Q: Let y = y(x) be the solution of the differential equation ( x)^ 1/2 \,dy = ( ^3 x - ( x)^ 3/2 y)\,dx, 0 < x < 2 , y ( 4 ) = 6 2 5 . If y ( 3 ) = 4 5 , then ^4 equals _______.

The given differential equation can be rewritten as: dy dx + ( x)y = ^3 x ( x)^ 1/2 This is a linear differential equation of the form dy dx + P(x)y = Q(x). Integrating Factor (IF) = e^ x \, dx = e^ ( x) = x Multiplying both sides by the IF and integrating, we get: y x = ^4 x ( x)^ 1/2 \, dx + C y x = (1 + ^2 x) ^2 x ( x)^ 1/2 \, dx + C Let x = t, then ^2 x \, dx = dt. y x = 1 + t^2 t^ 1/2 \, dt + C y x = (t^ -1/2 + t^ 3/2 ) \, dt + C y x = 2t^ 1/2 + 2 5 t^ 5/2 + C y x = 2( x)^ 1/2 + 2 5 ( x)^ 5

Q: Let y = y(x) be the solution of the differential equation x ( y x )dy = (y ( y x ) - x )dx, y(1) = 2 and let = ( y(e^ 12 ) e^ 12 ). Then the number of integral values of p, for which the equation x^2

Given the differential equation: x ( y x )dy = (y ( y x ) - x )dx Rearranging the terms, we get: x ( y x )dy - y ( y x )dx = -xdx Dividing both sides by x^2: ( y x ) ( xdy - ydx x^2 ) = - dx x ( y x )d ( y x ) = - dx x Integrating both sides: - ( y x ) = - |x| - C ( y x ) = |x| + C Using the initial condition y(1) = 2 : ( 2 ) = (1) + C C = 0 Thus, the solution to the differential equation is: ( y x ) = |x| We need to find = ( y(e^ 12 ) e^ 12 ). Substituting x = e^ 12 : = (e^ 12 ) = 12 The given

Q: Let y=y(x) be the solution of the differential equation: dy dx + ( 6x^2+(3x^2+2x^3+4)e^ -2x (x^3+2)(2+e^ -2x ) )y=2+e^ -2x , x (-1,2), satisfying y(0)= 3 2 . If y(1)= (2+e^ -2 ), then is equal to:

The given differential equation is a linear differential equation of the form dy dx + P(x)y = Q(x), where P(x) = 6x^2 + (3x^2 + 2x^3 + 4)e^ -2x (x^3+2)(2+e^ -2x ) and Q(x) = 2 + e^ -2x . We can rewrite P(x) as: P(x) = 3x^2(2+e^ -2x ) + 2e^ -2x (x^3+2) (x^3+2)(2+e^ -2x ) = 3x^2 x^3+2 + 2e^ -2x 2+e^ -2x The integrating factor (IF) is given by: IF = e^ P(x) dx = e^ ( 3x^2 x^3+2 + 2e^ -2x 2+e^ -2x ) dx IF = e^ (x^3+2) - (2+e^ -2x ) = x^3+2 2+e^ -2x The general solution of the differential equation i

Q: Let y = y(x) be the solution of the differential equation dy dx = (1 + x + x^2)(1 - y + y^2), y(0) = 1 2 . Then (2y(1) - 1) is equal to:

The given differential equation is dy dx = (1 + x + x^2)(1 - y + y^2). Separating the variables: dy y^2 - y + 1 = (x^2 + x + 1) dx Integrating both sides: dy (y - 1 2 )^2 + 3 4 = (x^2 + x + 1) dx 2 3 ^ -1 ( y - 1 2 3 2 ) = x^3 3 + x^2 2 + x + C 2 3 ^ -1 ( 2y - 1 3 ) = x^3 3 + x^2 2 + x + C Given y(0) = 1 2 , substituting x = 0 and y = 1 2 : 2 3 ^ -1 (0) = C C = 0 Substituting x = 1 to find y(1): 2 3 ^ -1 ( 2y(1) - 1 3 ) = 1^3 3 + 1^2 2 + 1 2 3 ^ -1 ( 2y(1) - 1 3 ) = 11 6 ^ -1 ( 2y(1) - 1 3 ) = 1

Question 1

Let y=y(x) be the solution of the differential equation x 1-x^2 \,dy + (y 1-x^2 - x ^ -1 x )dx = 0, x (0, 1), _ x 1^- y(x) = 1. Then y ( 1 2 ) equals:

Accepted Answer

The given differential equation is: x 1-x^2 \,dy + (y 1-x^2 - x ^ -1 x )dx = 0 Rearranging the terms, we get: x 1-x^2 \,dy + y 1-x^2 \,dx = x ^ -1 x\,dx Dividing the entire equation by 1-x^2 : x\,dy + y\,dx = x ^ -1 x 1-x^2 \,dx The left side is the exact differential of xy: d(xy) = x ^ -1 x 1-x^2 \,dx Integrating both sides: xy = x ^ -1 x 1-x^2 \,dx + C To evaluate the integral, use integration by parts. Let u = ^ -1 x and dv = x 1-x^2 \,dx. du = - 1 1-x^2 \,dx and v = - 1-x^2 u\,dv = uv - v\,d

Question 2

Let y = y(x) be the solution of the differential equation (x^2 - x x^2 - 1 )dy + (y(x - x^2 - 1 ) - x)dx = 0, x 1. If y(1) = 1, then the greatest integer less than y( 5 ) is _______.

Accepted Answer

The given differential equation is: (x^2 - x x^2 - 1 )dy + (y(x - x^2 - 1 ) - x)dx = 0 Dividing the entire equation by dx and rearranging, we get: x(x - x^2 - 1 ) dy dx + y(x - x^2 - 1 ) = x Dividing by x(x - x^2 - 1 ), we obtain a linear differential equation: dy dx + 1 x y = 1 x - x^2 - 1 The integrating factor (I.F.) is: I.F. = e^ 1 x dx = e^ x = x Multiplying the differential equation by the integrating factor x, we get: x dy dx + y = x x - x^2 - 1 d dx (xy) = x(x + x^2 - 1 ) (x - x^2 - 1 )(

Question 3

Let y = y(x) be the solution of the differential equation ( x)^ 1/2 \,dy = ( ^3 x - ( x)^ 3/2 y)\,dx, 0 < x < 2 , y ( 4 ) = 6 2 5 . If y ( 3 ) = 4 5 , then ^4 equals _______.

Accepted Answer

The given differential equation can be rewritten as: dy dx + ( x)y = ^3 x ( x)^ 1/2 This is a linear differential equation of the form dy dx + P(x)y = Q(x). Integrating Factor (IF) = e^ x \, dx = e^ ( x) = x Multiplying both sides by the IF and integrating, we get: y x = ^4 x ( x)^ 1/2 \, dx + C y x = (1 + ^2 x) ^2 x ( x)^ 1/2 \, dx + C Let x = t, then ^2 x \, dx = dt. y x = 1 + t^2 t^ 1/2 \, dt + C y x = (t^ -1/2 + t^ 3/2 ) \, dt + C y x = 2t^ 1/2 + 2 5 t^ 5/2 + C y x = 2( x)^ 1/2 + 2 5 ( x)^ 5

Question 4

Let y = y(x) be the solution of the differential equation x ( y x )dy = (y ( y x ) - x )dx, y(1) = 2 and let = ( y(e^ 12 ) e^ 12 ). Then the number of integral values of p, for which the equation x^2

Accepted Answer

Given the differential equation: x ( y x )dy = (y ( y x ) - x )dx Rearranging the terms, we get: x ( y x )dy - y ( y x )dx = -xdx Dividing both sides by x^2: ( y x ) ( xdy - ydx x^2 ) = - dx x ( y x )d ( y x ) = - dx x Integrating both sides: - ( y x ) = - |x| - C ( y x ) = |x| + C Using the initial condition y(1) = 2 : ( 2 ) = (1) + C C = 0 Thus, the solution to the differential equation is: ( y x ) = |x| We need to find = ( y(e^ 12 ) e^ 12 ). Substituting x = e^ 12 : = (e^ 12 ) = 12 The given

Question 5

Let y=y(x) be the solution of the differential equation: dy dx + ( 6x^2+(3x^2+2x^3+4)e^ -2x (x^3+2)(2+e^ -2x ) )y=2+e^ -2x , x (-1,2), satisfying y(0)= 3 2 . If y(1)= (2+e^ -2 ), then is equal to:

Accepted Answer

The given differential equation is a linear differential equation of the form dy dx + P(x)y = Q(x), where P(x) = 6x^2 + (3x^2 + 2x^3 + 4)e^ -2x (x^3+2)(2+e^ -2x ) and Q(x) = 2 + e^ -2x . We can rewrite P(x) as: P(x) = 3x^2(2+e^ -2x ) + 2e^ -2x (x^3+2) (x^3+2)(2+e^ -2x ) = 3x^2 x^3+2 + 2e^ -2x 2+e^ -2x The integrating factor (IF) is given by: IF = e^ P(x) dx = e^ ( 3x^2 x^3+2 + 2e^ -2x 2+e^ -2x ) dx IF = e^ (x^3+2) - (2+e^ -2x ) = x^3+2 2+e^ -2x The general solution of the differential equation i

Question 6

Let y = y(x) be the solution of the differential equation dy dx = (1 + x + x^2)(1 - y + y^2), y(0) = 1 2 . Then (2y(1) - 1) is equal to:

Accepted Answer

The given differential equation is dy dx = (1 + x + x^2)(1 - y + y^2). Separating the variables: dy y^2 - y + 1 = (x^2 + x + 1) dx Integrating both sides: dy (y - 1 2 )^2 + 3 4 = (x^2 + x + 1) dx 2 3 ^ -1 ( y - 1 2 3 2 ) = x^3 3 + x^2 2 + x + C 2 3 ^ -1 ( 2y - 1 3 ) = x^3 3 + x^2 2 + x + C Given y(0) = 1 2 , substituting x = 0 and y = 1 2 : 2 3 ^ -1 (0) = C C = 0 Substituting x = 1 to find y(1): 2 3 ^ -1 ( 2y(1) - 1 3 ) = 1^3 3 + 1^2 2 + 1 2 3 ^ -1 ( 2y(1) - 1 3 ) = 11 6 ^ -1 ( 2y(1) - 1 3 ) = 1

Question 7

Let x = x(y) be the solution of the differential equation 2y^2 dx dy - 2xy + x^2 = 0, y > 1, x(e) = e. Then x(e^2) is equal to:

Accepted Answer

The given differential equation is 2y^2 dx dy - 2xy + x^2 = 0. Dividing by 2y^2, we get: dx dy - x y + x^2 2y^2 = 0 This is a homogeneous differential equation. Let x = vy, then dx dy = v + y dv dy . Substituting these into the equation: v + y dv dy - v + v^2 2 = 0 y dv dy = - v^2 2 Separating the variables: dv v^2 = - dy 2y Integrating both sides: v^ -2 dv = - 1 2 dy y - 1 v = - 1 2 y - C Substituting v = x y : y x = 1 2 y + C Given x(e) = e, substituting x = e and y = e: e e = 1 2 e + C 1 = 1

Question 8

If the curve y = f(x) passes through the point (1, e) and satisfies the differential equation dy = y(2 + _e x)\,dx, x > 0, then f(e) is equal to :

Accepted Answer

Given differential equation is dy y = (2 + _e x) dx Integrating both sides, we get: dy y = (2 + _e x) dx _e y = 2x + x _e x - x + C _e y = x + x _e x + C Since the curve passes through (1, e), substitute x = 1 and y = e: _e e = 1 + 1 _e 1 + C 1 = 1 + 0 + C C = 0 The equation of the curve is _e y = x + x _e x To find f(e), substitute x = e: _e y = e + e _e e _e y = e + e(1) = 2e y = e^ 2e Answer: e^ 2e

Question 9

Let y = y(x) be the solution curve of the differential equation (1 + x) dy dx + (y+1) x = 0, y(0) = 0. If the curve y = y(x) passes through the point ( , -1 2 ), then a value of is :

Accepted Answer

The given differential equation can be written as: (1 + x)dy + (y+1) x dx = 0 d((y+1)(1 + x)) = 0 Integrating both sides: (y+1)(1 + x) = C Given y(0) = 0, substituting x = 0 and y = 0: (0+1)(1 + 0) = C C = 1 The equation of the curve is (y+1)(1 + x) = 1 Since the curve passes through ( , -1 2 ), substituting x = and y = -1 2 : ( -1 2 + 1 )(1 + ) = 1 1 2 (1 + ) = 1 1 + = 2 = 1 = 2 Answer: 2

Question 10

Let y=y(x) be the solution of the differential equation x ~d y ~d x -y=x^ 2 x, x (0, ). If y ( 2 )= 2 , then 6 y ( 6 )-8 y ( 4 ) is equal to :

Accepted Answer

The differential equation x dy dx - y = x^2 x can be rewritten as dy dx - y x = x x. The integrating factor is (x) = 1 x . Multiplying by (x) gives d dx ( y x ) = x. Integrating: y x = | x| + C, so y = x ( x) + Cx. Using y ( 2 ) = 2 : we get 2 = 0 + C 2 , thus C = 1. Therefore y = x(1 + ( x)). Now: y ( 6 ) = 6 (1 - 2) and y ( 4 ) = 4 (1 - 2 2 ). Thus 6y ( 6 ) - 8y ( 4 ) = (1 - 2) - 2 (1 - 2 2 ) = - .

Question 11

Let y=y(x) be the solution of the differential equation x d y d x - 2 y=x^ 3 (2-x^ 3 ) ^ 2 y, x 0. If y(2)=0, then (y(1)) is equal to

Accepted Answer

Divide the equation by ^2 y: x ^2 y dy dx - 2 y = x^3(2-x^3). Let v = y, so ^2 y dy dx = dv dx . The equation becomes x dv dx - 2v = x^3(2-x^3). Dividing by x: dv dx - 2v x = x^2(2-x^3). Using integrating factor = x^ -2 : d dx (x^ -2 v) = 2 - x^3. Integrating: x^ -2 v = 2x - x^4 4 + C. From y(2) = 0, we have 0 = 0 = 16 - 16 + 4C, giving C = 0. Therefore y = 2x^3 - x^6 4 . At x = 1: (y(1)) = 2 - 1 4 = 7 4 .

Question 12

Let y=y(x) be a differentiable function in the interval (0, ) such that y(1)=2, and _ t x ( t^ 2 y(x)-x^ 2 y(t) x-t )=3 for each x>0. Then 2 y(2) is equal to

Accepted Answer

Given the limit _ t x t^2 y(x) - x^2 y(t) x - t = 3. The limit is in 0 0 form. Applying L'Hopital's Rule with respect to t: _ t x d dt (t^2 y(x) - x^2 y(t)) d dt (x - t) = 3 _ t x 2t y(x) - x^2 y'(t) -1 = 3 Substituting t = x: 2x y(x) - x^2 y'(x) -1 = 3 x^2 y'(x) - 2x y(x) = 3 Dividing by x^4 to make it a linear differential equation or recognizing the quotient rule form: x^2 y'(x) - 2x y(x) x^4 = 3 x^4 d dx ( y(x) x^2 ) = 3 x^4 Integrating both sides with respect to x: y(x) x^2 = 3x^ -4 dx = 3x

Question 13

If the solution curve y=f(x) of the differential equation (x^ 2 -4 ) y^ -2 x y+2 x (4-x^ 2 )^ 2 =0, x>2, passes through the point (3,15), then the local maximum value of f is \_\_\_\_.

Accepted Answer

Rewriting: dy dx - 2x x^2-4 y = -2x(x^2-4). IF = e^ - 2x x^2-4 dx = 1 x^2-4 . d dx ( y x^2-4 ) = -2x y x^2-4 = -x^2 + C. y = (x^2-4)(C-x^2). At (3,15): 15 = 5(C-9) C = 12. y = (x^2-4)(12-x^2). y' = 4x(8-x^2) = 0 x = 2 2 (for x > 2). y'' = 32 - 12x^2 = -64 f(2 2 ) = (8-4)(12-8) = 16.

Question 14

Let y=y(x) be the solution of the differential equation x^ 4 ~d y+ (4 x^ 3 y+2 x ) d x=0, x>0, y ( 2 )=0. Then ^ 4 y ( 3 ) is equal to :

Accepted Answer

x^4\,dy + 4x^3 y\,dx + 2 x\,dx = 0 d(x^4 y) + 2 x\,dx = 0. Integrating: x^4 y - 2 x = C. y( /2) = 0: ( /2)^4 0 - 2 ( /2) = 0 C = 0. x^4 y = 2 x. At x = /3: ( /3)^4 y( /3) = 2 ( /3) = 1. ^4 y( /3) = 81 ( /3)^4 y( /3) 1 ( /3)^4 ^4 81 ... Directly: y( /3) = 81 ^4 , so ^4 y( /3) = 81.

Question 15

Let f be a twice differentiable non-negative function such that (f(x))^ 2 =25+ _ 0 ^ x ((f( t ))^ 2 + (f^ ( t ) )^ 2 ) dt . Then the mean of f ( _ e (1) ), f ( _ e (2) ), .., f ( _ e (625) ) is equal

Accepted Answer

Differentiating (f(x))^2 = 25 + _0^x((f(t))^2 + (f'(t))^2)\,dt: 2ff' = f^2 + (f')^2 (f - f')^2 = 0 f' = f. f(x) = Ce^x. At x = 0: f(0)^2 = 25 f(0) = 5 (non-negative), so f(x) = 5e^x. f( _e k) = 5k for k = 1, 2, , 625. Mean = 5(1+2+ +625) 625 = 5 625 626 2 625 = 5 626 2 = 1565.

Question 16

If y=y(x) satisfies the differential equation 16( x+9 x )(4+ 9+ x ) y ~d y=(1+2 y) d x, x>0 and y(256)= 2 , y(49)= , then 2 is equal to :

Accepted Answer

y 1+2 y d y= d x 16( 9 x +x )(4+ 9+ x ) 4+ 9+ x =t 1 2 9+ x d x 2 x =1 d x 1 2 n|1+2 y|= 4 d t 16 t +C . 1 2 n|1+2 y|= 1 4 n \, 4+ 9+ x +C 1 2 n(2 y+1)= 1 4 n|4+ 9+ x |+C Substituting (256, 2 ) 1 2 n 3= 1 2 n 3+C C=0 Substituting (49, ) 1 2 n (2 +1)= 1 4 n 8 n (2 +1)= 1 2 n 8 n (2 +1)= n 2 2 2 +1=2 2 2 =2 2 -1

Question 17

Let the solution curve of the differential equation x d y-y d x= x^ 2 +y^ 2 d x, x>0, y(1)=0, be y=y(x). Then y(3) is equal to

Accepted Answer

xdy - ydx = x^2+y^2 \,dx. Put y = vx: x^2 dv = x 1+v^2 \,dx. dv 1+v^2 = dx x . Integrating: (v+ 1+v^2 ) = x + C. y(1) = 0 v = 0 at x = 1: C = 0. v + 1+v^2 = x y + x^2+y^2 = x^2. At x = 3: y + 9+y^2 = 9 9+y^2 = (9-y)^2 18y = 72 y = 4.

Question 18

Let f:[1, ) R be a differentiable function. If 6 _ 1 ^ x f(t) d t=3 x f(x)+x^ 3 -4 for all x 1, then the value of f(2)-f(3) is

Accepted Answer

Differentiating 6 _1^x f(t)dt = 3xf(x) + x^3 - 4: 6f(x) = 3f(x) + 3xf'(x) + 3x^2 f(x) - xf'(x) = x^2. d dx ( f(x) x ) = xf'(x)-f(x) x^2 = -1. Integrating: f(x) x = -x + C f(x) = -x^2 + Cx. At x = 1: 0 = 3f(1) - 3 f(1) = 1. So -1 + C = 1 C = 2. f(x) = -x^2 + 2x. f(2) = 0, f(3) = -3. f(2) - f(3) = 3.

Question 19

Let y=y(x) be the solution of the differential equation x ~d y ~d x -2 y=2+3 x, x (- 2 , 2 ), y(0)=- 7 4 . Then y ( 6 ) is equal to :

Accepted Answer

The differential equation is x dy dx - 2y = 2 + 3 x, or dy dx - 2 x y = (2 + 3 x) x. This is a linear ODE with integrating factor (x) = e^ -2 x . Multiplying both sides and integrating: d dx [ye^ -2 x ] = (2 + 3 x) x e^ -2 x Computing the integral: 2 x e^ -2 x dx = -e^ -2 x For the second term, using integration by parts: 3 x x e^ -2 x dx = 3 4 e^ -2 x (-2 x - 1) Thus: ye^ -2 x = -e^ -2 x + 3 4 e^ -2 x (-2 x - 1) + C Simplifying: y = - 7 4 - 3 2 x + Ce^ 2 x Using y(0) = - 7 4 gives C = 0. Theref

Question 20

Let y=y(x) be the solution curve of the differential equation (1+x^ 2 ) d y+ (y- ^ -1 x ) d x=0, y(0)=1. Then the value of y(1) is :

Accepted Answer

Rewrite as dy dx + y 1+x^2 = ^ -1 x 1+x^2 . IF = e^ ^ -1 x . y e^ ^ -1 x = ^ -1 x 1+x^2 e^ ^ -1 x dx. Let t = ^ -1 x: = te^t dt = e^t(t-1) + C. y = ^ -1 x - 1 + Ce^ - ^ -1 x . Using y(0) = 1: C = 2. y(1) = 4 - 1 + 2 e^ /4 .

Differential Equations — Previous Year Questions & Solutions