Module–1

Ordinary Differential Equations-1

Question Bank

Question-1: Find the particular solution of the differential equation 5, given the boundary condition y = 1 , when x = 2.

Solution: rearrange the diff. Equation,

Which is the general solution.

Put the boundary conditions to find c,

-   , which gives, c = 1

   Hence the particular solution is,

                                     y = - + 1.

(2) The solution of equation of the form  ,

Question-2: Find the general solution of   = 3 + 2y

Solution:      here,   = 3 + 2y gives,

,

Integrating both sides,

,

By substitution,  u = (3 +2y),

                                             X =  In(3 + 2y) + c.

Question-3: Determine the particular solution of  (y² - 1)3y  given that y =1  when  x = 2.

Solution:  It gives,

When putting the values, y =1 , x = 2 ,

The particular solution will be,

(3) The solution of equation of the form 

Question-4: Solve the equation 4xy = y² - 1

Solution:  on separating variables, we get

                                               () dy  = dx

=  

   Using substitution, u = y² - 1

                                               2In(y² - 1) = In x + c.

Question-5: Solve-

Sol. We can write the given equation as-

So that-

I.F. =

The solution of equation (1) will be-

Or

Or

Or

Question-6: 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

Question-7: 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
 

Question-8: 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

Question-9: 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.

Question-10: Solve:

Sol.

We have-

Now differentiate (1) with respect to ‘x’, we get-

On integrating, we get p = c

Putting these values in equation (1)-

Question-11: Solve-

Sol.

On solving for x, we get-

On differentiating with respect to y,

Or

Or

Or

Which gives-

Integrating py = c

Thus, eliminate from the given equation and (1), we get-

Which is the required solution.

Question-12: Solve

Sol.

Its auxiliary equation is-

Where-

Therefore, the complete solution is-

Question-13: Find the P.I. Of (D + 2)

Sol.

P.I. =

Now we will evaluate each term separately-

And

Therefore-

Question-14: Find the P.I. Of (D + 1) (D + D’ – 1)z = sin (x + 2y)

Sol.

Question-15: Find P.I. Of

Sol. P.I =

Replace D by D+1

Put

Question-16: Solve-

Sol.

The given equation can be written as-

Its auxiliary equation is-

We get-

So that the C.F. Will be-

Now we will find P.I.-

Therefore, the complete solution is-

Question-17:

Ans. Putting,

AE is

CS = CF + PI

Question-18: Solve 

Ans. Let,

AE is

y= CF + PI

Question-19: Solve-

Sol.

Here we

Operate equation (2) by , we get-

Subtract (3) from (1)-

A.E-

From equation (2), we get-

These are the required solutions.

Question-20: The equations of motions of an object are given by-

Find the path of the object.

Sol.

Put d/dt = D, then the equations become-

On multiplying (1) by and (2) by D, we get-

Add (3) and (4), we get-

Now we need to solve (5) to get the value of y-

A.E.-

So that-

Now, on putting the value of Dy in (2), we get-