First-order equations (linear and nonlinear) | Engineering Mathematics for Electrical Engineering - Electrical Engineering (EE) PDF Download

First Order Linear Differential Equations

Solved Example 1: Solve
Solution: Here P = cot x
Q = sin 2x

General solution is y sinx

 Put sin x = t
cos x dx = dt 

Solved Example 2: Solve:
Solution: The equation is

General solution is

First Order Differential Equations reducible to Linear form

where P and Q are constants or functions of x alone can be reduced to linear form as follows:
Putting f(y) = v so that  ,
Above equation becomes
 

which is linear in v and x and its solution can be obtained by using working rule as for first order linear differential equation. Thus, we have I.F. = e^{∫P dx} 

Solution is 

Finally, replacing v by f(y) will give solution in terms of x and y alone.
An equation of the form where P₁ and Q₁ are constants or functions of y alone can be reduced to linear form in the same way as describe above by putting f(x) = v.

Bernoulli’s Equation 

Let  y^1-n  = z

Diff. w.r.t. x, (1- n).

The equation becomes 

which is linear in z and x.

Differential Equations of first order and higher degree

where p = dy/dx and P₁, P₂, … P_n are functions of x and y.

Equations solvable for p 
The left hand side of (1) can be factorized into factors of the first degree then (1) becomes

obtain a solution f_ix, y ,c = 0 corresponding to the equation p - R_i = 0 for i = 1, 2, … n.  
Thus the general solution of (1) is given by

Equations solvable for y 
The equation can be put in the form y = f(x, p) …(2)
Differentiating w.r.t. x we get  
which is a first order and first degree differential equation with variables p and x.
On solving equation (3) we get

Eliminating p from (1) and (3) we get the required solution.

Equations solvable for x
The equation can be put in the form x = f(y, p)
Differentiating w.r.t. y we get

This is first order and first degree differential equation with variables p and y.
On solving equation (4) we get 

Then eliminating p from (1) and (5) we get the required solution.

Note: The factor which does not involve a derivative of p with respect to x or y will always lead to singular solution. Hence such a factor can be omitted.  

Clairaut’s Equation

Now dp/dx = 0
∴ p = c (a constant)

Hence the general solution of (1) is
G.S. = y = cx + f(c)     …(2)
if x + fc(p) = 0 we use equation (2) and (1) to obtain a solution. This solution is not included in the general solution (2). Such a solution is called a singular solution.

Note : Clairaut’s equation always has a singular solution

Solved Example 1 :Solve p² - 5p + 6
Solution : Solving for p

Where,
 
∴ y = 3x + C      [Integrating]
When,

∴ y = 2x + C     [Integrating]
∴ The solution is (y - 3x - C) (y - 2x - C)

Solved Example 2 : Solve xp² - 2py  + x = 0 
Solution : Solving for p

This is a homogenous equation in x and y
Put y = vx

Separating the variables we get