + Q(y)dy = 0$$

The solution is given by

$$F(x, y) = int P(X)dx + int Q(y)dy = C$$

A function G(x, y) is homogenous of degree n if

$$G(tx, ty) = t^nG(x, y)$$

A differential equation $Pdx + Qdy = 0$ is said to be homogenous if both of the coefficients P and Q are homogeneous of the same degree

Substitute y = xv

Then $P(x, y)dx + Q(x, y)dy = 0$ turns into $P(x, xv)dx + Q(x, xv)(vdx + xdv) = 0$

$$P(x, xv) = x^nP(1, v)\ Q(x, xv) = x^nQ(1, v)$$

Dividing $x^n$, and collecting terms

$$(P(1, v) + vQ(1, v))dx + xQ(1, v)dv = 0$$

The integrating factor is

$$frac{1}{x(P(1, v) + vQ(1, v))}$$

Separate variables, and solve the equation

$$frac{dx}{x} + frac{Q(1, v)dv}{P(1, v) + vQ(1, v)} = 0$$