dy / dx + p (x) y = q (x) (1)

Where p (x) and q (x) are continuous Functions of x. A continuous function is a function having no holes or breaks in it. Now consider an another function μ ( x ) that is known as integrating factor, on multiplying it by the equation (1), we get:

μ (x) dy / dx + μ (x) p (x) y = μ (x) q (x) (2)

Now assume that the function μ (x) satisfies the following statement:

μ (x) p (x) = μ' (x) (3).

On substituting the equation (3) in the equation (2) we get:

μ (x) dy / dx + μ' ( x ) y = μ (x) q (x) (4)

Here the left side of the equation (4) can be written as:

μ (x) dy / dx + μ' ( x ) y = μ (x) y (x)'

Thus the left side of the expression (4) can be written as:

μ (x) y (x)' = μ (x) q (x) (5)

On integrating both sides of the expression (5) we get

∫ μ ( x ) y ( x ) ' dx = ∫ μ ( x ) q ( x ) dx

μ ( x ) y ( x ) + C = ∫ μ ( x ) q ( x ) dx (6)

Now the final step is to solve the equation y (x)

μ (x) y (x) = ∫ μ (x) q (x) dx – C

y (x) = ∫ μ ( x ) q (x) dx – C / μ (x)

Here C is an integral constant and if we take the negative sign to positive then that will not affect the final answer because it’s a constant. Thus with the change the expression we get:

y (x) = ∫ μ ( x ) q ( x ) dx + C / μ (x) (7)

This is the general solution of the equation (1) i.e. of the first order linear differential equation. Now on getting back to the function μ ( x ), from equation (3)

μ ( x ) p ( x ) = μ' ( x )

On dividing both sides by the function μ (x)

p ( x ) = μ' ( x ) / μ ( x )

Thus

log μ ( x ) ' = p ( x )

on integrating both sides, we get

log μ ( x ) + k = ∫ p ( x ) dx

log μ ( x ) = ∫ p ( x ) dx + k

To get the value of the function μ ( x ), do exponentiation of the expression given above:

μ ( x ) = exp [ ∫ p ( x ) dx + k ]

This can also be written as:

μ ( x ) = e

^{k}e∫ p ( x ) dx

[ Since x

^{a + b}= x

^{a}x

^{b}]

Here k is also an integral constant. C is already used before so k is used here. Ek is also a constant so it can be written as another constant.

μ (x) = K e∫ p (x) dx ( 8 )

The solution of the first order linear differential equation is:

y ( x ) = ∫ μ ( x ) q ( x ) dx + C / μ ( x ) ( 9 )

where μ ( x ) = e∫ p ( x ) dx ( 10 )

Here the constant K is removed by the substitution of the expression ( 8 ) into ( 7 ).

Thus the steps to solve the first order linear Differential Equations are:

Make the correct form of the differential equation as in (1).

Then determine the integrating factor μ ( x ) by using equation (10).

After that multiply each term with the function μ ( x ).

Then integrate both sides and properly refers the integral constant.

Then get the solution y (x).

If an IVP is given then use the initial condition to get the value of the constant.

The nonlinear first order differential equation can be expressed in the form:

N (y) dy / dx = M (x)

This form is also known as separable differential equation.

The solution of the non linear differential is quite simple. Rewrite the equation

N (y) dy = M ( x ) dx

On integrating both sides

N (y) dy = M ( x ) dx.

This Integration will provide an implicit solution and from that explicit solution can be obtained i.e. y ( x ). Implicit solution is the solution that is not in the form y = y ( x ). The interval of validity is The Range of the variable which is independent on which the solution is valid. Mean to say we need to avoid division by zero, and complex Numbers.

The first order linear differential equation can be summarized as:

Separable variables can be represented in the form:

M (y) dx + N (y) dy = 0.

First order homogeneous differential equation:

M (x,y) dx + N (x,y) dy = 0.

Where M and N are nth degree homogeneous.

Linear equation is written as:

y' + P (x) y = Q (x).

4. Bernoulli equation is expressed as:

y' + P (x) y = Q (x) y

^{n}.