Unit – 5

First order ordinary Differential Equations

Q1) What do you understand by exact differential equations?

A1)

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
is the differential co-efficient of M with respect to y keeping x constant and
is the differential co-efficient of N with respect to x keeping y constant.

Q2) Solve .

A2)

Here M = and N =

Then the equation is exact and its solution is-

Q3) Solve-

A3)

We can write the equation as below-

Here M = and N =

So that-

The equation is exact and its solution will be-

Or

Q4) Solve (5 + 3 – 2x) dx + (2y – 3 – 5) dy = 0

A4)

Here, M = 5 + 3 – 2x, N = 2y – 3 – 5

Since,
the given equation is exact.

Now (terms is not containing ) (y constant)

Q5) Solve

A5)

We

)

Given equation is exact.

Its solution is  (terms not containing )

Q6) Solve)

A6)

Which is in the form

and

and

Hence the given differential equation is exact.

Solution is (terms not containing x)

... (1)

Now,

Equation (1) becomes

Q7) Solve-

A7)

Here we have-
 

Now divide by xy, we get-

Multiply (1) by , we get-

Which is an exact differential equation-

Q8) Solve-

A8)

Here given,

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

Then-

Here,

Then,

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

Q9) Solve

A9)

Dividing throughout by (x+1),given equation becomes

which is Leibnitz’s equation

Thus the solution of (1) is

Q10) Solve

A10)

Putting the given equation becomes

Which is Leibnitz’s equation in z.

Thus the solution of (1) is

Hence the solution of the given equation is

Q11) Solve

A11)

This equation contains and is therefore not a linear in y, but since only z occurs, it can be written as

Which is a Leibnitz’s equation in z.

Thus the solution is

Q12) Solve

A12)

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
 

Q13) Solve

A13)

We have,

The given equation reduces to a linear differential equation in z.

Hence the solution is

Q14) Solve

A14)

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-

Q15) Use Euler’s method to  find y(0.4) from the differential equation

    with h=0.1

A15)

Given  equation  

Here

We  break  the  interval in four steps.

So that

By Euler’s  formula

, n=0,1,2,3   ……(i)

For n=0 in equation (i) we  get

For n=1 in equation (i) we  get

.01

For n=2 in equation (i) we  get

For n=3 in equation (i) we  get

Hence y(0.4)  =1.061106.

Q16) Given  with the initial condition  y=1 at   x=0.Find  y for x=0.1 by  Euler’s   method(five  steps).

A16)

Given equation is  

Here   

No. Of steps  n=5 and so that

So that

Also

By Euler’s  formula

, n=0,1,2,3,4   ……(i)

For n=0 in equation (i) we  get

For n=1 in equation (i) we  get

For n=2 in equation (i) we  get

For n=3 in equation (i) we  get

For n=4 in equation (i) we  get

Hence  

Q17) Use modified Euler’s method to compute y  for  x=0.05.  Given that

Result correct to three decimal places.

A17)

Given equation  

Here

Take  h  =  = 0.05

By modified Euler’s  formula the  initial iteration is

)

The iteration  formula by  modified Euler’s method  is

-----(i)

For n=0  in equation  (i)  we get

Where   and  as  above

For n=1  in equation  (i)  we get

For n=3  in equation  (i)  we get

Since  third and  fourth approximation are equal .

Hence y=1.0526  at x =  0.05  correct  to  three decimal places.

Q18) Solve

A18)

Given equation is where

Factorizing,

Thus we have

From (1)

From (2)

Integrating, Thus x y=c or , constitute the required solution.

Otherwise combining these into one, the required solution can be written as

Q19) Solve

A19)

Differentiating it with respect to x we get

This is Leibnitz’s linear equation in x and p. Here

Therefore the solution of (ii) is

Substituting this value of x in (i) we get

The equations (iii) and (iv) taken together with parameter p constitute the general solution (i).

Q20) Solve

A20)

Differentiating (2) with respect to x we get

Putting the value of p in (1)