Linear Differential Equations In this lecture we discuss the methods of solving first order linear differential equations and reduction of order.

Linear Equations A linear first order equation is an equation that can be expressed in the form where a1(x), a0(x), and b(x) depend only on the independent variable x, not on y.

We assume that the function a1(x), a0(x), and b(x) are continuous on an interval and a1(x)  0 on that interval. Then, on dividing by a1(x), we can rewrite equation (1) in the standard form where P(x), Q(x) are continuous functions on the interval.

Let’s express equation (2) in the differential form If we test this equation for exactness, we find Consequently, equation(3) is exact only when P(x) = 0. It turns out that an integrating factor , which depends only on x, can easily obtained the general solution of (3).

Multiply (3) by a function (x) and try to determine (x) so that the resulting equation is exact. We see that (4) is exact if  satisfies the DE Which is our desired IF

and so (7) can be written in the form In (2), we multiply by (x) defined in (6) to obtain We know from (5) and so (7) can be written in the form

Integrating (8) w.r.t. x gives and solving for y yields

Working Rule to solve a LDE: 1. Write the equation in the standard form 2. Calculate the IF (x) by the formula 3. Multiply the equation in standard form by (x) and recalling that the LHS is just obtain

4. Integrate the last equation and solve for y by dividing by (x).

Dividing by x cos x, throughout, we get Example 1. Solve Dividing by x cos x, throughout, we get

yields Multiply by Integrate both side we get

Problem (2g p. 62): Find the general solution of the equation Ans.:

The usual notation implies that x is independent variable & y the dependent variable. Sometimes it is helpful to replace x by y and y by x & work on the resulting equation. * When diff equation is of the form

Q. 4 (b) Solve

A first order equation that can be written in the form is known as Bernoulli’s equation . It is clearly linear when n=0 or 1. For n > 1, it can be reduced to a linear equation by change of variable

Divide equation (*) be yn yields Taking z = y1-n, we find that and so (**) becomes Since 1/(1-n) is just a constant,

Example: Solve the DE This is a Bernoulli equation with n =3, P(x) = -5, and Q(x) = -5x/2 We first divide by y3 to obtain Next we make the substitution z = y-2 Since dz/dx = -2y-3 dy/dx, then above equation reduces to

or which is linear. Hence Substituting z = y-2 gives the solution

Reduction of Order: A general second order DE has the form In this section we consider two special types of second order equations that can be solved by first order methods.

Type A: Dependent variable missing. When y is not explicitly present, (1) can be written as Then (2) transforms into If we can solve (3) for p, then (2) can be solved for y.

Problem (1 (g) p. 65): Solve The variable y is missing

which is linear,

Problem (1b p. 65): Solve the DE Ans.:

Type B: Dependent variable missing. When x is not explicitly present, then (1) can be written as Then (4) becomes If we can solve (5) for p, then (4) can be solved for y.

Problem (1(e), P 65): Solve Here x is not explicitly present  (1) can be writen as,

Problem (2b p. 65): Find the specified particular solution of the DE Ans.: