Ordinary Differential Equation - Assignment

# Ordinary Differential Equation - Assignment | Mathematical Models - Physics PDF Download

Q.1. x (y2 +1) dx + y (x2 +l) dy = 0, y (1) = 2
Ans. Dividing both sides by (v2 +1 y2 +1) , we have

Integrating both sides gives

Writingln A in place of c1 the above solution can be written in the form
(x2+1)(y2+1) = A
Using, y (l) = 2 ⇒ (1 +1)(4 +1) = A ⇒ A = 10
Thus the required solution is (x2 +1)( y2 +1) = 10.

Q.2. (x2 - yx2) dy + (y2 + xy2) dx = 0
Ans. The given differential equation can be written as
x2 (1 - y) dy + y2 (1 + x) dx = 0
Dividing both sides by x2 y2 , we obtain

Integrating both sides gives

Q.3.
Ans. This equation can be written as

This is a homogeneous differential equation. Letting y = xv

Where we have used the fact that

Where we have written, c02 = c .

Q.4.
Ans. We can write,
This is a linear differential equation but with integrating factor

Q.5.
Ans. The given equation can be written as

If we write c for 3c0 then

Q.6.
Ans. This is Bernoulli’s differential equation. Hence letting v = y1-2 = y-1 we obtain
This is linear differential equation in dependent variable v and independent variable x.
Hence,

Using the given condition y (1) = 2 we obtain

Thus, y = 2 x

Q.7.
Ans. The given differential equation can be written as

Putting, y = xv , we obtain

Q.8. (2y sin x cos x + y2sin x) dx + (sin2 x - 2y cos x)dy = 0 , y (0) = 3
Ans. M = 2y sin x • cos x + y2 sin x
N = sin2 x - 2 y cos x

Thus the given differential equation is exact. Hence there exists a function u (x, y) such that
(I)
and, (II)
Integrating equation (I) partially with respect to x .
(III)
Differentiating equation (III) with respect to y , we obtain
(IV)
From equation (II) and (IV)

Since, y (0) = 3, c1 - 9
Hence Solution ,

Q.9.
Ans.This equation is exact since

Hence there exists a function u (x, y) such that
(I)
(II)
Hence using equation (I), we obtain
(III)
Differentiating this equation with respect to y , we obtain
(IV)
From equations (II) and (IV)

Hence the required solution is

Using the given initial condition

Thus the required solution is

Q.10.
Ans. This equation can be written as

This is a Bernoulli’s differential equation.
Using, v = y1-n = y1-4 = y-3 we obtain

This is a linear differential equation in dependent variable v and independent variable x .
Hence,
Hence,

Since, we obtain

Using the given condition y (1) = 1/2

Thus the required solution is

Q.11. Find the particular and general solution of the equation

Ans. The particular solution is

The general solution of the corresponding homogeneous differential equation is
yh = c1e-2x + c2 e-3x
Thus the general solution of the given nonhomogeneous differential equation is
y = yh + yp
⇒ y = c1e -2x + c2e -3x + x (e -2x - e -3x) -( cos x - sin x) .

Q.12. Given a differential equation (D2 - 1) y = e2x sin 2x

(a) Write the particular solution.

(b) Write the general solution.
Ans.

Thus, the general solution is y = c1ex + c2e -x -

Q.13. Write the particular and general solution of (D2 + 3D + 2) y = x sin 2x
Ans. (D2 + 3D + 2) y = x sin2x

If the nonhomogeneous term r (x) of the differential equation

(D2 + aD + b y = r (x) ,

is of the form r (x) = xv (x) , then the particular solution is given by

Here we see that v (x) = sin2x hence

The general solution of the corresponding nonhomogeneous equation is
yh = c1e-x + c2 e-2x
hence the general solution of the given nonhomogeneous differential equation is

Q.14. Solve (D2 - 4D + 3) y = 2xex
Ans.

Thus the general solution of the given nonhomogeneous equation is

Q.15. Solve (D2 +1) y = cos2 x
Ans.

The general solution of the corresponding homogeneous differential equation is
yh = c1 cos x + c2 sin x
Hence the general solution of the nonhomogeneous equation is
y = yh + yP

Q.16. (1 + y2) dx + (1 + x2) dy = 0,    y (0) = 1
Ans. Dividing by (1 + x2) (1+y2), we get
Integrating both sides gives; tan -1 x + tan -1 y = c From the given condition
From the given condition
y(0) = 1 and tan -1 0 + tan -11 = c ⇒ c = ⇒/4
Thus, tan -1 x + tan -1 y = π/4.

Q.17. sec2 x tan ydx + sec2 y tan xdy = 0
Ans. This equation is exact since
Hence there exists a function u (x, y) such that
(i)
and (ii)
Integrating equation (i) partially with respect to x we get
u (x, y) = tan x • tan y + h (y )    (iii)
Differentiating it with respect to y , we obtain

From equations (ii) and (iv)

Thus, u (x, y ) = tan x tan y + c0
Hence the solution of the given differential equation is
u (x, y ) = c1 ⇒ tan x tan y = c (where c1 - c0 = c)

Q.18.
Ans. The given differential equation can be written as

Letting y = xv, we obtain

From the given condition y (0) = 0, we obtain k = 0
Thus, x2 - 2xy - y2 = 0

Q.19.
Ans. This is a linear differential equation with integrating factor
(I)

From the given condition

Q.20.
Ans.
This is not a linear differential equation but when we write
We obtain a linear differential equation

Hence the solution is

⇒ x = -( y +1) + cey

Q.21.
Ans. Letting v = y1 -1/2 = y1/2 , we obtain

This is a linear equation in dependent variable v and independent variable x .
Hence,
In order to evaluate the above integral we write

Hence,
Using the given initial condition

We obtain,
Hence,

Q.22. Find the equation of the curve passing through the point (-2,3) given that the tangent to the curve at any point (x, y) is
Ans. We know that the slope of tangent to any curve is
Thus,

Putting the value of initial conditions

Q.23.
Ans. The given differential equation can be written as

Assuming, y = xv we obtain

Q.24. (y sec2 x + sec x tan x) dx + (tan x + 2 y ) dy = 0
Ans. This equation is exact since

Hence the exists a function u (x, y) such that
(I)
(II)
Integrating equation (I) partially with respect to x .
u (x, y) = y tan x + sec x + h (y)    (III)
Differentiating equation (III) with respect to y we obtain
(IV)
From equations (II) and (IV)

Hence, u (x, y) = y tan x + sec x + y2 + c0
Hence the required solution is
y tan x + sec x + y2 + c0 = c1 ⇒ y tan x + sec x + y2 = c

Q.25.
Ans. Letting v = y1-n = y1-(-3) = y4
We obtain

This equation is a linear differential equation in dependent variable v and Independent variable x.
Hence,
Hence,

Using the given condition
y (1)=2
We obtain
16 = 1 + c ⇒ c = 15
Thus, y4 = x4 + I5x-2

Q.26. Given a nonhomogeneous differential equation

(D2-1) y = ex + e -x

(a) Find the particular solution of this differential equation.

(b) If the particular solution is written as x [f (x)] then find the value of [f'(x)2-[f (x)]2 .

Write the general solution of this differential equation.
Ans. (a)

Hence f ( x) = sinh x

f'(x) = cosh x
[f'(x)]2- f(x)2 = cos2 hx - sin2 hx
⇒ [f'(x)]- [f(x)]2 = 1
(c) The general solution of the corresponding homogeneous differential equation
(D2 -1)y = 0 , is

yh = C1ex + C2e-x
Thus the general solution of the given nonhomogeneous equation is
y = yh + yp

⇒ y = c1ex + c2e-x + x sinh x

Q.27. Given a differential equation

(a) Write the particular solution

(b) Write the general solution.
(c) Find the solution that passes though point (0,0) and has the slope 1 at x = 0.
Ans. (a)
= (l + D2)- (x + x2) =(l-D2)(x + x2)
yp = x + x2 - 2

(b) Hence the general solution is

y = c1 sinx + c2 cosx + (x + x2 - 2)

(c) Given that y (0) = 0 and y'( 0) = 1

0 = c1 . 0 + c2 . 1 + (0 + 0 - 2) ⇒ C2 = 2

y' = c1 cos x - c2 sin x + (1 + 2x) ⇒ 1 = c1.1 - c2 . 0 + (1 + 2 . 0) ⇒ c1 = 0
Hence the solution with the required properties is y = 2 cos x + (x + x2 - 2)

Q.28. Find the particular and general solution of (D2 - 4) y = x2e3x
Ans.

The general solution of the corresponding homogeneous equation is

yh = C1e2x + C2e-2x

Thus the general solution of the corresponding homogeneous equation is
y = yh + yp ⇒ y = cie2x + c2e-2x +

Q.29. Solve (D2 - 4D + 3) y = 3ex cos2x
Ans.

The general solution of the corresponding homogeneous differential equation is
yh = c1ex + c2 e3x
Thus the general solution of the given nonhomogeneous equation is

Q.30. Solve (D2 - 9) y = x2e4x
Ans.

Hence, the general solution is

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## FAQs on Ordinary Differential Equation - Assignment - Mathematical Models - Physics

 1. What is an ordinary differential equation (ODE)?
Ans. An ordinary differential equation (ODE) is a type of differential equation that involves one or more unknown functions and their derivatives with respect to a single independent variable. It represents a relationship between the function, its derivatives, and the independent variable.
 2. What is the significance of ordinary differential equations in mathematics and physics?
Ans. Ordinary differential equations have significant importance in mathematics and physics as they help in modeling and solving various dynamic systems. They are used to describe phenomena involving rates of change, motion, growth, decay, and many other physical processes.
 3. How are ordinary differential equations classified based on their order and linearity?
Ans. Ordinary differential equations can be classified based on their order and linearity. The order of an ODE is determined by the highest derivative present in the equation. If the equation contains only the derivatives of the unknown function, it is called a linear ODE; otherwise, it is a nonlinear ODE.
 4. What are initial value problems (IVPs) and boundary value problems (BVPs) in the context of ordinary differential equations?
Ans. In the context of ordinary differential equations, an initial value problem (IVP) is a type of problem where the values of the unknown function and its derivatives are specified at a single point. The goal is to find a solution that satisfies the given conditions. On the other hand, a boundary value problem (BVP) involves specifying the values or conditions at multiple points, and the solution must satisfy those conditions.
 5. What are some common methods for solving ordinary differential equations?
Ans. Several methods are commonly used to solve ordinary differential equations, depending on the characteristics of the equation. Some of the popular methods include separation of variables, integrating factors, substitution, power series method, Laplace transforms, and numerical methods such as Euler's method or Runge-Kutta methods. The choice of method depends on the complexity and nature of the equation.

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