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

2. 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.

3. 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.

4. 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).

5. 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

6. 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

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

8. 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

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

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

11. yields
Multiply by
Integrate both side we get

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

13. 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

14. Q. 4 (b) Solve

16. 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

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

18. 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

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

20. 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.

21. 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.

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

23. which is linear,

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

26. 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.

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

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