Differential Equations

Statement for Linked Answer Questions 2 and 3. 3. Which of the following is a solution of the

differential equation

5. The solution of the differential equation

x2 6. The partial differential equation

( (A) second order non-linear ordinary

differential equation

(B) third order non-linear ordinary differential equation

(D) mixed order non-linear ordinary differential equation

ME – 2011

12. Consider the differential equation

y x. The general solution with constant c is

(A) y t n t n

(B) y t n ( ) (C) y t n ( ) (D) y t n ( )

ME – 2012

13. Consider the differential equation x

y with the boundary conditions of y(0) =0 and y(1) = 1. The complete solution of the differential equation is

14. The partial differential equation

subjected to the boundary conditions u(0) = 0 and u(L) = U, is

18. The general solution of the differential equation os x y with c as a

2. The solution of

y y , ( ) in the range x is given by

(A) ( os x s n x) (B) ( os x s n x) (C) ( os x s n x) (D) ( os x s n x)

CE – 2006

3. A spherical naphthalene ball expanded to the atmosphere losses volume at a rate proportional to its instantaneous surface area due to evaporation. If the initial diameter of the ball is 2 cm and the diameter reduces to 1 cm after 3 months, the ball completely evaporates in

(A) 6 months (B) 9 months

4. The solution of the differential equation x

xy x given that at x = 1, y = 0 is

(A)

(B)

(C)

(D)

5. The differential equation = 0.25 y is to be solv us ng t b w r mpl t Eul r’s method with the boundary condition y = 1 at x = 0 and with a step size of 1. What would be the value of y at x = 1?

(A) 1.33 (B) 1.67

6. The degree of the differential equation

+ 2x = 0 is (A) 0

(B) 1

7. The solution for the differential equation

= x y with the condition that y = 1 at x = 0 is

(A) y =

(B) In(y) = + 4

8. A body originally at 60⁰C cools down to C in 15 minutes when kept in air at a temperature of 25⁰C. What will be the temperature of the body at the end of 30 minutes?

(A) 35.2⁰C (B) 31.5⁰C

9. The general solution of + y = 0 is (A) y = P cos x + Q sin x

(B) y = P cos x (C) y = P sin x (D) y = P s n x 10. Solution of

= at x = 1 and y = √ is (A) x y

(B) x y

11. Solution of the differential equation 3y

+ 2x = 0 represents a family of (A) Ellipses

(B) Parabolas

12. The order and degree of the differential equation + 4 √() y = 0 are respectively

(A) 3 and 2 (B) 2 and 3

13. The solution to the ordinary differential equation + 6y = 0 is

(A) y = + (B) y = + (C) y = + (D) y = +

14. The partial differential equation that can equation is given by

2. The following differential equation has

2 3

the boundary conditions are (i) y=0 for x=0 and (ii) y=0 for x=a

The form of non-zero solutions of y (where m varies over all integers) are y ∑ s nm x

y ∑ osm x y ∑ x y ∑

ECE – 2007

4. The solution of the differential equation

2 2

5. Which of the following is a solution to the differential equation x t x t

7. Match each differential equation in Group I to its family of solution curves from Group II.

Group I Group II P. dy y

dx x 1. Circles Q. dy y

dx  x 2. Straight Lines R. dy x

dx y 3. Hyperbolas S. dy x

dx  y

(A) P-2, Q-3, R-3, S-1 (B) P-1, Q-3, R-2, S-1 (C) P-2, Q-1, R-3, S-3 (D) P-3, Q-2, R-1, S-2 ECE – 2010

8. Consider a differential equation

y x x with the initial condition y s ng Eul r’s rst or r m t o with a step size of 0.1, the value of y is

(A) 0.01 (B) 0.031

9. A function n x satisfies the differential equation where L is a constant. The boundary conditions are:

n and n . The solution to this equation is

(A) n x xp x (B) n x xp x √ (C) n x xp x (D) n x xp x ECE– 2011

10. The solution of the differential equation

y y is (A) x (B) x

ECE\EE\IN – 2012

11. With initial condition x(1) = 0.5, the solution of the differential equation, t x t is

(A) x t (B) x t

ECE\IN – 2012

12. Consider the differential equation y t

t y t

t y t t w t y t | n |

num r l v lu o y

s (A)

(B)

13. A system described by a linear, constant coefficient, ordinary, first order differential equation has an exact solution given by y t or t when the forcing function is x(t) and the initial condition is y(0). If one wishes to modify the system so that the solution becomes – 2y(t) for t > 0, we need to

(A) Change the initial condition to – y(0) and the forcing function to 2x(t) (B) Change the initial condition to 2y(0)

and the forcing function to –x(t) (C) Change the initial condition to

j√ y(0) and the forcing function to j√ x(t)

(D) Change the initial condition to – 2y (0) and the forcing function to – 2x(t)

ECE – 2014

14. If the characteristic equation of the differential equation

y has two equal roots, t n t v lu s o r

(A) ±1 (B) 0,0

15. Which ONE of the following is a linear non-homogeneous differential equation, where x and y are the independent and dependent variables respectively?

(A)

xy (B) xy

(C)

xy (D)

16. If z xy ln xy then (A) x y (B) y x

17. If a and b are constants, the most general solution of the differential equation

x t x

t x s (A)

(B) bt

18. With initial values y(0) = y (0) = 1, the solution of the differential equation

y t x s EE – 2005

1. The solution of the first order differential qu t on x’ t 3x(t), x (0) = x is (A) x (t) = x

(B) x (t) = x

2. For the differential equation

x with initial conditions x n |

t , the solution is (A) x t

(B) x t (C) x t (D) x t

EE – 2011

3. With K as a constant, the possible solution for the first order differential equation

is (A) (B)

EE – 2013

4. A function y x x is defined over an open interval x = (1,2). At least at one point in this interval , is exactly

(A) 20 (B) 25

5. The solution for the differential equation x

t x w t n t l on t ons x n x

s (A) t t (B) s n t os t (C) s n t os t (D) os t t

6. Consider the differential equation x x y

Which of the following is a solution to this differential equation for x > 0?

(A) (B) x

1. The general solution of the differential equation (D² 4D +4)y = 0, is of the form (given D = d/dx), and C1 and C2 are constants

(A) C1e^2x

(B) C1e^2x + C2

2. urv s or w t urv tur ρ t

3. For an initial value problem ÿ ẏ y y n y ̇

various solutions are written in the following groups. Match the type of solution with the correct expression.

Group 1 Group 2

6. Consider the differential equation

7. Consider the differential equation

8. Consider the differential equation ÿ ẏ y with boundary conditions 10. The maximum value of the solution y(t) of

the differential equation y t ÿ t

IN– 2014

11. The figure shows the plot of y as a function of x

The function shown is the solution of the differential equation (assuming all initial conditions to be zero) is :

(A) (B) x

(C)

x (D) |x|

y x

Answer Keys and Explanations

1. [Ans. D]

x y xy lnx x y

xy lnx x

omp r ng w g t

x lnx x ow ^∫ ^∫ x y(I.F.) = ∫ x x y ∫ x x

olv ng bov n t v lu o

utt ng x t n n t v lu o y t x w s y

2. [Ans. C]

Given equation is x p y

x qy p q y p q ts solut on s y

um o roots p p

ro u t o roots q q

3. [Ans. C]

Given equation is y

x p y

x q p q y ut p n q

y y x

4. [Ans. B]

The given differential equation may be written as

y y y ux l ry qu t on s

Substituting D=2, we get (

) 5. [Ans. B]

First order equation, dy

Py Q,

dx 

Where P = 2x and Q =

Since P and Q are functions of x, then Integrating factor,

I.F. = e^{ }^Pdxe^^x² Solution is

y ∫ x y ∫ x

ye  x c

Since, y ,  c = 1 y (1 + x) e^^x²

6. [Ans. A]

Order: The order of a differential equation is the order of the highest derivative appears in the equation

Degree: The degree of a differential equation is the degree of the highest order differential coefficient or derivative, when the differential coefficients are free from radicals and fraction.

The general solution of differential qu t on o or r ‘n’ must nvolv ‘n’

arbitrary constant.

7. [Ans. C]

v n y x y y

y x nt gr t ng

y x y

x n x y

y x x x

x and x 8. [Ans. A]

y y y A.E is, D²+2D+1 =0 ²=0

m 1

The C.F. is (C1+C2x)e^-x P.I. = 0

ow y ₁

n y ₂ ₂

y 9. [Ans. D]

ẍ x

Auxiliary equation is m² + 3 = 0

i.e. m = ±√

x os√ t s n √ t ẋ √ os√ t s n√ t At t = 0

1 = A 0 = B x = os √ t

x os √ t 10. [Ans. A]

Given differential equation is x y

x y x

y x (

x) y x … Standard form

x y …

Where P and Q function of x only and solution is given by

∫ x

Where, integrating factor (I.F) ^∫ r

x n x ^∫ x olut on y x ∫ x x x yx x

Given condition y

m ns t x y r or yx x

y x

x 11. [Ans. B]

is third order ( ) and it is linear, since the product is not allowed in linear differential equation 12. [Ans. D]

x y x

∫ y

y ∫ x t ny x

y t n .x

13. [Ans. A]

x y

x x y

x y y n y

Choice (A) satisfies the initial condition as well as equation as shown below

y x

y n y lso y

x x y

x y

x x y x y x x x x x x x

o y x is the solution to this equation with given boundary conditions

14. [Ans. D]

15. [Ans. B]

m m

u u

At x=0, At x=L,

( )

n u x

Solving we get u = U()

16. [Ans. A]

t x y y

t x y

So by observation it is understood that, t,x

y- * + ,x y-

17. [Ans. *] Range 34 to 36 y

x y x y x

t x y t x y

x y x

t x y 18. [Ans. D]

x os x y Let x y z y

x z z x

x os z z

x os z os (z ) s (z

) z x Integrating t n (z

) x t n (z

) x t n (x y

) x

19. [Ans. A]

Since the determinant of wronskian matrix is constant values for, therefore it is same for both t=0 and t=

t x t x t

t x t x t t

20. [Ans. B]

∫ y

y ∫ x x ln y x ln ln (y

) x y

v n y n y

1. [Ans. A]

Given + p(t) y = q(t) y ; n > 0 Given, v = y

t n y y y t

t n y v t

Substitution in the differential equation we get

+ p(t)y = q(t) y Multiplying by (1 n) y we get

t p t n y q t n Now since y = v we get

t n pv n q Where p is p (t) and q is q(t) 2. [Ans. A]

y x y

x y y

x( )

This is a linear differential equation

y ( ) os x s n x

os x s n x os x

s n x t n y os x s n x n y

os s n

s n x os x

os x s n x s n x

os x

x t x s

y ( os x s n x )

3. [Ans. A]

t … Where, V = r r

t r r t Substituting in (i) we get 4 r

= r ) r

 dr = kdt Integration we get r = kt + C At t = 0, r = 1

 1 = k × 0 + C

 C = 1 r = kt + 1

Now at t = 3 months r = 0.5 cm 0.5 = k × 3 + 1

 r

utt ng r n solv ng g v s

 t = 6 months 4. [Ans. A]

Given x y

x xy – x x y x y

x xy x Dividing by x

y x (

x) y (x x )

Which is a linear first order differential of the form

x y

Integrating factor = I.F = ^∫

= ^∫ x y × I.F = ∫ .(I.F)dx

yx ∫ (x

x ) x x ∫ x x Now at x = 1, y = 0

i.e. 0 ×

C x y x

– x y

x x 5. [Ans. C]

x y y t x h = 1

Iterative equation for backward (implicit) Euler methods for above equation would be

y y x y y y

 y y y

 0.25hy y + y = 0 Putting k = 0 in above equation 0.25h y y + y = 0

Since, y = 1 and h = 1

0.25 y y + 1 = 0 ^±√

y = 2 6. [Ans. B]

Degree of a differential equation is the power of its highest order derivative after the differential equation is made free of radicals and fractions if any, in derivative power.

Hence, here the degree is 1, which is power of

7. [Ans. D]

y x x y

This is variable separable form

= x dx

∫ y

y ∫ x x log y x

t x y log y x

y 8. [Ans. B]

= θ θ0) (Newton’s law of cooling) θ

θ θ t

∫ = ∫ t

 ln θ θ = kt +

 θ θ C.

θ θ C.

Given θ = 25⁰C Now t t θ 60 = 25 + C.e⁰ C = 35

At t m nut s θ ⁰C 40 = 25 + 35 =

Now at t = 30 minutes Θ

= 25 + 35 ( )

= 25 + 35 × ( ) (s n )

= 31.

≈ C

9. [Ans. A]

+ y = 0 + 1 = 0

E s m  m ± General solution is

y = [ cos (1 × x) + sin (1 × x)]

= cosx + sinx

= P cosx + Q sinx

Where P and Q are some constants 10. [Ans. D]

y x

x y

y dy = x dx ∫ y y ∫ x x y x

At x = 1, y = √ √

C = 2

Solution is y x

x + y = 4 11. [Ans. A]

3y x y

x x

y y y x x ∫ y y ∫ x x

y x y x x

( ) y

( ) x

( ) y

( )

Which is the equation of a family of ellipses

12. [Ans. A]

x √( y

x) y Removing radicals we get . y

x / 0( y

x) y 1

The order is 3 since highest differential is

The degree is 2 since power of highest differential is 2

13. [Ans. C]

y x

x Auxiliary equation is

+ D – 6 = 0 (D 2) = 0 D = 3 or D = 2

Solution is y = + 14. [Ans. C]

Z = ax + by + ab … p z

x q z

y b

Substituting a and b in (i) in terms of p and q we get z = px + qy + pq

15. [Ans. D]

y x

x x n y

This is a linear differential equation of the form

x y w t

x n x IF = Integrations factor

^∫ ^∫ x Solution is

y (IF) = ∫ x

 y. x = ∫ xx x

 yx = ∫ x x

 yx = + C

 y = + Order of highest derivative=2

Hence, most appropriate answer is (B) 3. [Ans. A]

Given, Differential equation,

2 2

d y k y 0

dx  

Auxilary equation is y

B0 otherwise y=0 always

 sin ka=0 some boundary / initial condition)

6. [Ans. B]

The order of a differential equation is the order of the highest derivative involving in equation, so answer is 2.

The degree of a differential equation is the degree of the highest derivative involving in equation, so answer is 1.

7. [Ans. A]

∫ ∫ log y log x log

y x w s qu t on o str g t l n

Q. ∫ ∫

Applying Laplace Transform on both sides

13. [Ans. D]

Let the differential equation be y t

t y t x t

Apply Laplace transform on both sides 2 y t

t y t 3 {x t } sy s y y s x s s y s x s y y s x s

y s

Taking inverse Laplace on both sides

{y s } 2x s

s 3 y { s } y t x t y

So if we want y t as a solution both x(t) and y(0) has to be multiplied by . Hence change x(t) by x t and y(0) by y

14. [Ans. A]

x y

x y The auxiliary equation is m m ± then either

m or m i.e., roots of the equation are equal to or

15. [Ans. A]

xy is a first order linear equation non omog n ous

xy 0 is a first order linear equation (homogeneous)

r non l n r qu t ons 16. [Ans. C]

z xy ln xy z

x y ln xy xy

xy y y ln xy y z

y x ln xy xy

xyx x ln xy x o x z

x xy ln xy xy

y z

y xy ln xy xy o x z

x y z y 17. [Ans. B]

x t x

t x Pre auxiliary equation is m m

Pre roots of AE are m . Repeated roots are present.

So, most general solution in n t bt

18. [Ans. *] Range 0.53 to 0.55

E m m m olut ons s y bx … y bx

b … s ng y

y n g v s n b

y x

t x y EE

1. [Ans. A]

v n x’ t x (t) i.e. x

∫ x

x ∫ t lnx = t

x Putting

Now putting initial condition x(0) = x x

Solution is x = x i.e. x(t) = x 2. [Ans. B]

x t x

t x

Auxiliary equation m m

 m m m (m+4)(m+2)=0

m= 2, 4

x(t) =

x(0) = 1 1= … (1)



 …

On solving (1) & (2), we get and

x(t)= 2 3. [Ans. A]

y x Integrate on both sides y

4. [Ans. B]

x x

p n nt rv l x x y

x x y

Value is in between 20 and 30 So it is 25

5. [Ans. C]

t x g v n x os t s n t x n x

t s n t os t x

x os t s n t

6. [Ans. C]

x y xy y y y

x y x

w subst tut y x t n x x x (

x ) x n y

xs t s s IN

1. [Ans. C]

y m m m

Since there is double root at 2, so general solution of the given differential equation would be

+ 2. [Ans. B]

v n ρ os θ … n ρ y

y … now y’ t nθ …

Equating equations (i) and (ii) and using equation (iii) in equation (ii), we get y os θ= os θ

y= .x

Which is equation of a parabola.

3. [Ans. A]

A.E.

D= 1+ 10i

C.F = (A cos10 x + B sin 10 x)

x x 4. [Ans. C]

5. [Ans. C]

6. [Ans. A]

Given = 1 + y² Integrating ∫ = ∫ x Or t ny = x + c Or y = t n x

7. [Ans. C]

x y

Auxiliary equation, m + 1 = 0 m = 1

C.F =

y =

y y

8. [Ans. C]

The solution for the differential equation is

y x

Now, y and y , placing these values

We get, and y

9. [Ans. A]

Given partial differential equation is x

t x

t We know that

(x y ) is said to be

Parabolic if Hyperbolic if El ps

Compare the given differential equation with standard from A = 1, B = 0, C = 0

Parabolic

10. [Ans. D]

y t ÿ t ±

y os x s n x y

ẏ s n x os x ẏ

So, y os x s n x or m x m

y s n x os x s n x os x

y os x s n x

y or x m x m y m x os s n

√ √ √ √ 11. [Ans. D]

By back tracking, from option (D) y

x |x| x or x = x or x Integrating

∫ y

x ∫ x x or x ∫ x x or x y x

or x or x

In document GQB_CE.pdf (Page 118-137)