First order linear differential equation

First order differential equations: A first order differential equation is a relation dy/dx = f(x, y).....(1)in which f(x, y) is a function of two variables defined on a region in the xy-plane. A solution of the equation (1) is a differential function y = y(x) defined on an interval of x-values such that d/dx y(x) = f(x, y(x)) on that interval.
The initial condition that y(x0) = y0 amounts to requiring the solution curve y = y(x) to pass through the point (x0, y0).The general form of a first order and first degree differential equation is f(x, y, dy/dx) =0....(i).we know that the tangent of the direction of a curve in Cartesian rectangular coordinates at any point is given by dy/dx .

so the equation in (i)  can be known as the equation which establish the relationship between the coordinates of a point and the slope of the tangent ie dy/dx to the integral curve at a point .Solving the differential equation  given by (i) means finding those curves for which the direction of tangent at each point coincides with the direction of the field.All the curves represented by the general solution when takes together will give the locus of the differential equation .Since there is one arbitrary constant in the general  solution of the equation of the first order. The locus of the equation can be said to be made up of single infinity of curve.

First Order Linear Differential Equation A first order differential equation that can be written in the form  dy/dx + P(x) y = Q(x),where P and Q are functions of x, is a linear first order equation. The above equation is the standard form.  Let us understand First Order Differential Equation Solver using some example. Suppose we have the equation dy/dx = 1 – y/x is a first order differential equation in which f(x, y) = 1 – (y/x).
now let us take one more example show that the function y = 1/x + x/2 is a solution of the initial value problem dy/dx = 1 – y/x,y(2) = 3/2 . The given function satisfies the initial condition because y(2) = (1/x + x/2)x = 2= ½ + 2/2 = 3/2To show that it satisfies the differential equation, we show that the two sides of the equation agree when we substitute (1/x) + (x/2) for y.On the left: dy/dx = d/dx(1/x + x/2) = -1/x^2 + ½.On the right: 1 – y/x = 1 – 1/x(1/x + x/2) = 1 – 1/x^2 – ½ = -1/x^2 + ½.The function y = (1/x) + (x/2) satisfies both the differential equation and the initial condition, which is what we needed to show.