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-