Q: Let f : [ 1 . ∞ ) → ℝ be a differentiable function such that f 1 = 1 3 and 3 ∫ 1 x f t d t = x f x - x 3 3 , x ∈ [ 1 , ∞ ) . Let e denote the base of the natu

Given, 3 ∫ 1 x f t d t = x f x - x 3 3 Now differentiating both side we get, ⇒ 3 f x = f x + x f ' x − x 2 ⇒ x f ' x − 2 f x = x 2 ⇒ f ' x − 2 x f x = x Which is a linear differential equation, ⇒ I . F . = e − 2 x d x = 1 x 2 Now solution is given by, ⇒ y 1 x 2 = ∫ x × 1 x 2 d x = ln ⁡ x + C ⇒ y = x 2 ( ln ⁡ x + C ) ⇒ f x = x 2 ( ln x + C ) Now using given value, f 1 = 1 3 we ge

Q: For x ∈ ℝ , let y x be a solution of the differential equation x 2 - 5 d y d x - 2 x y = - 2 x x 2 - 5 2 such that y 2 = 7 . Then the maximum value of the function y x is

Given, x 2 - 5 d y d x - 2 x y = - 2 x x 2 - 5 2 ⇒ d y d x + - 2 x x 2 - 5 y = - 2 x x 2 - 5 Which is a linear differential equation, So, integrating factor, I.F. = e ∫ - 2 x x 2 - 5 d x = 1 x 2 − 5 Now solution of differential equation is given by, y ⋅ 1 x 2 − 5 = ∫ − 2 x ⋅ x 2 − 5 x 2 − 5 d x ⇒ y x 2 − 5 = x 2 − 5 x 2 − 5 − x 2 + C Now using the given value, y ( 2 ) = 7 ⇒ C = 3 ⇒ y

Q: If y x is the solution of the differential equation x d y - y 2 - 4 y d x = 0 for x > 0 , y 1 = 2 , and the slope of the curve y = y x is never zero, then the value of 10 y 2 is _____

Given differential equation is d y y 2 - 4 y = d x x ⇒ 1 4 1 y - 4 - 1 y d y = d x x Integrating both sides, we get, 1 4 ln y - 4 y = ln x + c Given, y 1 = 2 i.e. c = 0 So, y - 4 y = ± x 4 or, y = 4 1 ± x 4 Considering y = 4 1 + x 4 So, 10 y 2 = 10 4 1 + 4 = 8

Q: For x ∈ ℝ , let the function y x be the solution of the differential equation d y d x + 12 y = cos π 12 x , y 0 = 0 . Then, which of the following statements is/are TRUE?

Given, d y d x + 12 y = cos π x 12 I . F = e 12 x Now solution of the differential equation is given by, ⇒ y · e 12 x = ∫ e 12 x · cos π x 12 d x + C ⇒ y · e 12 x = e 12 x 12 2 + π 12 2 12 cos π x 12 + π 12 sin π x 12 + C Now given curve is passing through 0 , 0 So, y 0 = 0 ⇒ C = - 12 12 2 + π 12 2 So y = 1 λ 12 cos π x 12 + π 12 sin π x 12 ⏟ f 1 x - 12 e - 12 x Where λ = 1

Q: For any real numbers α and β , let y α , β x , x ∈ R , be the solution of the differential equation d y d x + α y = x e β x , y 1 = 1 Let&nb

$ d y d x +a y=x e^ x $ Integrating factor (I.F.) $= e ^ adx = e ^ ax $ So, the solution is $y e^ a x = x e^ x e^ a x d x$ $ y e^ a x = x e^ ( + ) x d x$ If $ + 0$ $ y e^ a x =x e^ ( + ) x ( + ) - e^ ( + ) x ( + )^2 +C$ $ y= x e^ x ( + ) - e^ x ( + )^2 +C e^ -a x $ $ y= e^ x ( + ) (x- 1 a+ )+C e^ - x $ Put $ = =1$ in (1) $y= e^x 2 (x- 1 2 )+C e^ -x $ $y(1)=1$ $1= e 2 1 2 + C e c=e- e^2 4 $ So, $y= e^x 2 (x- 1 2 )+ (e- e^2 4 ) e^ -x $ If $ + =0 \& =1$ \( aligned & d y

Q: Let b be a nonzero real number. Suppose f : ℝ → ℝ is a differentiable function such that f 0 = 1 . If the derivative f ' of f satisfies the equation f ' x =

f ' x = f x b 2 + x 2 ⇒ f ' x f x = 1 b 2 + x 2 ⇒ l n f x = 1 b tan - 1 x b + C ∴ f 0 = 1 ∴ 0 = 0 + C ⇒ C = 0 ∴ l n f x = 1 b tan - 1 x b ⇒ f x = e 1 b tan - 1 x b ∴ f ' x = e 1 b tan - 1 x b . 1 b 2 + x 2 > 0 for x ∈ R ⇒ f x is increasing for x ∈ R . f x . f - x = e 1 b tan - 1 x b · e - 1 b tan - 1 x b = e 0 = 1 .

Question 1

Let y : (- , ) (0, ) be the solution of the differential equation dy dx = e^ 5x y^3 + y^3 e^x + e^x y^4 , satisfying y(0) = 1 2 . Then the value of y( _e 2) is

Accepted Answer

The given differential equation is dy dx = e^ 5x y^3 + y^3 e^x + e^x y^4 Factoring the numerator and denominator, we get dy dx = y^3 (e^ 5x + 1) e^x (1 + y^4) Separating the variables x and y, we obtain 1 + y^4 y^3 dy = e^ 5x + 1 e^x dx ( 1 y^3 + y ) dy = (e^ 4x + e^ -x ) dx Integrating both sides yields ( y^ -3 + y ) dy = (e^ 4x + e^ -x ) dx - 1 2y^2 + y^2 2 = e^ 4x 4 - e^ -x + C Multiplying by 2, we get y^2 - 1 y^2 = e^ 4x 2 - 2e^ -x + 2C Using the initial condition y(0) = 1 2 , we substitute

Question 2

Let y = f(x) be the real valued function defined on the interval (0, ), satisfying y(1) = 0 and the differential equation x dy dx = y - x^3. Then which of the following statements is (are) TRUE ?

Accepted Answer

The given differential equation is x dy dx = y - x^3. Dividing by x, we get: dy dx - 1 x y = -x^2 This is a linear differential equation of the form dy dx + P(x)y = Q(x), where P(x) = - 1 x and Q(x) = -x^2. Integrating Factor (IF) = e^ - 1 x dx = e^ - x = 1 x . Multiplying the differential equation by the IF: d dx ( y 1 x ) = -x Integrating both sides with respect to x: y x = - x^2 2 + C y = - x^3 2 + Cx Given y(1) = 0: 0 = - 1 2 + C C = 1 2 Thus, f(x) = x 2 - x^3 2 . To find local extrema, we f

Question 3

For all x>0, let y_1(x), y_2(x), and y_3(x) be the functions satisfying aligned & d y_1 d x -( x)^2 y_1=0, y_1(1)=5 \ & d y_2 d x -( x)^2 y_2=0, y_2(1)= 1 3 \ & d y_3 d x - ( 2-x^3 x^3 ) y_3=0, y_3(

Accepted Answer

aligned & d y_1 y_1 + d y_2 y_2 + d y_3 y_3 = ( ^2 x+ ^2 x+ 2-x^3 x^3 ) d x \ & (y_1 y_2 y_3 )= -1 x^2 +C \ & (y_1(x) y_2(x) y_3(x) )= -1 x^2 +C \ & (5 1 3 3 5 e )= -1 x^2 +C \ & C=0 aligned aligned & y_1(x) y_2(x) y_3(x)=e^ -1 x^2 \ & _ x 0 e^ -1 x^2 +2 x e^ 3 x x \ & _ x 0 1 e^ 3 x+ 1 x^2 x + _ x 0 2 x e^ 3 x x aligned _ x 0 1 e^ 3 x x x e^ 1 x^2 1 x +2 =0+2 \ _ y e ^ y ^2 y = e ^ y ^2 2 y 1 = \ =2

Question 4

Let y(x) be the solution of the differential equation x^2 d y d x +x y=x^2+y^2, x> 1 e satisfying y(1)=0. Then the value of 2 (y(e))^2 y (e^2 ) is _____ .

Accepted Answer

Put y = vx d y d x =v+x d v d x D.E. aligned & x ^2 (v+ x ~d v dx )+ x ^2 v= x ^2 (1+v^2 ) \ & v+x d v d x +v=1+v^2 \ & x d v d x =1+v^2-2 v \ & d v (v-1)^2 = d x x \ & - 1 v-1 = |x|+ C aligned x x - y = |x|+ C = x + C ( . Since . x > 1 e ) Given y(1)=0 C=1 So x x-y = (e x) Now y(e)= e 2 and y (e^2 )= 2 e^2 3 2(y(e))^2 y (e^2 ) = 2 e^2 4 2 e^2 3 = 3 4 =0.75

Question 5

Let f : [ 1 . &#8734; ) &#8594; &#8477; be a differentiable function such that f 1 = 1 3 and 3 &#8747; 1 x f t d t = x f x - x 3 3 , x &#8712; [ 1 , &#8734; ) . Let e  denote the base of the natu

Accepted Answer

Given, 3 &#8747; 1 x f t d t = x f x - x 3 3 Now differentiating both side we get, &#8658; 3 f x = f x + x f ' x &#8722; x 2 &#8658; x f ' x &#8722; 2 f x = x 2 &#8658; f ' x &#8722; 2 x f x = x Which is a linear differential equation, &#8658; I . F . = e &#8722; 2 x d x = 1 x 2 Now solution is given by, &#8658; y 1 x 2 = &#8747; x &#215; 1 x 2 d x = ln &#8289; x + C &#8658; y = x 2 ( ln &#8289; x + C ) &#8658; f x = x 2 ( ln x + C ) Now using given value,  f 1 = 1 3  we ge

Question 6

For x &#8712; &#8477; , let y x  be a solution of the differential equation x 2 - 5 d y d x - 2 x y = - 2 x x 2 - 5 2 such that y 2 = 7 . Then the maximum value of the function y x  is

Accepted Answer

Given, x 2 - 5 d y d x - 2 x y = - 2 x x 2 - 5 2 &#8658; d y d x + - 2 x x 2 - 5 y = - 2 x x 2 - 5 Which is a linear differential equation, So, integrating factor, &#160;I.F. = e &#8747; - 2 x x 2 - 5 d x = 1 x 2 &#8722; 5 Now solution of differential equation is given by, y &#8901; 1 x 2 &#8722; 5 = &#8747; &#8722; 2 x &#8901; x 2 &#8722; 5 x 2 &#8722; 5 d x &#8658; y x 2 &#8722; 5 = x 2 &#8722; 5 x 2 &#8722; 5 &#8722; x 2 + C Now using the given value,  y ( 2 ) = 7 &#8658; C = 3 &#8658; y

Question 7

If y x is the solution of the differential equation x d y - y 2 - 4 y d x = 0  for x &#62; 0 , &#160; y 1 = 2 , and the slope of the curve y = y x is never zero, then the value of 10 y 2 is _____

Accepted Answer

Given differential equation is  d y y 2 - 4 y = d x x &#8658; 1 4 1 y - 4 - 1 y d y = d x x Integrating both sides, we get,  1 4 ln y - 4 y = ln x + c Given, y 1 = 2 i.e.  c = 0 So, y - 4 y = &#177; x 4  or, y = 4 1 &#177; x 4 Considering y = 4 1 + x 4 So, 10 y 2 = 10 4 1 + 4 = 8

Question 8

For x &#8712; &#8477; , let the function y x be the solution of the differential equation  d y d x + 12 y = cos &#960; 12 x , y 0 = 0 .  Then, which of the following statements is/are TRUE?

Accepted Answer

Given, d y d x + 12 y = cos &#960; x 12 I . F = e 12 x Now solution of the differential equation is given by, &#8658; y &#183; e 12 x = &#8747; e 12 x &#183; cos &#960; x 12 d x + C &#8658; y &#183; e 12 x = e 12 x 12 2 + &#960; 12 2 12 cos &#960; x 12 + &#960; 12 sin &#960; x 12 + C Now given curve is passing through  0 , 0 So,  y 0 = 0 &#8658; C = - 12 12 2 + &#960; 12 2 So y = 1 &#955; 12 cos &#960; x 12 + &#960; 12 sin &#960; x 12 &#9183; f 1 x - 12 e - 12 x Where  &#955; = 1

Question 9

For any real numbers  &#945;  and  &#946; , let  y &#945; , &#946; x , x &#8712; R , be the solution of the differential equation d y d x + &#945; y = x e &#946; x , y 1 = 1 Let&nb

Accepted Answer

$ d y d x +a y=x e^ x $ Integrating factor (I.F.) $= e ^ adx = e ^ ax $ So, the solution is $y e^ a x = x e^ x e^ a x d x$ $ y e^ a x = x e^ ( + ) x d x$ If $ + 0$ $ y e^ a x =x e^ ( + ) x ( + ) - e^ ( + ) x ( + )^2 +C$ $ y= x e^ x ( + ) - e^ x ( + )^2 +C e^ -a x $ $ y= e^ x ( + ) (x- 1 a+ )+C e^ - x $ Put $ = =1$ in (1) $y= e^x 2 (x- 1 2 )+C e^ -x $ $y(1)=1$ $1= e 2 1 2 + C e c=e- e^2 4 $ So, $y= e^x 2 (x- 1 2 )+ (e- e^2 4 ) e^ -x $ If $ + =0 \& =1$ \( aligned & d y

Question 10

Let b be a nonzero real number. Suppose  f : &#8477; &#8594; &#8477; is a differentiable function such that  f 0 = 1 . If the derivative f ' of f satisfies the equation  f ' x =

Accepted Answer

f ' x = f x b 2 + x 2 &#8658; f ' x f x = 1 b 2 + x 2 &#8658; l n f x = 1 b tan - 1 x b + C &#8756; f 0 = 1 &#160; &#160; &#160; &#160; &#8756; 0 = 0 + C &#8658; C = 0 &#8756; &#160; l n f x = 1 b tan - 1 x b &#8658; f x = e 1 b tan - 1 x b &#8756; f ' x = e 1 b tan - 1 x b . 1 b 2 + x 2 &#62; 0  for  x &#8712; R &#8658; f x is increasing for  x &#8712; R . f x . f - x = e 1 b tan - 1 x b &#183; e - 1 b tan - 1 x b = e 0 = 1 .

Question 11

Let T denote a curve y = y x which is in the first quadrant and let the point 1 , &#160; 0 lie on it. Let the tangent to T at a point P intersect the y-axis at Y P . If P Y P has length 1 for each poi

Accepted Answer

Let point P is ( h , k ) Now, equation of tangent at P ( y - k ) = m T ( x - h ) ...(i) ( m T = slope of tangent at P ) For Y P , put x = 0 in.....(i) Y P 0 , k - h m T Now, as given P Y P = 1 h 2 + h 2 m T 2 = 1 h 2 1 + m T 2 = 1 1 + m T 2 = 1 h 2 m T 2 = 1 h 2 - 1 m T 2 = 1 - h 2 h 2 m T = ± 1 - h 2 h Put m T = d y d x a n d h = x d y d x = ± 1 - x 2 x ...(i) y = ± - ln ⁡ 1 + 1 - x 2 x + 1 - x 2 As the curve lies in 1 s t quadrant y must be positive. Hence, y = ln ⁡ 1 + 1 - x 2 x - 1 - x 2 Als

Differential Equations — Previous Year Questions & Solutions