First-degree differential equation: Difference between revisions

Latest revision as of 20:49, 9 July 2012

Definition

A first-degree differential equation is a differential equation that is linear in its highest-order derivative. Explicitly, of the independent variable is $x$ , the dependent variable is $y$ , and the order is $k$ it has the form:

$P (x, y, y^{'}, \dots, y^{(k - 1)}) y^{(k)} = Q (x, y, y^{'}, \dots, y^{(k - 1)})$

Relation with explicit differential equations

First-degree differential equations are typically solved by converting them to explicit differential equations. This simply involves dividing by $P$ . The explicit differential equation form is:

$y^{(k)} = \frac{Q (x, y, y^{'}, \dots, y^{(k - 1)})}{P (x, y, y^{'}, \dots, y^{(k - 1)})}$

However, when doing this division, we may throw away some solutions where $P (x, y, y^{'}, \dots, y^{(k - 1)}) = Q (x, y, y^{'}, \dots, y^{(k - 1)}) = 0$ . These solutions can be checked for separately, but we can note that these are differential equations of a lower order (specifically, order at most $k - 1$ ) hence presumably easier to solve.

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 A '''first-degree differential equation''' is a [[differential equation]] that is linear in its highest-order derivative. Explicitly, of the independent variable is <math>x</math>, the dependent variable is <math>y</math>, and the order is <math>k</math>it has the form:
-<math>p_1(x,y,y',\dots,y^{(k-1)})y^{(k)} + p_2(x,y,y',\dots,y^{(k-1)}) = 0</math>
+<math>P(x,y,y',\dots,y^{(k-1)})y^{(k)} = Q(x,y,y',\dots,y^{(k-1)})</math>
 ===Relation with explicit differential equations===
-First-degree differential equations are typically solved by converting them to [[explicit differential equation]]s. This simply involves dividing by <math>p_1</math> and rearranging. The explicit differential equation form is:
+First-degree differential equations are typically solved by converting them to [[explicit differential equation]]s. This simply involves dividing by <math>P</math>. The explicit differential equation form is:
-<math>y^{(k)} = \frac{-p_2(x,y,y',\dots,y^{(k-1)})}{p_1(x,y,y',\dots,y^{(k-1)})}</math>
+<math>y^{(k)} = \frac{Q(x,y,y',\dots,y^{(k-1)})}{P(x,y,y',\dots,y^{(k-1)})}</math>
-However, when doing this division, we may throw away some solutions where <math>p_1(x,y,y',\dots,y^{(k-1)}) = p_2(x,y,y',\dots,y^{(k-1)}) = 0</math>. These solutions can be checked for separately, but we can note that these are differential equations of a lower order (specifically, order at most <math>k - 1</math>) hence presumably easier to solve.
+However, when doing this division, we may throw away some solutions where <math>P(x,y,y',\dots,y^{(k-1)}) = Q(x,y,y',\dots,y^{(k-1)}) = 0</math>. These solutions can be checked for separately, but we can note that these are differential equations of a lower order (specifically, order at most <math>k - 1</math>) hence presumably easier to solve.