Unit – 5

First order ordinary differential equations

Q1) Solve-

A1) We can write the given equation as-

So that-

I.F. =

The solution of equation (1) will be-

Or

Or

Or

Q2) Solve sin x )

A2) 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.

Q3) Solve

A3) 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
 

Q4) Solve

A4) 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-

Q5) Solve-

A5) We can write the equation as below-

Here M = and N =

So that-

The equation is exact and its solution will be-

Or

Q6) Solve-

A6)

We can write the given equation as-

Here,

M =

Multiply equation (1) by we get-

This is an exact differential equation-

Q7) Solve-

A7) 

Here given,

M = 2y and N = 2x log x - xy

Then-

Here,

Then,

Now multiplying equation (1) by 1/x, we get-

Q8) Solve-

A8) 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-

Q9) Solve

A9)

Put,

AE is

Q10) Solve

A10) Let,

AE is

y= CF + PI

Q11) Solve

A11) 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

Q12) Solve-

A12) 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.