Ordinary Differential Equation - Assignment | Mathematical Methods - 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 = y^1-2 = 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 = y^1-n = y^1-4 = 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
y_h = c₁e^-2x + c² e-3x
Thus the general solution of the given nonhomogeneous differential equation is
y = y_h + y_p
⇒ y = c₁e ^-2x + c₂e ^-3x + x (e ^-2x - e ^-3x) -( cos x - sin x) .

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

(a) Write the particular solution.

(b) Write the general solution.
Ans. 




Thus, the general solution is y = c₁e^x + 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
y_h = c₁e^-x + c₂ e^-2x
hence the general solution of the given nonhomogeneous differential equation is


Q.14. Solve (D² - 4D + 3) y = 2xe^x
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
y_h = c₁ cos x + c₂ sin x
Hence the general solution of the nonhomogeneous equation is
y = y_h + y_P


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) + ce^y

Q.21.
Ans. Letting v = y^{1 -1/2} = y^1/2 , 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 = y^1-n = y^1-(-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 = e^x + 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)]² - [f(x)]² = 1
(c) The general solution of the corresponding homogeneous differential equation
(D² -1)y = 0 , is

y_h = C₁e^x + C₂e^-x
Thus the general solution of the given nonhomogeneous equation is
y = y_h + y_p

⇒ y = c₁e^x + 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)
y_p = 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 = x²e^3x
Ans.



The general solution of the corresponding homogeneous equation is

y_h = C₁e^2x + C₂e^-2x

Thus the general solution of the corresponding homogeneous equation is
y = y_h + y_p ⇒ y = c_ie^2x + c₂e^-2x +

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




The general solution of the corresponding homogeneous differential equation is
y_h = c₁e^x + c₂ e^3x
Thus the general solution of the given nonhomogeneous equation is


Q.30. Solve (D² - 9) y = x²e^4x
Ans.



Hence, the general solution is