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Homogeneous and Heterogeneous Linear ODES

A linear differential equation is one that does not contain any powers (greater than one) of the function or its derivatives. The most general form is:

$$Q(x) \ensuremath{\frac{d{y}}{d{x}}} + P(x) y = R(x)$$ (20-14)

Equation 20-15 can always be reduced to a simpler form by defining $p=P/Q$ and $r=R/Q$ :

$$\ensuremath{\frac{d{y}}{d{x}}} + p(x) y = r(x)$$ (20-15)

If $r(x) = 0$ , Eq. 20-15 is said to be a homogeneous linear first-order ODE; otherwise Eq. 20-15 is a heterogeneous linear first-order ODE.

The reason that the homogeneous equation is linear is because solutions can superimposed--that is, if $y_1(x)$ and $y_2(x)$ are solutions to Eq. 20-15, then $y_1(x) + y_2(x)$ is also a solution to Eq. 20-15. This is the case if the first derivative and the function are themselves linear. The heterogeneous equation is also called linear in this case, but it is important to remember that sums and/or multiples of heterogeneous solutions are also solutions to the heterogeneous equation.

The homogeneous equation has a solution of the form

$$y(x) = e^{-\int p(x) dx}$$ (20-16)

MATHEMATICA$^{\text{\scriptsize {\textregistered }}}$ Example

(notebook Lecture-20)

(html Lecture-20)

(xml+mathml Lecture-20)

DSolve in Homogeneous and Heterogeneous ODEs

1. Show how DSolve solves the homogeneous equation
   $$\ensuremath{\frac{d{y}}{d{x}}} + p(x) y = 0$$

2. Show how DSolve solves the heterogeneous equation
   $$\ensuremath{\frac{d{y}}{d{x}}} + p(x) y = r(x)$$

3. Note how the homogeneous solution is one of the terms in the sum for the heterogenous solution.

A trick (or, an integrating factor which amounts to the same thing) can be employed to find the solution to the heterogeneous equation. Multiply both sides of the heterogeneous equation by $e^{\int p(x)}$ :¹

$$e^{\int_a^x p(z) dz } \ensuremath{\frac{d{y}}{d{x}}} + e^{\int_a^x p(z) dz} p(x) y = e^{\int_a^x p(z) dz} r(x)$$ (20-17)

Notice that the left-hand-side can be written as a derivative of a simple expression

$$e^{\int_a^x p(z) dz } \ensuremath{\frac{d{y}}{d{x}}} + e^{\int_a^... ...dz} p(x) y = \ensuremath{\frac{d{}}{d{x}}}( e^{\int_a^x p(z) dz } y ) therefore$$ (20-18)

$$\ensuremath{\frac{d{}}{d{x}}}( e^{\int_a^x p(z) dz } y ) = e^{p(x)} r(x)$$ (20-19)

which can be integrated and then multiplied on both sides by $e^{-\int_a^x p(z) dz}$ :

$$y(x) = e^{-\int_a^x p(z) dz} \left[ \int_b^x r(z) \left( e^{\int_a^z p(\eta)d \eta}\right) dz \right]$$ (20-20)