A differential equation is an equation which involves one or more derivatives of a function of a single independent variable.

Example: $\frac{dy}{dx}$ + 3y = e$^{x}$

We can use the same method to work out derivatives of other functions (like sine, cosine, logarithms, etc). The next set of functions that we want to take a look at are exponential, trigonometric functions etc., which are discussed in detail below. 

This is a special case of first order differential equations. The common form of such equation is:

$\frac{dy}{dx}$ + P(x) y = Q(x)

Here, P and Q are continuous functions in ‘x’. Any equation reducible to this form can be solved as below :

1) Firstly we obtain P and Q.

2) Then we find an integrating factor, IF = e$^{(\int P(x)dx)}$

3) Now, the solution of the linear differential equation in this case if given by y. IF = $\int$ IF Q(x) dx + C. Where, C is the constant of integration.

Example: 

Find the solution to the equation: $\frac{dy}{dx}$ = 9 - 196y

Solution:

$\frac{dy}{dx}$ + 196y = 9

 P (x) = 196 and Q (x) = 9

IF = e$^{(\int 196 dx)}$ = e$^{(196 x)}$

$\therefore$ solution is e$^{(196 x)}$  y = $\int$ 9 e$^{(196 x)}$ dx + c

 y e$^{(196 x)}$ = $\frac{(9e^{(196x)})}{196 + c}$
The differential equations of the following type are known as exact differential equations.

f (x, y) + g (x, y) $\frac{dy}{dx}$ = 0

It is an important condition to be noted that $\frac{d}{dy}$ (f (x, y)) with ‘x’ being constant, $\frac{d}{dx}$ (g (x, y)) with ‘y’ being constant, must be  equal along with the formation above for the differential equation to be exact.

To solve such differential equations we do the following.

1) Determine f (x, y) and g (x, y) and check if $\frac{d}{dy}$ (f (x, y)) = $\frac{d}{dx}$ (g (x, y)). If they are equal then we proceed with these steps.

Case 1:

a) Find $\int$f (x, y) dx = M + h (y) OR Find $\int$g (x, y)dy = N + l (x).

b) Find $\frac{d}{dy}$ (M + h (y)) and compare it with g (x, y) to find h’ (y).

c) Then find h (y) by integrating it.

d) Then the solution is given by M + h (y) = C 

Case 2:

a) Find $\int$g (x, y)dy = N + l (x).

b) Find $\frac{d}{dx}$ (N + l (x)) and compare it with f (x, y) to find l’ (x).

c) Then find l (x) by integrating it.

d) Then the solution is given by N + l (x) = C

We could use either of the two cases but which one to use depends upon the fact how easy it is to integrate the given functions. The one that is easier to integrate is been chosen and then the steps are followed accordingly.

A separable as the name suggests are the one in which we can easily separate the variables as g (y) $\frac{dy}{dx}$ = f(x)

To solve such differential equations we:

1) Obtain g (y) and f (x) and rewrite it in the form g (y) dy = f (x) dx

2) Integrate both sides and obtain the general solution of the given differential equation.

Example:

Solve the equation: $\frac{dy}{dx}$ = 6 y x$^{2}$

Solution:

Given differential equation is,

$\frac{1}{y}$ dy = 6 x$^{2}$ dx

Integrating both sides we get

Log y = 6 $\frac {x^{3}}{3}$ + C

 Log y = 2 x$^{3}$ + C

This is the required solution.
The differential equations of the form $\frac{dy}{dx}$ + P y = Q y$^{n}$

Where P and Q are functions in ‘x’ and ‘n’ is a real number, comes under this category. If n = 0 or n = 1 then the equation is linear and we have discussed how to solve such equations. So we will bother here about equations with n >= 2.

We perform the following steps

1) Rewrite the given equation by dividing it by y$^{n}$ we get y$^{(-n)}$ $\frac{dy}{dx}$  + P y$^{(1 - n)}$ = Q

2) Substitute u = y$^{(1 - n)}$

    du = (1 - n) y$^{(- n)}$ dy

3) Then we will have $\frac{1}{(1 - n)}$ $\frac{du}{dx}$ + P u = Q

4) This is now a linear differential equation which can be solved easily in terms of u and x.

5) Now substitute back the value of u to get the final solution in terms of y and x.

Example:

Solve: $\frac{dy}{dx}$ + 4xy = x$^{2}$ y$^{2}$

Solution:

Rewriting the given equation after dividing by y$^{2}$ we get

y$^{(-2)}$ $\frac{dy}{dx}$ + $\frac{4x}{y}$ = x$^{2}$

Let u = $\frac{1}{y}$

du = - y$^{(-2)}$ dy

So we have

$\frac{-du}{dx}$ + 4x u = x$^{2}$

 $\frac{du}{dx}$ - 4x u = x$^{2}$

IF = e$^{(\int P dx)}$ = e$^{(\int-4x dx)}$ = e$^{(-2x^{2})}$

Solution in u and x is 

e$^{(-2x^{2})}$. u = $\int$ e$^{(-2x^{2})}$ x$^{2}$ dx + C

$\frac{e^{(-2x^{2})}}{y}$ = $\int$ e$^{(-2x^{2})}$ x$^{2}$ dx + C

This can be solved further to find the final solution after integration as below.

= $\frac{e^{(-2x^{2})}}{y}$ = $\frac{[e^{-2x^{2}}]}{-4x}$ x$^{2}$ dx + c 

= $\frac{e^{(-2x^{2})}}{y}$ = $\frac{-x e^{(-2x^{2})}}{2}$ dx + C