DIFFERENTIAL EQUATIONS

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9.1 Overview

(i) An equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation. (ii) A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation and a differential equation involving derivatives with respect to more than one independent variables is called a partial differential equation. (iii) Order of a differential equation is the order of the highest order derivative occurring in the differential equation. (iv) Degree of a differential equation is defined if it is a polynomial equation in its derivatives. (v) Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it. (vi) A relation between involved variables, which satisfy the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called the general solution and the solution free from arbitrary constants is called particular solution. (vii) To form a differential equation from a given function, we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants. (viii) The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves. (ix) 'Variable separable method" is used to solve such an equation in which variables can be separated completely, i.e., terms containingxshould remain withdxand terms containingy should remain withdy.Chapter 9

DIFFERENTIAL EQUATIONS

180 MATHEMATICS(x) A function F (x,y) is said to be a homogeneous function of degreen if

F (lx, ly)= ln F (x,y) for some non-zero constantl. (xi) A differential equation which can be expressed in the formdy dx= F (x,y) ordx dy = G (x,y), where F (x,y) and G (x,y) are homogeneous functions of degree zero, is called a homogeneous differential equation. (xii) To solve a homogeneous differential equation of the typedy dx = F (x,y), we make substitutiony =vx and to solve a homogeneous differential equation of the typedx dy = G (x,y), we make substitutionx =vy. (xiii) A differential equation of the formdy dx + Py = Q, where P and Q are constants or functions ofx only is known as a first order linear differential equation. Solution of such a differential equation is given byy (I.F.) =( )Q I.F.dx´ò+ C, where

I.F. (Integrating Factor) =Pdxeò.

(xiv) Another form of first order linear differential equation isdx dy + P1x = Q1, where P

1 and Q1 are constants or functions ofy only. Solution of such a differential

equation is given byx (I.F.) =( )1Q × I.F.dyò+ C, where I.F. =1Pdyeò.

9.2 Solved Examples

Short Answer (S.A.)

Example 1Find the differential equation of the family of curvesy = Ae2x + B.e-2x.

Solutiony = Ae2x + B.e-2x

DIFFERENTIAL EQUATIONS 181dy

dx = 2Ae2x - 2 B.e-2x and2 2d y dx = 4Ae2x + 4Be-2x Thus2 2d y dx = 4yi.e.,2 2d y dx- 4y =0. Example 2Find the general solution of the differential equationdy dx=y x.

Solutiondy

dx=y xÞdy y =dx xÞdy yò=dx xòÞ logy = logx + logcÞ y =cx

Example 3Given thatdy

dx=yex andx = 0,y =e.Find the value ofy whenx = 1.

Solutiondy

dx=yex Þ dy yò =xe dxò Þlogy =ex +c Substitutingx = 0 andy =e,we get loge= e0+ c, i.e., c = 0 (loge = 1)

Therefore, logy= ex.

Now, substitutingx = 1 in the above, we get logy = eÞ y=ee.

Example 4Solve the differential equationdy

dx +y x=x2. SolutionThe equation is of the type+P = Qdyydx, which is a linear differential equation.

Now I.F. =1dxxò=elogx =x.

Therefore, solution of the given differential equation is

182 MATHEMATICSy.x =2xx dxò, i.e. yx =4

4xc+Hencey =3

4x c x+. Example 5Find the differential equation of the family of lines through the origin. SolutionLety =mx be the family of lines through origin. Therefore,dy dx=m

Eliminatingm, we gety =dy

dx.x or xdy dx -y = 0. Example 6Find the differential equation of all non-horizontal lines in a plane. SolutionThe general equation of all non-horizontal lines in a plane is ax +by =c, wherea¹0.

Therefore,dxa bdy+= 0.

Again, differentiating both sides w.r.t.y, we get2

2d xady = 0Þ2

2d x dy= 0. Example 7Find the equation of a curve whose tangent at any point on it, different from origin, has slopeyyx+. logy =x + logx +cÞlogy

DIFFERENTIAL EQUATIONS 183Þ y

x=ex +c= ex.ecÞy x=k . ex y=kx . ex.

Long Answer (L.A.)

Example 8Find 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 thex - axis. SolutionLet the equation of normal at P(x,y) be Y -y =( )-X -dxxdy,i.e.,

Y +Xdx

Therefore, the length of perpendicular from origin to (1) is2 1dx y xdy dx dy+ +ç ÷è ø...(2) Also distance between P andx-axis is |y|. Thus, we get2 1dx y xdy dx dy+ +ç ÷è ø= |y|

21dxydyé ù

dy=ordx dy=2 22 -xy y x

184 MATHEMATICSCase I:dx

dy =0Þ dx =0 Integrating both sides, we getx = k,Substituting x =1, we getk =1. Therefore,x = 1 is the equation of curve (not possible, so rejected).

Case II:dx

dy=2 2 2 22

2x y dy y x

dx xy y x-Þ =-.Substituting y = vx, we get2 2 2 2

2dv v x xv xdxvx-+ =Þ21.2dv vxvdx v-= -=2(1 )

2v v- +Þ22

1v dxdvxv-=+Integrating both sides, we get

log (1 +v2) = - logx + logcÞlog (1 +v2) (x) = logcÞ (1 +v2) x=c Þx2 +y2 =cx. Substitutingx = 1,y = 1, we getc = 2. Therefore,x2 +y2 - 2x = 0 is the required equation. Example 9Find the equation of a curve passing through1,4p tangent to the curve at any point P (x,y) is2cosy y x x-. SolutionAccording to the given condition2cosdy y y dx x x= -... (i) This is a homogeneous differential equation. Substitutingy =vx, we get v +xdv dx = v -cos2vÞdvxdx = - cos2v

DIFFERENTIAL EQUATIONS 185Þsec2v dv =dx

x-Þtanv= - logx+ c

Þtanlogyx cx+ =...(ii)

Substitutingx = 1,y =4p, we get.c = 1. Thus, we get tany

Example 10Solve2dyx xydx- = 1 + cosy

Þ2 2 1

2cos2dy

x xydx y x-

2sec212y

dy ç ÷è øé ù- =ê úë ûDividing both sides byx3 , we get2

2 3sec

12 2y dyx yxdx é ù-ç ÷ê úè ø=ê úê úë ûÞ31tan2d y

9.1 Overview

(i) An equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation. (ii) A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation and a differential equation involving derivatives with respect to more than one independent variables is called a partial differential equation. (iii) Order of a differential equation is the order of the highest order derivative occurring in the differential equation. (iv) Degree of a differential equation is defined if it is a polynomial equation in its derivatives. (v) Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it. (vi) A relation between involved variables, which satisfy the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called the general solution and the solution free from arbitrary constants is called particular solution. (vii) To form a differential equation from a given function, we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants. (viii) The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves. (ix) 'Variable separable method" is used to solve such an equation in which variables can be separated completely, i.e., terms containingxshould remain withdxand terms containingy should remain withdy.Chapter 9

DIFFERENTIAL EQUATIONS

180 MATHEMATICS(x) A function F (x,y) is said to be a homogeneous function of degreen if

F (lx, ly)= ln F (x,y) for some non-zero constantl. (xi) A differential equation which can be expressed in the formdy dx= F (x,y) ordx dy = G (x,y), where F (x,y) and G (x,y) are homogeneous functions of degree zero, is called a homogeneous differential equation. (xii) To solve a homogeneous differential equation of the typedy dx = F (x,y), we make substitutiony =vx and to solve a homogeneous differential equation of the typedx dy = G (x,y), we make substitutionx =vy. (xiii) A differential equation of the formdy dx + Py = Q, where P and Q are constants or functions ofx only is known as a first order linear differential equation. Solution of such a differential equation is given byy (I.F.) =( )Q I.F.dx´ò+ C, where

I.F. (Integrating Factor) =Pdxeò.

(xiv) Another form of first order linear differential equation isdx dy + P1x = Q1, where P

1 and Q1 are constants or functions ofy only. Solution of such a differential

equation is given byx (I.F.) =( )1Q × I.F.dyò+ C, where I.F. =1Pdyeò.

9.2 Solved Examples

Short Answer (S.A.)

Example 1Find the differential equation of the family of curvesy = Ae2x + B.e-2x.

Solutiony = Ae2x + B.e-2x

DIFFERENTIAL EQUATIONS 181dy

dx = 2Ae2x - 2 B.e-2x and2 2d y dx = 4Ae2x + 4Be-2x Thus2 2d y dx = 4yi.e.,2 2d y dx- 4y =0. Example 2Find the general solution of the differential equationdy dx=y x.

Solutiondy

dx=y xÞdy y =dx xÞdy yò=dx xòÞ logy = logx + logcÞ y =cx

Example 3Given thatdy

dx=yex andx = 0,y =e.Find the value ofy whenx = 1.

Solutiondy

dx=yex Þ dy yò =xe dxò Þlogy =ex +c Substitutingx = 0 andy =e,we get loge= e0+ c, i.e., c = 0 (loge = 1)

Therefore, logy= ex.

Now, substitutingx = 1 in the above, we get logy = eÞ y=ee.

Example 4Solve the differential equationdy

dx +y x=x2. SolutionThe equation is of the type+P = Qdyydx, which is a linear differential equation.

Now I.F. =1dxxò=elogx =x.

Therefore, solution of the given differential equation is

182 MATHEMATICSy.x =2xx dxò, i.e. yx =4

4xc+Hencey =3

4x c x+. Example 5Find the differential equation of the family of lines through the origin. SolutionLety =mx be the family of lines through origin. Therefore,dy dx=m

Eliminatingm, we gety =dy

dx.x or xdy dx -y = 0. Example 6Find the differential equation of all non-horizontal lines in a plane. SolutionThe general equation of all non-horizontal lines in a plane is ax +by =c, wherea¹0.

Therefore,dxa bdy+= 0.

Again, differentiating both sides w.r.t.y, we get2

2d xady = 0Þ2

2d x dy= 0. Example 7Find the equation of a curve whose tangent at any point on it, different from origin, has slopeyyx+. logy =x + logx +cÞlogy

DIFFERENTIAL EQUATIONS 183Þ y

x=ex +c= ex.ecÞy x=k . ex y=kx . ex.

Long Answer (L.A.)

Example 8Find 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 thex - axis. SolutionLet the equation of normal at P(x,y) be Y -y =( )-X -dxxdy,i.e.,

Y +Xdx

Therefore, the length of perpendicular from origin to (1) is2 1dx y xdy dx dy+ +ç ÷è ø...(2) Also distance between P andx-axis is |y|. Thus, we get2 1dx y xdy dx dy+ +ç ÷è ø= |y|

21dxydyé ù

dy=ordx dy=2 22 -xy y x

184 MATHEMATICSCase I:dx

dy =0Þ dx =0 Integrating both sides, we getx = k,Substituting x =1, we getk =1. Therefore,x = 1 is the equation of curve (not possible, so rejected).

Case II:dx

dy=2 2 2 22

2x y dy y x

dx xy y x-Þ =-.Substituting y = vx, we get2 2 2 2

2dv v x xv xdxvx-+ =Þ21.2dv vxvdx v-= -=2(1 )

2v v- +Þ22

1v dxdvxv-=+Integrating both sides, we get

log (1 +v2) = - logx + logcÞlog (1 +v2) (x) = logcÞ (1 +v2) x=c Þx2 +y2 =cx. Substitutingx = 1,y = 1, we getc = 2. Therefore,x2 +y2 - 2x = 0 is the required equation. Example 9Find the equation of a curve passing through1,4p tangent to the curve at any point P (x,y) is2cosy y x x-. SolutionAccording to the given condition2cosdy y y dx x x= -... (i) This is a homogeneous differential equation. Substitutingy =vx, we get v +xdv dx = v -cos2vÞdvxdx = - cos2v

DIFFERENTIAL EQUATIONS 185Þsec2v dv =dx

x-Þtanv= - logx+ c

Þtanlogyx cx+ =...(ii)

Substitutingx = 1,y =4p, we get.c = 1. Thus, we get tany

Example 10Solve2dyx xydx- = 1 + cosy

Þ2 2 1

2cos2dy

x xydx y x-

2sec212y

dy ç ÷è øé ù- =ê úë ûDividing both sides byx3 , we get2

2 3sec

12 2y dyx yxdx é ù-ç ÷ê úè ø=ê úê úë ûÞ31tan2d y