Definition
An exact differential equation is formed by differentiating its solution directly without any other process,
Is called an exact differential equation if it satisfies the following condition Here 
Step by step method to solve an exact differential equation
1. Integrate M w.r.t. x keeping y constant.
2. Integrate with respect to y, those terms of N which do not contain x.
3. Add the above two results as below
Example1: Solve 
Sol. Here M = and N = Then the equation is exact and its solution is 
Example2: Solve
Sol. We can write the equation as below Here M = and N = So that The equation is exact and its solution will be Or 
Example3: Determine whether the differential function ydx –xdy = 0 is exact or not.
Solution
. Here the equation is the form of M(x , y)dx + N(x , y)dy = 0 But, we will check for exactness, 
These are not equal results, so we can say that the given diff. eq. is not exact.
Equation reducible to exact form
1. If M dx + N dy = 0 be a homogenous equation in x and y, then 1/ (Mx + Ny) is an integrating factor. 
Example: Solve Sol. We can write the given equation as Here, M = Multiply equation (1) by we get This is an exact differential equation 
2. I.F. for an equation of the type IF the equation Mdx + Ndy = 0 be this form, then 1/ (Mx – Ny) is an integrating factor. 
Example: Solve Sol. Here we have Now divide by xy, we get Multiply (1) by , we get Which is an exact differential equation 
3. In the equation M dx + N dy = 0, (i) If be a function of x only = f(x), then is an integrating factor. (ii) If be a function of y only = F(x), then is an integrating factor. 
Example: Solve Sol. Here given, M = 2y and N = 2x log x  xy Then Here, Then, Now multiplying equation (1) by 1/x, we get 
4. For the following type of equation An I.F. is Where 
Example: Solve Sol. We can write the equation as below Now comparing with We get a = b = 1, m = n = 1, a’ = b’ = 2, m’ = 2, n’ = 1 I.F. = Where On solving we get h = k = 3
Multiply the equation by , we get It is an exact equation. So that the solution is 
Key takeaways
Is called an exact differential equation if it satisfies the following condition 2. If M dx + N dy = 0 be a homogenous equation in x and y, then 1/ (Mx + Ny) is an integrating factor. 3. I.F. for an equation of the type IF the equation Mdx + Ndy = 0 be this form, then 1/(Mx – Ny) is an integrating factor. 
A differential equation of the form
Is called linear differential equation.
It is also called Leibnitz’s linear equation.
Here P and Q are the function of x
Working rule
(1)Convert the equation to the standard form (2) Find the integrating factor. (3) Then the solution will be y (I.F) = 
Note is called the integrating factor.
Example1: Solve
Sol. We can write the given equation as So that I.F. = The solution of equation (1) will be Or Or Or 
Example2: Solve 
Sol. We can write the equation as We see that it is a Leibnitz’s equation in x So that Therefore the solution of equation (1) will be Or 
Example3: Solve sin x ) 
Solution: here we have, sin x ) which is the linear form, Now, Put tan so that sec² dx = dt, we get Which is the required solution. 
Bernoulli’s equation
The equation
Is reducible to the Leibnitz’s linear equation and is usually called Bernoulli’s equation.
Working procedure to solve the Bernoulli’s linear equation
Divide both sides of the equation  By, so that Put so that Then equation (1) becomes ) 
Here we see that it is a Leibnitz’s linear equations which can be solved easily.
Example: Solve 
Sol. We can write the equation as On dividing by , we get Put so that Equation (1) becomes, Here, Therefore the solution is Or Now put Integrate by parts Or 
Example: Solve 
Sol. here given, Now let z = sec y, so that dz/dx = sec y tan y dy/dx Then the equation becomes Here, Then the solution will be 
Example: Solve 
Sol. here given We can rewrite this as Which is a linear differential equation The solution will be Put 
Key takeaways
Is called linear differential equation 2. is called the integrating factor. 3. y (I.F) = 4. The equation called Bernoulli’s equation. 
Example Solve 
Sol. Here we have or On integrating, we get 
Equation solvable for y
Steps
First differentiate the given equation w.r.t. x.
Second Eliminate p from the given equation, then the eliminant is the required solution.
Example: Solve 
Sol. Here we have Now differentiate it with respect to x, we get Or This is the Leibnitz’s linear equation in x and p, here Then the solution of (2) is Or Or Put this value of x in (1), we get 
Equation solvable for x
Example: Solve 
Sol. Here we have On solving for x, it becomes Differentiating w.r.t. y, we get or On solving it becomes Which gives Or On integrating Thus eliminating from the given equation and (1), we get Which is the required solution 
.Clairaut’s equation
An equation
y = px + f(p) ...... (2) 
is known as Clairaut’s equation.
Differentiating (1) w.r.t. x, we get Put the value of p in (1) we get y = ax + f(a) Which is the required solution. 
Example: Solve
Sol. Put So that Then the given equation becomes Or Or Which is the Clairaut’s form. Its solution is i.e. 
Key takeaways

References
 E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
 P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
 S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
 W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
 N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
 B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
 T. Veerarajan, “Engineering Mathematics”, Tata McgrawHill, New Delhi, 2010
 Higher engineering mathematics, HK Dass
 BV ramana, higher engineering mathematics