Chapter 9 – Differential Equations

Find the differential equation of the family of curves y = Ae2x + B.e–2x. Solution y = Ae2x + B.e–2x DIFFERENTIAL EQUATIONS 181 dy d2y = 2Ae2x – 2 B.e–2x and = 4Ae2x + 4Be–2x dx dx 2 d2y d2y Thus = 4y i.e., 2 – 4y = 0. dx 2 dx

Find the general solution of the differential equation = . dy y dy dx dy dx Solution = ⇒ = ⇒ ∫ = ∫ dx x y x y x ⇒ logy = logx + logc ⇒ y = cx

Given that = yex and x = 0, y = e. Find the value of y when x = 1. dy dy Solution = yex ⇒ ∫ = ∫ e dx ⇒ logy = ex + c dx y Substituting x = 0 and y = e,we get loge = e0 + c, i.e., c = 0 ( loge = 1) Therefore, log y = ex. Now, substituting x = 1 in the above, we get log y = e ⇒ y = ee. dy y

Solve the differential equation + = x2. dx x Solution The equation is of the type + Py = Q , which is a linear differential equation. Now I.F. = ∫ x dx = elogx = x. Therefore, solution of the given differential equation is 182 MATHEMATICS x4 y.x = ∫ x x 2 dx , i.e. yx = +c x3 c Hence y = + .

Find the differential equation of the family of lines through the origin. Solution Let y = mx be the family of lines through origin. Therefore, =m dy dy Eliminating m, we get y = . x or x – y = 0. dx dx

Find the differential equation of all non-horizontal lines in a plane. Solution The general equation of all non-horizontal lines in a plane is ax + by = c, where a ≠ 0. Therefore, a + b = 0. Again, differentiating both sides w.r.t. y, we get d 2x d 2x a =0⇒ = 0. dy 2 dy 2

Find the equation of a curve whose tangent at any point on it, different from origin, has slope y + . dy y  1 Solution Given = y+ = y 1 +  dx x x dy  1  ⇒ = 1+  dx y  x Integrating both sides, we get  y logy = x + logx + c ⇒ log   = x + c  x DIFFERENTIAL EQUATIONS 183 y y ⇒ = ex + c = ex.ec ⇒ = k . ex x x ⇒ y = kx . ex.

Find the equation of a curve passing through the point (1, 1) if the perpendicular distance of the origin from the normal at any point P(x, y) of the curve is equal to the distance of P from the x – axis. – dx Solution Let the equation of normal at P(x, y) be Y – y = dy ( X – x ) ,i.e., dx  dx  Y+ X – y + x  =0 ...(1) dy  dy  Therefore, the length of perpendicular from origin to (1) is y+ x ...(2)  dx  1+    dy  Also distance between P and x-axis is |y|. Thus, we get y+ x = |y|  dx  1+    dy   dx    dx  2  dx  dx 2  ⇒  y + x  = y 1+    ⇒  dy    dy    dy  dy  ( ) x – y 2 + 2 xy  = 0 ⇒ =0 dx 2 xy or = 2 2 dy y –x 184 MATHEMATICS Case I: = 0 ⇒ dx = 0 Integrating both sides, we get x = k, Substituting x = 1, we get k = 1. Therefore, x = 1 is the equation of curve (not possible, so rejected). dx 2x y dy y 2 − x 2 Case II: = 2 2 ⇒ = . Substituting y = vx, we get dy y −x dx 2 xy dv v 2 x 2 − x 2 dv v 2 − 1 v+ x = ⇒ x. = −v dx 2vx 2 dx 2v −(1 + v 2 ) 2v − dx = ⇒ dv = 2v 1+ v 2 Integrating both sides, we get log (1 + v2) = – logx + logc ⇒ log (1 + v2) (x) = log c ⇒ (1 + v2) x = c ⇒ x2 + y2 = cx. Substituting x = 1, y = 1, we get c = 2. Therefore, x2 + y2 – 2x = 0 is the required equation.  π

Find the equation of a curve passing through 1,  if the slope of the y y tangent to the curve at any point P (x, y) is − cos 2 . x x Solution According to the given condition dy y y = − cos 2 ... (i) dx x x This is a homogeneous differential equation. Substituting y = vx, we get dv dv v+x = v – cos2v ⇒ x = – cos2v dx dx DIFFERENTIAL EQUATIONS 185 ⇒ sec2v dv = − ⇒ tan v = – logx + c ⇒ tan + log x = c ...(ii) π Substituting x = 1, y = , we get. c = 1. Thus, we get  y tan   + log x = 1, which is the required equation. 2 dy  y π

Solve x − xy = 1 + cos   , x ≠ 0 and x = 1, y = dx x 2 Solution Given equation can be written as dy  y  x2 − xy = 2cos2   , x ≠ 0. dx  2x  x2 − xy  y dx =1 sec 2   ⇒  y ⇒  2 x   2 dy  2cos 2    x dx − xy  = 1  2x  2 Dividing both sides by x3 , we get  y  sec2    x dy − y   2 x   dx  1 d   y  1  = 3 ⇒ tan   = dx   2 x   x 3 2  x  x   Integrating both sides, we get  y  −1 tan   = 2 + k .  2x  2x 186 MATHEMATICS π Substituting x = 1, y = , we get 3  y 1 3 k= , therefore, tan   = − 2 + is the required solution. 2 2x 2x 2

State the type of the differential equation for the equation. xdy – ydx = x 2 + y 2 dx and solve it. Solution Given equation can be written as xdy = ( x + y + y) dx , i.e., 2 2 dy x2 + y 2 + y = ... (1) dx x Clearly RHS of (1) is a homogeneous function of degree zero. Therefore, the given equation is a homogeneous differential equation. Substituting y = vx, we get from (1) dv x 2 + v 2 x 2 + vx dv v+ x = i.e. v + x = 1+ v 2 + v dx x dx dv dx = 1+ v 2 ⇒ = ... (2) dx 1+ v 2 x Integrating both sides of (2), we get log (v + 1 + v 2 ) = logx + logc ⇒ v + 1 + v 2 = cx y y2 ⇒ + 1 + 2 = cx ⇒ y+ x 2 + y 2 = cx2 x x DIFFERENTIAL EQUATIONS 187

The degree of the differential equation 1 +  =  is  dx   dx 2 

• (A)1
• (B)2
• (C)3
• (D)4 Solution The correct answer is (B).

The degree of the differential equation d2y  dy   d2y  + 3   = x 2 log  2  is dx 2  dx   dx 

• (A)1
• (B)2
• (C)3
• (D)not defined Solution Correct answer is (D). The given differential equation is not a polynomial equation in terms of its derivatives, so its degree is not defined.   dy  2  d 2 y

The order and degree of the differential equation 1 +    = 2   dx   dx respectively, are

• (A)1, 2
• (B)2, 2
• (C)2, 1
• (D)4, 2 Solution Correct answer is (C).

The order of the differential equation of all circles of given radius a is:

• (A)1
• (B)2
• (C)3
• (D)4 Solution Correct answer is (B). Let the equation of given family be (x – h)2 + (y – k)2 = a2 . It has two orbitrary constants h and k. Threrefore, the order of the given differential equation will be 2.

The solution of the differential equation 2 x . – y = 3 represents a family of

• (A)straight lines
• (B)circles
• (C)parabolas
• (D)ellipses 188 MATHEMATICS Solution Correct answer is (C). Given equation can be written as 2dy dx = y + 3 x ⇒ 2log (y + 3) = logx + logc ⇒ (y + 3)2 = cx which represents the family of parabolas

The integrating factor of the differential equation (x log x) + y = 2logx is

• (A)ex
• (B)log x
• (C)log (log x)
• (D)x dy y 2 Solution Correct answer is (B). Given equation can be written as dx + x log x = x . I.F. = ∫ x log x Therefore, = elog (logx) = log x.  dy  dy

A solution of the differential equation   − x + y = 0 is  dx  dx

• (A)y = 2
• (B)y = 2x
• (C)y = 2x – 4
• (D)y = 2x2 – 4 Solution Correct answer is (C).

Which of the following is not a homogeneous function of x and y. 2  y y

• (A)x2 + 2xy
• (B)2x – y
• (C)cos   +
• (D)sinx – cosy x x Solution Correct answer is (D).

Solution of the differential equation x + y = 0 is 1 1

• (A)x + y = c
• (B)logx . logy = c
• (C)xy = c
• (D)x + y = c Solution Correct answer is (C). From the given equation, we get logx + logy = logc giving xy = c. DIFFERENTIAL EQUATIONS 189

The solution of the differential equation x + 2 y = x 2 is x2 + c x2 x4 + c x4 + c

• (A)y =
• (B)y = +c
• (C)y =
• (D)y = 4 x2 4 x2 4 x2 Solution Correct answer is (D). I.F. = e∫ x dx = e2log x = elog x2 = x 2 . Therefore, the solution x4 x +c is y . x = ∫ x .xdx = + k , i.e., y = 2 2 . 4 4 x2

Fill in the blanks of the following:

• (i)Order of the differential equation representing the family of parabolas y2 = 4ax is __________ .  dy   d y  3 2
• (ii)The degree of the differential equation    2  = 0 is ________ . +  dx   dx 
• (iii)The number of arbitrary constants in a particular solution of the differential equation tan x dx + tan y dy = 0 is __________ . x2 + y 2 + y
• (iv)F (x, y) = is a homogeneous function of degree__________ . (v) An appropriate substitution to solve the differential equation  x x 2 log   − x 2 dx  y = is__________ . dy  x xy log    y (vi) Integrating factor of the differential equation x − y = sinx is __________ . (vii) The general solution of the differential equation = e x − y is __________ . 190 MATHEMATICS (viii) The general solution of the differential equation + =1 is __________ . (ix) The differential equation representing the family of curves y = A sinx + B cosx is __________ .  e −2 x y  dx dy (x)  −  = 1( x ≠ 0) when written in the form + Py = Q , then  x x  dy dx P = __________ . Solution (i) One; a is the only arbitrary constant. (ii) Two; since the degree of the highest order derivative is two. (iii) Zero; any particular solution of a differential equation has no arbitrary constant. (iv) Zero. (v) x = vy. 1 dy y sin x (vi) ; given differential equation can be written as − = and therefore x dx x x 1 1 I.F. = e∫ − x dx = e–logx = . (vii) ey = ex + c from given equation, we have eydy = exdx. x2 x2 + c ; I.F. = ∫ x dx = elogx = x and the solution is y . x = ∫ x .1 dx = (viii) xy = +C. 2 e 2 d2y (ix) + y = 0; Differentiating the given function w.r.t. x successively, we get dx 2 dy d2y = Acosx – Bsinx and = –Asinx – Bcosx dx dx 2 d2y ⇒ + y = 0 is the differential equation. dx 2 (x) ; the given equation can be written as DIFFERENTIAL EQUATIONS 191 dy e –2 x y dy y e –2 x = − i.e. + = dx x x dx x x This is a differential equation of the type + Py = Q.

State whether the following statements are True or False.

• (i)Order of the differential equation representing the family of ellipses having centre at origin and foci on x-axis is two. d2y dy
• (ii)Degree of the differential equation 1 + 2 =x+ is not defined. dx dx dy dy
• (iii)+ y = 5 is a differential equation of the type + Py = Q but it can be solved dx dx using variable separable method also.  y y cos   + x x
• (iv)F(x, y) =  y  is not a homogeneous function. x cos   x x2 + y 2 (v) F(x, y) = is a homogeneous function of degree 1. x− y (vi) Integrating factor of the differential equation − y = cos x is ex. (vii) The general solution of the differential equation x(1 + y2)dx + y (1 + x2)dy = 0 is (1 + x2) (1 + y2) = k. (viii) The general solution of the differential equation + y sec x = tanx is y (secx – tanx) = secx – tanx + x + k. (ix) x + y = tan–1y is a solution of the differential equation y2 + y 2 + 1= 0 192 MATHEMATICS d 2 y 2 dy (x) y = x is a particular solution of the differential equation −x + xy = x . dx 2 dx Solution x2 y 2 (i) True, since the equation representing the given family is + = 1 , which a 2 b2 has two arbitrary constants. (ii) True, because it is not a polynomial equation in its derivatives. (iii) True (iv) True, because f ( λx, λy) = λ° f (x, y). (v) True, because f ( λx, λy) = λ1 f (x, y). (vi) False, because I.F = e ∫ −1dx = e – x . (vii) True, because given equation can be written as 2x −2 y dx = dy 1+ x 2 1+ y 2 ⇒ log (1 + x2) = – log (1 + y2) + log k ⇒ (1 + x2) (1 + y2) = k (viii) False, since I.F. = e ∫ sec xdx = elog(sec x + tan x ) = secx + tanx, the solution is, y (secx + tanx) = ∫ (sec x + tan x) tan xdx = ∫ ( sec x tan x + sec x − 1) dx = secx + tanx – x +k dy 1 dy (ix) True, x + y = tan–1y ⇒ 1+ = dx 1+ y 2 dx dy  1  dy − (1 + y 2 ) ⇒  – 1  = 1 , i.e., = which satisfies the given equation. dx  1 + y 2  dx y2 DIFFERENTIAL EQUATIONS 193 (x) False, because y = x does not satisfy the given differential equation.