First Order Differential Equations

LECTURE NOTES ON

ORDINARY DIFFERENTIAL EQUATIONS based on the course-contents of MMB-352 (A course of Mathematics (Main/Subsidiary)

prescribed for B. Sc. (Hons.) III Semester)

UNIT-1

First Order Differential Equations

Preliminaries (A topic of Unit I)

In this lecture, we shall revise the basic notions regarding to this course, which we have already discussed in +2 level Mathematics.

Differential Equations: An equation containing the derivatives or dif- ferentials of one or more dependent variables with respect to one or more independent variables is called a differential equation.

Ordinary Differential Equations: A differential equation in which each involved dependent variable is a function of a single independent variable is known as an ordinary differential equation.

Throughout the course, we shall study of different methods for solving ordinary differential equations. But for the sake of simplicity, we shall use the term ‘differential equation’ instead of ‘ordinary differential equation’.

Order of a Differential Equation: The order of a differential equation is the order of highest order derivative appearing in equation.

Degree of a Differential Equation: The degree of a differential equation is the power of highest order derivative occurring in the equation, when differential coefficients are made free from radicals and fractions.

1

2

eg, the differential equation _dx^d3y3−6 ^dy_dx2

−4y= 0 is of order 3 and degree 1.

Classification of Differential Equations: Differential equations are clas- sified into linear and nonlinear differential equations. Ann^th order differential equation is called linearif it can be expressed in the form:

P₀dⁿy

dxⁿ +P₁dⁿ⁻¹y

dxⁿ⁻¹ +P₂dⁿ⁻²y

dxⁿ⁻² +...Pn−1

dy

dx +P_ny=Q

where P₀, P₁, ..., P_n and Q are either constants or functions of the variablex and P₀ 6= 0.

A differential equation, which is not linear is called a nonlinear differen- tial equation.

Linear differential equations are further classified intohomogeneous and nonhomogeneous equations according to Q ≡ 0 and Q 6≡ 0, respectively.

In this regards, we shall study deeply later in Unit II.

Formation of a Differential Equation: Usually, differential equations are derived to eliminate arbitrary constants (parameters) involved in a family of curves. For illustration, consider the two-parameter family of straight lines

y =mx+c (1)

having the slop m and passing through the point (0, c). Differentiating Eq.

(2), we get

dy dx =m.

Differentiating again, we get

d²y

dx² = 0 (2)

which is a second order differential equation.

Remark 1: To eliminaten arbitrary constants from n-parameter family of curves, we obtain a differential equation of order n.

Solution of a Differential Equation: A solution (or integral/primitive) of a differential equation is an explicit or implicit relation between the vari- ables involved that does not contain derivatives and satisfies the differential equation.

Naturally, the solution of a differential equation in one dependent variable y and one independent variable x are of the form y = f(x) (explicit form) or of the form φ(x, y) = 0 (implicit form). Geometrically speaking, both forms represents the curves in xy-plane. Henceforth, the solutions of such a differential equation are also called ‘integral curves’ of the equation.

For example, Eq. (1) forms an integral curve of differential equation (2).

General and Particular Solutions: A solution of a differential equation which contains a number of arbitrary constants equal to the order of the dif- ferential equation is called the general solution or thecomplete solution of the differential equation.

For example, y=acosx+bsinxis the general solution of the differential equation y⁰⁰+y= 0. Although, y =acosx and y=bsinx both satisfy the given equation yet they are not general solutions as each of them contains only one arbitrary constant.

In lieu of the concept of general solution, it can be highlighted that a single differential equation can possess an infinite number of solutions corresponding to the unlimited number of choices for the arbitrary constants. A solution of a differential equation obtained by giving particular values to arbitrary constants in the general solution is called a particular solution.

Elementary Methods of Solving of Differential Equations of First Order and First Degree: In class 12, we have studied the techniques of solving the following three types of differential equations of order one and degree one:

1. Differential Equations with variable separable

4

2. Homogeneous Differential Equations 3. Linear Differential Equations

As a continuation, we shall discuss several another methods for solving the differential equations of order one and degree one in our upcoming lectures.

Bernoulli Equation (A topic of Unit I)

Recall that the general solution of linear differential equation:

dy

dx+P(x)y =Q(x) is

y×I.F.= Z

(Q×I.F.)dx+c

where cis an arbitrary constant and I.F.=e^{RP dx} (called integrating factor).

Sometimes certain nonlinear differential equations can be reduced to linear forms by making suitable substitutions and hence can be solved easily.

Bernoulli equation is one of such equations. Indeed, an equation of the form dy

dx +P(x)y=Q(x)yⁿ (1)

(where n6= 0 and n6= 1) is called Bernoulli equation.

Dividing Eq. (1) by yⁿ, we get y⁻ⁿdy

dx+P(x)y⁻ⁿ⁺¹ =Q(x). (2) Put y⁻ⁿ⁺¹ =u so that (1−n)y^{−n dy}_dx = ^du_dx. Thus, Eq. (2) reduces to

1 1−n

du

dx+P(x)u=Q(x) 1

2 or,

du

dx+ (1−n)P(x)u= (1−n)Q(x)

which is a linear differential equation in dependent variableuand independent variable x. After solving it and then replacing uby y⁻ⁿ⁺¹, we can obtain the general solution of Eq. (1).

Example 2.24/p.42: Solve (1−x²)_dx^dy +xy =xy². Solution. Dividing given equation byy², we get

(1−x²)y⁻²dy

dx +xy⁻¹ =x which can be written as

y⁻²dy

dx + x

1−x²y⁻¹ = x 1−x².

Puty⁻¹ =u so that−y^−2dy_dx = ^du_dx. Then, above equation reduces to du

dx − x

1−x²u=− x 1−x²

which is a linear differential equation in dependent variableuand independent variable x. Its integrating factor is

I.F. = e^{RP dx} =e⁻

R _x

1−x2dx

= e¹²

R −2x 1−x2dx

=e(1/2) log(1−x²)

= (1−x²)^1/2. Therefore, the solution is

u×I.F.= Z

(Q×I.F.)dx+c

or,

u(1−x²)^1/2 = Z

− x

1−x²(1−x²)^1/2dx+c

= Z

−x(1−x²)^−1/2dx+c

= Z

t^−1/2.1

2dt+c (put t = 1−x² so that−2xdx=dt)

= 1 2

t^1/2

1/2+c=t^1/2+c

= (1−x²)^1/2+c.

Putting u=y⁻¹ in above equation, we get

y⁻¹(1−x²)^1/2 = (1−x²)^1/2+c

or, √

1−x² =y√

1−x²+cy or,

(1−y)√

1−x² =cy

which is the required solution of given differential equation.

Example 2.25/p.42: Solve x^dy_dx+y=y²logx.

Solution. Dividing given equation byy², we get xy⁻²dy

dx +y⁻¹ = logx which can be written as

y⁻²dy dx + 1

xy⁻¹ = logx x .

Put y⁻¹ =uso that −y^−2dy_dx = ^du_dx. Then, above equation reduces to

−du dx + 1

xu= logx x

4 or,

du dx − 1

xu=−logx x

which is a linear differential equation in dependent variableuand independent variable x. Its integrating factor is

I.F.=e^{−R dxx} =e^−logx = 1 x. Therefore, the solution is

u×I.F.= Z

(Q×I.F.)dx+c or,

u.1

x = −

Z logx x

1

xdx+c

= −

Z

te^−tdt+c (put t = logx so that _x¹dx=dt and x=e^t)

= −[−te^−t−e^−t] +c= (t+ 1)e^−t+c

= (logx+ 1).1 x +c which can be written as

u= logx+ 1 +cx.

Puttingu=y⁻¹ in above equation, we get 1

y = logx+ 1 +cx or,

(1 + logx+cx)y= 1

which is the required solution of given differential equation.

Home Assignments:

Q.33/p.61: ^dy_dx = 2ytanx+y²tan²x. Ans. −¹_ysec²x=c+¹₂tan³x.

Q.34/p.61: cosxdy =y(sinx−y)dx. Ans. y⁻¹ =ccosx+ sinx.

Q.36/p.61: ^dy_dx =x³y³−xy. Ans. _y¹2 = 1 +x²+ce^x2. Q.38/p.61: (x²y³+xy)dy−dx= 0. Ans. ¹_x = 2−y² +e^−y2/2

Exact Differential Equation (A topic of Unit I)

A differential equation is said to be exact if it can be derived from its primitive (general solution) directly by differentiation, without any subsequent multiplication, elimination, etc.

Theorem 1. The necessary and sufficient condition for the differential equa- tion M dx+N dy= 0 to be exact is that

∂M

∂y = ∂N

∂x

In order to obtain the solution of an exact differential equation, we have to proceed as follows:

1. IntegrateM with respect to x, keeping y as constant.

2. Integrate with respect toy only those terms of N which do not contain x.

3. Add the two expressions obtained in (1) and (2) above and equate the result to an arbitrary constant.

In other words, the solution of an exact differential equation is Z

(y=constant)

M dx+

Z

(terms not having x)

N dy=c.

1

2

Ex.40 / P.61 Solveysin 2xdx−(1 +y²+ cos²x)dy= 0 Solution. The given differential equation is

ysin 2xdx−(1 +y²+ cos²x)dy= 0 compare the above equation with

M dx+N dy= 0, we get

M = ysin 2x and N =−(1 +y²+ cos²x)

⇒ ∂M

∂y = sin 2x and ∂N

∂x = 2 cosxsinx= sin 2x i.e.,

∂M

∂y = ∂N

∂x

The given equation is exact. Now the solution of the differential equation is Z

(y=constant)

M dx+

Z

(terms not having x)

N dy = c Z

ysin 2xdx+ Z

(−1−y²)dy = c

−ycos 2x

2 +

−y− y³ 3

= c So, the solution is

ycos 2x+ 2y+2y³ 3 =c

Ex.41 / P.61 Solve

y 1 + ¹_x

+ cosy

dx+ (x+ logx−xsiny)dy= 0.

Solution. The given differential equation is

y

1 + 1 x

+ cosy

dx+ (x+ logx−xsiny)dy = 0 compare the above equation with

M dx+N dy= 0, we get

M =

y

1 + 1

x

+ cosy

and N = (x+ logx−xsiny)

⇒ ∂M

∂y =

1 + 1 x

−siny and ∂N

∂x = 1 + 1

x −siny i.e.,

∂M

∂y = ∂N

∂x

The given equation is exact. Now the solution of the differential equation is Z

(y=constant)

M dx+

Z

(terms not having x)

N dy = c Z

y

1 + 1 x

+ cosy

dx+ Z

0dy = c y(x+ logx) +xcosy = c So, the solution is

xy+ylogx+xcosy=c.

Ex.42 / P.61 Solve (e^y + 1) cosxdx+e^ysinxdy= 0.

Solution. The given differential equation is

(e^y+ 1) cosxdx+e^ysinxdy= 0 compare the above equation with

M dx+N dy= 0,

4 we get

M = (e^y+ 1) cosx and N =e^ysinx

⇒ ∂M

∂y = e^ycosx and ∂N

∂x =e^ycosx i.e.,

∂M

∂y = ∂N

∂x

The given equation is exact. Now the solution of the differential equation is Z

(y=constant)

M dx+

Z

(terms not having x)

N dy = c Z

(e^y+ 1) cosxdx+ Z

0dy = c (e^y+ 1) sinx = c So, the solution is

(e^y + 1) sinx=c.

Home Assignment

1. (Exp.2.29/P.46) Solve xdx+ydy+^xdy−ydx_x2+y² = 0.

[Sol:x²−2tan^{−1 x}_y +y² =2c]

2. (Exp.2.30/P.46) Solve (1 +e^xy)dx+e^xy

1− ^x_y

dy= 0.

[Sol:x+ye^xy =c]

3. (Exp.2.31/P.47) Solve (sinxcosy+e^2x)dx+(cosxsiny+tany)dy = 0.

[Sol: ¹₂e^2x−cosxcosy+ log secy=c]

Integrating Factor: Part-I (A topic of Unit I)

The equation

ydx+ (x²y−x)dy = 0 (1)

is easily seen to non exact because ^∂M_∂y = 1 and ^∂N_∂x = 2xy−1.

However, if we multiply both sides of equation (1) by the factor _x¹2, then equation (1) becomes

y

x²dx+ (x²y−x)

x² dy = 0, (2)

which is exact because

∂M

∂y = 1

x² = ∂N

∂x.

To what extent can other non exact equations be made exact in this way?.

In other way, if

M(x, y)dx+N(x, y)dy= 0 (3)

is not exact, under what conditions can a function µ(x, y) be found with the property that

µ(M dx+N dy) = 0

is exact? Any function µ that act in this way is called an integrating factor (in short, I.F.) for (8). Therefore _x¹2 is an integrating factor for (1).

A differential equation of the type (8) always has an integrating factor but there is no general method of finding them. Here we shall discuss some of the methods for finding the integrating factors.

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2

Rule I. If the differential equation (8) is homogeneous and M x+N y 6= 0, then _{M x+N y}¹ is the integrating factor.

Example 2.35/P.50 Solve

x²ydx−(x³+y³)dy= 0 (4) Solution. From (4), we have M =x²y and N =−(x³+y³), so

M x+N y =x³y−x³y−y⁴ =−y⁴ 6= 0,

and the equation (4) is homogeneous. Therefore the integrating factor is −_y¹4. Multiplying equation (4) by this integrating factor, we have

−x² y³dx+

x³ y⁴ +1

y

dy= 0.

Now

M =−x²

y³ N = x³ y⁴ + 1

y and hence

∂M

∂y = 3x²

y⁴ = ∂N

∂x. Therefore, the differential equation is exact and hence

Z

y=constantM dx=− Z x²

y³dx=− x³ 3y³ Z

terms not having xN dy= Z dy

y = logy.

Thus the required solution is Z

M dx+ Z

N dy=c

− x³

3y³ + logy=c i.e.,

x³ = 3y³(logy−c).

Home Assignment:

1. (Q.8/P.60) Solve x²ydx−(x³+y³)dy = 0.

[Ans:y=ce^x3/3y3.]

2. (Q.39/P.61) Solve (x²−2xy−y²)dx−(x+y)²dy = 0.

[Ans:x³−y³+ 3xy(x−y) =C]

Rule II. If in the differential equation (8), M =yf₁(xy) and N =xf₂(xy).

Then _{(M x−N y)}¹ is an integrating factor.

Example 2.37/P.51 Solve

(xy+ 2x²y²)ydx+ (xy−x²y²)xdy= 0 (5) Solution. From equation (5), we have M = (xy+ 2x²y²)y = yf₁(xy) and N = (xy−x²y²)x=xf2(xy). Therefore the integrating factor is

1

M x−N y = 1

x²y²+ 2x³y³−x²y²+x³y³ = 1 3x³y³. Multiplying (5) by this integrating factor, we have

y(xy+ 2x²y²)

3x³y³ dx+x(xy−x²y²)

3x³y³ dy = 0 or,

1 3

1 x²y + 2

x

dx+ 1 3

1 xy² − 1

y

dy = 0 Now,

∂M

∂y =− 1

x²y² = ∂N

∂x

and hence the above differential equation is exact. Therefore Z

y=constantM dx= 1 3

Z 1 x²y + 2

x

dx= 1 3

− 1

xy+ 2 logx

Z

terms not having xN dy=−1 3

Z 1 y =−1

3logy.

4

Thus the required solution is Z

M dx+ Z

N dy=c or,

1 3

− 1

xy + 2 logx

+1

3(−logy) = C or,

2 logx−logy= 1

xy +C₁, where C₁ = 3C Home Assignment:

1. (Q.48/P.61) Solve (x³y³+x²y²+xy+1)ydx+(x³y³−x²y²−xy+1)xdy= 0.

[Ans:xy− _xy¹ −logy² =C.]

2. (Q.51/P.61) Solve (xysinxy+ cosxy)ydx+ (xysinxy−cosxy)xdy = 0.

[Ans:Cycosxy=x]

Rule III.The equation (8) hase^Rf(x)dxas the integrating factor if _N¹

∂M

∂y − ^∂N_∂x is a function of x, say f(x).

Remark 1. If _N¹

∂M

∂y −^∂N_∂x

=k(a constant), then the integrating factor is e^Rkdx.

Example 2.39/P.52 Solve

(x²+y²)dx−2xydy = 0 (6)

Solution. From equation (6), we have M = (x² + y²) and N = −2xy.

Therefore ^∂M_∂y = 2y and ^∂N_∂x =−2y, and hence 1

N ∂M

∂y −∂N

∂x

=− 1

2xy(2y−(−2y)) = −2

x =f(x)(say).

Thus the integrating factor is

e^Rf(x)dx=e^{R −2x dx}

=e^{−2 logx}

= 1

x² (7)

Multiplying the equation (6) by this integrating factor, we obtain x²+y²

x²

dx− 2xy

x² dy= 0 or,

1 + y²

x²

dx− 2y

x dy= 0 or,

dx+d −y²

x

= 0 Now integrating term by term, we get

x− y² x =C.

which is the required solution.

Home Assignment:

1. (Q.52/P.61) Solve (x³−2y²)dx+ 2xydy = 0.

[Ans:x+_x^y22 =C]

2. (Q.53/P.61) Solve (x²+y²+ 2x)dx+ 2ydy= 0.

[Ans:e^x(x²+y²) =C]

Lecture No. 05

Integrating Factor: Part-II (A topic of Unit I)

In the previous lecture, we have discussed the integrating factor of the following non exact differential equation

M(x, y)dx+N(x, y)dy= 0 (8)

Now, we also include the following methods to find out the integrating factor of the differential equation (8).

Rule IV.The equation (8) hase^Rf(y)dyas the integrating factor if _M¹

∂N

∂x − ^∂M_∂y is a function of y, say f(y).

Remark 2. If _M¹

∂N

∂x − ^∂M_∂y

=k(a constant), then the integrating factor is e^Rkdy.

Example 2.41 /P.53 Solve

(xy³+y)dx+ 2(x²y²+x+y⁴)dy = 0. (9) Solution. From equation (9), we haveM = (xy³+y)andN = 2(x²y²+x+y⁴).

Therefore

∂M

∂y = 3xy²+ 1 and ∂N

∂x = 2(2xy²+ 1) 1

and hence 1 M

∂N

∂x − ∂M

∂y

= 1

xy³+y 4xy²+ 2−3xy²−1

= 1

xy³+y xy² + 1

= 1

y (10)

Thus the integrating factor is e^Rf(y)dy =e

R ₁

ydy

= e^logy =y. Multiplying (9) by y, we obtain the following exact differential equation

(xy⁴+y²)dx+ 2(x²y³+xy+y⁵)dy = 0 (11) Now,

M₁ = (xy⁴ +y²) and N₁ = 2(x²y³+xy+y⁵) and hence

Z

y=constantM₁dx= 1

2xy⁴+y²x

and Z

terms not having xN₁dy= 2 6y⁶. Thus the required solution is

1

2xy⁴ +y²x+ 2

6y⁶ =C or,

3xy⁴+ 6y²x+ 2y⁶ = 6C.

Home Assignment:

1. (Q.54/P.61) Solve (2x²y−3y²)dx+ (2x³−12xy+ logy)dy = 0.

[Ans: 6x³y³−27xy⁴−3y³logy−y³ =C.]

2. (Q.55/P.61) Solve (3x²y⁴+ 2xy)dx+ (2x³y³−x²)dy = 0.

[Ans:x³y³ +x² =Cy]

3

Rule V. If the differential equation (8) is of the form

x^ay^b(mydx+nxdy) +x^cy^d(pydx+qxdy) = 0,

where a, b, c, d, m, n, p and q are constants, then x^hy^k is the integrating factor of the given differential equation, where h, k are constants and can be obtained by applying the condition that after multiplication by x^hy^k the given equation is exact.

Example 2.43/P.54 Solve

(y²+ 2x²y)dx+ (2x³−xy)dy= 0. (12) Solution. Equation (12) can also be written as

y(y+ 2x²)dx+x(2x²−y)dy= 0

Let x^hy^k be the integrating factor. Multiplying (12) by this integrating factor, we have

(x^hy^k+2+ 2x^h+2y^k+1)dx+ (2x^h+2y^k−x^h+1y^k+1)dy= 0 (13) Now,

M =x^hy^k+2+ 2x^h+2y^k+1 and

N = 2x^h+2y^k−x^h+1y^k+1 If (13) is exact, then ∂M/∂y=∂N/∂x, that is,

(k+ 2)x^hy^k+1+ 2(k+ 1)x^h+2y^k= 2(h+ 3)x^h+2y^k−(h+ 1)x^hy^k+1 Equating the coefficients of x^hy^k+1 and x^h+2y^k on both sides and solving, we have

h=−5

2 and k =−1 2 Thus the integrating factor is

x^hy^k=x⁻⁵²y⁻¹²

Multiplying (12) by this integrating factor, we have

(x⁻⁵²y³² + 2x⁻¹²y¹²)dx+ (2x¹²y⁻¹² −x⁻³²y¹²)dy= 0 From this equation, we have

M₁ =x⁻⁵²y³² + 2x⁻¹²y¹² and

N₁ = 2x¹²y⁻¹² −x⁻³²y¹² Since the equation is exact, we have

Z

y=constantM₁dx=−2

3x⁻³²y³² + 4x¹²y¹²

and Z

terms not having xN₁dy= 0 Therefore the required solution is

Z

M₁dx+ Z

N₁dy=C or,

−2

3x⁻³²y³² + 4x¹²y¹² =C Home Assignment:

1. (Q.57/P.61) Solve (3x+ 2y²)ydx+ 2x(2x+ 3y²)dy = 0.

[Ans:x³y⁴ +x²y⁶ =C.]

2. (Q.60/P.61) Solve (2x²y−3y⁴)dx+ (3x³+ 2xy³)dy= 0.

[Ans: 12x^−10/13y^15/13+ 5x^−36/13y^24/13=C]

Change of Variables: If we are unable to apply the rules/methods, which we have already discussed to solve a given differential equation, then we shall reduce the given differential equation by suitable substitution. This procedure

5 of reducing the given differential equation by substitution is called the change of dependent / independent variable.

Example 2.45/P.55 Solve

xdx+ydy= a²(xdy−ydx)

x²+y² (14)

Solution. Let x =rcosθ and y =rsinθ, so r² = x²+y² and tanθ =y/x.

Differentiating the above relations, we have

dx=−rsinθdθ, dy=rcosθdθ and rdr =xdx+ydy.

From (14), we obtain

rdr = a²

r² r²cos²θdθ+r²sin²θdθ

= a² r²r²dθ

=a²dθ (15)

On integration, we have

r² = 2a²θ+C or,

x²+y² = 2a²tan⁻¹ y x +C which is the required solution.

Home Assignment:

1. (Q.2.46/P.56) Solve sec²y ^dy_dx

+ 2xtany=x³. [Ans: tany= ¹₂(x²−1) +Ce^−x2]

Orthogonal Trajectories (A topic of Unit I)

A family of curves whose members cut every member of another family of curves at right angles is called an orthogonal trajectories of second family of curves.

Formation of Orthogonal trajectories in Cartesian Coordinates: Let the given family of curves is represented by the equation

F(x, y, c) = 0 (16)

wherec is a parameter.

We know that differentiating equation (16) w. r. toxand then eliminating c with the help of (16), we get a first order ODE of the form

dy

dx =f(x, y) (17)

Equation (17) is called the differential equation of the family (16).

As ^dy_dx represents the slop of a curve of the family (16), therefore the slope of the curve perpendicular to this curve will be −_dy/dx¹ . Thus the equation of the orthogonal trajectories is

− 1

dy/dx =f(x, y) or dy

dx =− 1

f(x, y) (18)

1

2

Solving Differential equation (18), we obtain the orthogonal trajectories of the family (16) as desired.

Example 4.48/p.136 Find the orthogonal trajectories of x²+y² =cx.

Solution. The given family of curves is

x²+y² =cx (19)

Differentiating (19) w.r. to x, we get

2x+ 2ydy

dx =c (20)

Eliminating c between (19) and (20), we get

x²+y² =

2x+ 2ydy dx

x x²+y² = 2x²+ 2xydy

dx y²−x² = 2xydy

dx

⇒ dy

dx = y²−x²

2xy =f(x, y)

which is the differential equation of family (19). Hence Differential equation of orthogonal trajectories is

dy

dx =− 1

f(x, y) = 2xy

x²−y² (21)

which is a first order differential equation. Puttingy =vxso that ^dy_dx =v+x^dv_dx

in (21), we get

v +xdv

dx = 2x(vx) x²−(vx)² v +xdv

dx = 2v 1−v² xdv

dx = 2v 1−v² −v xdv

dx = 2v−v(1−v²)

1−v² = 2v−v+v³ 1−v² xdv

dx = v+v³ 1−v² xdv

dx = v(1 +v²) 1−v² 1−v²

v(1 +v²)dv = dx x [(1 +v²)−2v²]

v(1 +v²) dv = dx x dv

v − 2v

1 +v²dv = dx x On integrating, we get

logv−log(1 +v²) = logx+ loga v

1 +v² = ax Putting v = ^y_x in above equation, we get

y/x

1 + (y/x)² = ax

⇒ yx

x²+y² = ax

⇒ x²+y² = 1 ay or

x²+y² =by,

where ¹_a =b, which is the desired orthogonal trajectories.

4

Example 4.49/p.137 Find the orthogonal trajectories of the family y = x+ce^−x and determine that particular member of each family that passes through (0,3).

Solution. The given family of curves is

y=x+ce^−x (22)

Differentiating (22), we get dy

dx = 1−ce^−x (23)

From (22), ce^−x =y−x. Putting this value in (23), we get dy

dx = 1−(y−x) = 1 +x−y

which is Differential equation of family (22). Thus Differential equation of Orthogonal trajectories is

dy

dx = − 1

1 +x−y

⇒ dx

dy = −1−x+y

⇒ dx

dy +x = y−1 (24)

Equation (24) is linear differential equation in which x is dependent variable and y is independent variable. Hence

I.F.=e^{RP dy} =e^R1dy =e^y

The general solution is

x×I.F. = Z

(Q×I.F.)dy+a xe^y =

Z

((y−1)×e^y)dy+a

= Z

ye^ydy− Z

e^ydy+a

= (ye^y −e^y)−e^y +a xe^y = ye^y −2e^y+a

(x−y+ 2)e^y = a, (25)

which is orthogonal trajectories of (22). This orthogonal trajectories passes through the point (0,3). So, putting x= 0 and y= 3 in (25), we get

(x−y+ 2)e^y = a (0−3 + 2)e³ = a

⇒ a = −e³ Hence, (25) becomes

(x−y+ 2)e^y = a (x−y+ 2)e^y = −e³

(x−y+ 2) = −e^3−y or

x−y+ 2 +e^3−y = 0

which is the required curve passing through the point (0,3).

Home Assignment

1. (Q.48/P-157) Find the members of the orthogonal trajectories for x+y=ce^y which passes through (0,5).

[Ans : y=2−x+3e^−x]

6

2. (Que 49/P-157) Find the orthogonal trajectories of the following family of curves:

• c₁x²+y² = 1

[Ans: 2log|y|=x²+y²+c₂]

• 4y+x²+ 1 +c1e^2y = 0

[Ans: y= ¹₄ − ¹₆x²+c₂x⁻⁴]

Orthogonal Trajectories in Polar Coordinates (A topic of Unit I)

Consider the equation of the family of curves in polar form:

F(r, θ, c) = 0 (26)

wherec is a parameter. Differential equation of family (26) is of the form:

f

r, θ,dr dθ

= 0 (27)

Replacing ^dr_dθ by−r^2dθ_dr in (27), we obtain the differential equation of orthogonal trajectories of the family (26) given by

f

r, θ,−r²dθ dr

= 0 (28)

On solving (28), we get the equation of orthogonal trajectories of (26).

Example 4.51/p.138 Find the equation of the orthogonal trajectory of the family of circles having a polar equation r=f(θ) = 2acosθ.

Solution. The given family of curves is

r= 2acosθ (29)

Differentiating (29) w.r. to θ, we get dr

dθ =−2asinθ (30)

1

2

From (29), we have 2a= _cos^r_θ. Using this, (30) becomes dr

dθ =−2asinθ dr

dθ =− r cosθ

sinθ dr

dθ =−rtanθ which is differential equation of family of (29).

Hence, differential equation of orthogonal trajectories is

−r²dθ

dr = −rtanθ rdθ

dr = tanθ

⇒ dθ

tanθ = dr r

⇒ cosθ

sinθdθ = dr r On integrating, we get

Z cosθ

sinθdθ+c⁰ =

Z dr r

log sinθ+ log 2a = logr, where c⁰ = log 2a

⇒ r= 2asinθ

which is the required equation of orthogonal trajectories.

Example 4.52/p.139 Find the orthogonal trajectories ofr =c₁(1−sinθ).

Solution. The given family of curves is

r=c₁(1−sinθ) (31)

Differentiating (31) w.r. to θ, we get dr

dθ =−c₁cosθ (32)

Dividing (32) by (31) (to eliminate c), we have 1

r dr

dθ =− cosθ 1−sinθ

⇒ dr

dθ =− rcosθ 1−sinθ which is differential equation of family of (31).

Hence, the differential equation of orthogonal trajectories is

−r²dθ

dr = − rcosθ 1−sinθ

⇒ rdθ

dr = cosθ 1−sinθ

⇒ dr

r = 1−sinθ cosθ dθ

⇒ dr

r = (secθ−tanθ)dθ On integrating, we get

logr = log(secθ+ tanθ) + log cosθ+ loga logr = log[(secθ+ tanθ) cosθa]

⇒ logr = log[a(1 + sinθ)]

r=a(1 + sinθ)

which is the required equation of orthogonal trajectories of family (31).

Home Assignment

Find the orthogonal trajectories of the following family of curves:

1. (Q.) r² =csin 2θ [Ans : r² =acos2θ]

2. (Q.) r=c(secθ+ tanθ) [Ans : r=ae^−sinθ]

Lecture No. 08

Nonlinear ODEs solvable for p (A topic of Unit I)

Differential Equations of first order but not of the first degree: The general form of differential equation of first order and n^th(n > 1) degree, is

dy dx

n

+a₁ dy

dx ⁿ⁻¹

+a₂ dy

dx ⁿ⁻²

+· · ·+an−1

dy dx

+a_n = 0 (33) or pⁿ+a₁pⁿ⁻¹+a₂pⁿ⁻²+· · ·+a_n−1p+a_n = 0

where,p= ^dy_dx and a_a, a₂, cdots, an−1 and a_n are the functions of x and y.

The equation (33) can be written as f(x, y, p) = 0

Solution of the Differential Equation (33): We have the following two cases:

1. Differential Equation (33) can be written as product of first degree factors.

2. It can not be resolved into factors of first degree.

Case I: Equations solvable for p: In this case, Differential equation (33) is of the form

[p−f₁(x, y)][p−f₂(x, y)][p−f₃(x, y)]· · ·[p−f_n(x, y)] = 0. (34) 1

Now, equating each factor to zero,

p−f_i(x, y) = 0, i= 1,2, cdots, n Let the solutions of these factors be

φ(x, y, c_i) = 0, i= 1,2,· · · , n

Taking,c₁ =c₂ =c₃ =· · ·=c_n =c, hence required solution is φ₁(x, y, c)φ₂(x, y, c)· · ·φ_n(x, y, c) = 0

Remark. Since the differential equation (33) is of first order, therefore its solution contains only one arbitrary constant.

Example 3.1/p.64 Solve (p−xy)(p−x²)(p−y²) = 0.

Solution. Here

(p−xy) = 0 p = xy dy

dx = xy dy

y = xdx dy

y −xdx = 0 on integrating, we have

logy−x²

2 −c₁ = 0 And,

(p−x²) = 0 dy

dx −x² = 0 dy−x² dx = 0

⇒ y− x³

3 −c₂ = 0,

3 Also, from the third factor, we have

(p−y²) = 0 dy

dx −y² = 0 dy

y² −dx = 0

⇒ − 1

y−x−c₃ = 0

or 1

y +x+c₃ = 0

Taking, c₁ =c₂ =c₃ =c, hence solution of given differential equation is

logy− x²

2 −c y− x³

3 −c 1

y +x+c

= 0

Example 3.4/p.65 Solve x²p²+xyp−6y² = 0 Solution. Solving the given equation for p, we have

p= 2y

x , − 3y x If p= ^2y_x, then

logy = 2 logx+ logc₁ or y = c1x²

and if p=−^3y_x, then

yx³ = c₂ Therefore, the required solution is

(y−cx²)(yx³−c) = 0

Example 3.5/p.65 Solvexy²(p²+ 2) = 2py³+x³ Solution. The given equation can be written as

(yp−x)[xyp+ (x²−2y²)] = 0 If yp−x= 0, then integration yields

y² −x² =c₁ If xyp+x²−y² = 0, then

xyp+x² −y² = 0 xydy

dx +x²−2y² = 0 ydy

dx +x− 2y²

x = 0

2ydy dx − 4y²

x = −2x dv

dx − 4

xv = −2x, where v =y² ⇒ dv

dx = 2ydy dx This is a linear differential equation in v and its solution is

v

x⁴ = c₂+ 1 x² y² = c₂x⁴+x² Thus, the required solution is

(y²−x²−c)(y²−cx⁴−x²) = 0

Home Assignment

1. (Ex. 3.2/p.64)Solve (p+y+x)(xp+y+x)(p+ 2x) = 0 [Ans : (1−x−y−ce^−x)(2xy+x²−c)(y+x²−c) =0]

5 2. (Ex. 3.6/p.65) Solve x²p³+y(1 +x²y)p²+y³p= 0

[Ans : (y−c)(ye^−x1 −c)(xy+cy−1) = 0]

3. (Q. 1/p.73) Solve p²−7p+ 12 = 0 [Ans : (y−4x−c)(y−3x−c) =0]

4. (Q. 3/p.73) Solve p³+ 2xp² −y²p²−2xy²p= 0 [Ans : (y−c)(y+x²−c)(xy+cy+1) =0]

Nonlinear ODEs solvable for y and x (A topic of Unit I)

Case II:- If the differential equation f(x, y, p) = 0 (first order but not first degree) can not be resolved into linear factors in p.

In this case we will discuss the following two cases:

1. Differential equation solvable fory.

2. Differential equation solvable forx.

(a) Differential equation solvable for y:- In this case the differential equation

f(x, y, p) = 0 (35)

can be re-written as

y=g(x, p) (36)

Now differentiate with respect to x, we get dy

dx =p=φ

x, p,dp dx

(37) And suppose that the solution of Equation (37) is

F(x, p, c) = 0 (38)

Then, the elimination of pbetween (36) and (38) give the required solution.

If this elimination is not possible, then equation (36) and (38) together give the required solution.

1

2

Example 3.7 / p.67Solve y+px=x⁴p² Solution. Here

y=x⁴p²−px (39)

D.w.r. to x, we get

dy

dx =p = 4x³p²+ 2x⁴pdp

dx −p−xdp dx

⇒ (4x³p²−2p) +

2x⁴pdp

dx −xdp dx

= 0

⇒ (2x³p−1)

xdp dx + 2p

= 0 xdp

dx + 2p = 0, (We take the factor which contains derivative terms) dp

p + 2dx

x = 0 Integrating it, we get

logp+ 2 logx = logc

⇒ p = c x² put this value in equation (39),

y = x⁴p² −px

= x⁴ c

x² 2

− c x²x y = c²− c

x i.e.,

xy+c=c²x which is the required solution.

Example 3.8 / p.67Solve y= sinp−pcosp.

Solution. Given differential equation is

y= sinp−pcosp (40)

Differentiate w. r. to x, we get dy

dx =p = cospdp dx −

−psinpdp

dx + cospdp dx

or p = psinpdp dx p

sinpdp dx −1

= 0

⇒ sinpdp

dx −1 = 0 sinpdp=dx

On integrating, we have

−cosp+c = x

cosp = c−x (41)

From the given differential equation (40) p = sinp−y

cosp =

p1−cos²p−y cosp

=

p1−(c−x)²−y

c−x , (using(41))

=

√1−c²−x²+ 2cx−y

c−x ,

Put this value in equation (41), we have

c−x= cos √

1−c²−x²+ 2cx−y c−x

which is the required solution.

4

Home Assignment 1. (Q.8/p.74)Solve y= 2px+ tan⁻¹(xp²)

[Ans : y=2c√

x+ tan⁻¹c²] 2. (Ex. 3./p.xx)Solve f

[Ans : ]

(b) Differential equation solvable for x:- If the differential equation

f(x, y, p) = 0 (42)

is solvable forx, then we can write it as

x=h(y, p) (43)

Differentiate with respect toy, we get dx

dy = 1 p =ψ

y, p,dp dy

(44) If the solution of Equation (44) is

G(y, p, c) = 0 (45)

Then, the elimination of pfrom (43) and (45), we get the required solution.

If this elimination is not possible, then equation (43) and (45) together give the required solution.

Example 3.10 / p.68Solve y²logy=xyp+p² Solution. Given differential equation is

y²logy = xyp+p², (46)

or x = 1

pylogy− p

y (47)

Differentiate w. r. to y, we get dx dy = 1

p = − 1 p²

dp

dy(ylogy) + 1 p

1·logy+y· 1 y

− 1 y

dp dy + p

y² 1

p = − 1 p²

dp

dy ·ylogy+1

plogy+1 p − 1

y dp dy + p

y²

1− y p

dp dy

1

plogy+ p y²

1− y

p dp dy

= 0

1−y p

dp dy

logy p + p

y²

= 0 1− y

p · dp

dy = 0

⇒ dp

p = dy dy On integrating, we get

logp = logy+ logc

⇒ p = yc Put this value in (46), we have

y²logy = xyp+p², y²logy = xy(yc) + (yc)² Hence, solution is

logy =xc+c²

Example 3.11 / p.69 Solve xp³ =a+bp Solution. Here,

x= a p³ + b

p² (48)

6

Differentiate w.r. to y, we get dx dy = 1

p = −3a p⁴

dp dy +

−2b p³

dp dy

⇒ 1 = − 3a

p + 2b 1

p² dp dy dy+ 1

p² 3a

p + 2b

dp = 0 On integrating

y+

−3a 2p²

+

2b

−p

= c y = 3a

2p² +2b

p +c (49)

Here, it is not easy to eliminate p from (48) and (49). Hence, the equations (48) and (49) both constitute the solution.

Home Assignment 1. (Ex. 3.12/p.69)Solve x=y+alogp= 0

[Ans : x=c−alog(1−p) +alogp, y=c−alog(1−p)]

2. (Ex. 3./p.xx)Solve f [Ans : ]

Clairaut and Lagrange Equation (A topic of Unit I)

The most general form of a differential equation of the first order but not of the first degree (say n^th degree) is

dy dx

n

+P₁ dy

dx n−1

+P₂ dy

dx n−2

+· · ·+Pn−1

dy dx

+P_n= 0 (50) or pⁿ+P₁pⁿ⁻¹+P₂pⁿ⁻²+· · ·+Pn−1p+P_n= 0 where p= ^dy_dx and P₁, P₂,· · ·, P_n are functions ofx and y. This equation can also be written as

F(x, y, p) = 0. (51)

The above equation however can not be solved in this general form.

The Clairaut Equation

When the given equation (51) is of the first degree in x and y, and of the form

y=px+f(p) (52)

Equation (52) is known as Clairaut’s equation. To solve it, we differentiate with respect tox to obtain

p= dy

dx = [x+f⁰(p)]p⁰+p or [x+f⁰(p)]dp

dx = 0 (53)

1

2

Ifdp/dx= 0, thenp= c= constant. Eliminatingpbetween this and equation (52), we get

y=cx+f(c) (54)

which is required solution of Clairaut’s equation.

Remark 3. Sometimes by a suitable substitution, an equation can be reduced to Clairaut’s form.

Example (3.13/P-71)Solve (y−px)(p−1) =p.

Solution. The given equation can be written as y =xp+ p

p−1, which is a Clairaut equation. Hence the solution is

y = xc+ c c−1, y = xc(c−1) +c

(c−1) y(c−1) = xc(c−1) +c y(c−1)−xc(c−1) = c

i.e., which is the required solution.

Example (3.14/P-71)Solve p= log(px−y) Solution. The given equation is

y=px−e^p.

Which is a Clairaut equation. Hence the solution is y = cx−e^c e^c = cx−y, i.e.,

c= log(cx−y)

Example (3.17/P-73) Solve (px−y)(py+x) =h²p.

Solution. This equation can be written as

p²xy+px²−py²−xy = h²p p²xy−xy+px²−py²−h²p = 0

p²xy−xy+p(x²−y²−h²) = 0,

Putting x² =u⇒2xdx=du and y² =v ⇒2ydy =dv, and p= dy dx = x

y dv du the given equation takes the form

p²xy−xy+p(x²−y²−h²) = 0 x

y dv dv

2

xy−xy+x y

dv

du(u−v−h²) = 0 x²

y dv

dv 2

−y+1 y

dv

du(u−v−h²) = 0 x²

dv dv

2

−y²+ dv

du(u−v−h²) = 0 u

dv dv

2

−v+ dv

du(u−v−h²) = 0

or uP²+ (u−v−h²)P −v = 0, where P = dv du

or v = uP − h²P

P + 1 which is of Clairaut‘s form and has the solution as

v =uc− h²c c+ 1 where u=x² and v =y². i.e.,

y² =x²c− h²c c+ 1.

4

Home Assignment 1. (Example 3.18/P-73)Solve y= 2px+y²p³.

[Ans : y² =cx+¹₈c³]

2. (Que 21/P-74)sinpxcosy= cospxsiny+p, [Ans : y=cx−sin⁻¹c]

3. (Que 22/P-74)xy(y−px) = x+py, [Ans : y² =cx²+ (1+c)]

4. (Que 24/P-74) Solve x²p²+yp(2x+y) +y² = 0 by reducing it to Clairaut’s form by using the substitutiony =u and xy=v.

[Ans : xy =cy+c]

[RB1] The Lagrange Equation An equation of the form

y=xf₁(p) +f₂(p) (55)

is known as Lagrange’s Equation. To solve it, we differentiate with respect to xto obtain

p= dy

dx = f₁(p) +xf₁⁰(p)dp

dx +f₂⁰(p)dp dx p−f₁(p) = dp

dx[xf₁⁰(p) +f₂⁰(p)]

dx

dp = xf₁⁰(p) +f₂⁰(p) p−f₁(p)

or dx

dp − f₁⁰(p)

p−f₁(p)x = f₂⁰(p)

p−f₁(p) (56)

which is a linear equation in x, and hence can be solved in the form

x=φ(p, c) (57)

eliminatingp from equations (55) and (57), we get the required solution. If it is not possible to eliminatep, then the values of x andy in terms of pcan be found from equations (55) and (57), and these will constitute the required solution.

Note. The Clairaut Equation is the particular case of the Lagrange Equation.

Example 2/ P.46 Solve the equation y=xp²−_p¹.

Solution. This equation is Lagrange equation. Differentiating it w.r.t. x, we get

dy

dx =p = x

2pdp dx

+p²(1)−

−1 p²

dp dx p−p² =

2xp+ 1 p²

dp dx dx

dp = 2xp+ _p¹2

p−p² dx

dp = 2xp

p−p² + 1 p²(p−p²)

⇒ dx

dp + 2x

p−1 = − 1 p³(p−1)

This equation is linear in x [An equation is linear if it is of the form

dy

dx +P y =Q]. Therefore

I.F.=e

R ₂

p−1dp

=e^log(p−1)2 = (p−1)²

6

On integrating, we get

x.(I.F.) = Z

(I.F.)Qdp x(p−1)² =

Z

(p−1)²

− 1 p³(p−1)

dp x(p−1)² = −

Z p−1

p³

dp x(p−1)² = −

Z 1 p² − 1

p³

dp x(p−1)² = c₁+ 1

p− 1 2p² x(p−1)² = 2c₁p² + 2p−1

2p² x = 2c₁p² + 2p−1

2p²(p−1)² x = cp²+ 2p−1

2p²(p−1)² (58)

Substituting this value of x in the given equation, we get y = xp²−1

p y = cp²+ 2p−1

2(p−1)² −1

p (59)

Equations (58) and (59) together give the required solution. Hence,

x= cp²+ 2p−1

2p²(p−1)² , y= cp²+ 2p−1 2(p−1)² −1

p

Example 4 /P.47 Solve the equationy = 2px+pⁿ.

Solution. This equation is Lagrange equation. Differentiating it w.r.t. x, we

get

dy

dx =p = 2p.(1) +x.

2dp

dx

+npⁿ⁻¹dp dx p = 2p+ (2x+npⁿ⁻¹)dp

dx

−p = (2x+npⁿ⁻¹)dp dx dx

dp = 2x+npⁿ⁻¹

−p

⇒ dx dp + 2

px = −npⁿ⁻² This equation is linear in x. Therefore

I.F.=e^R(2/p)dp =e^{2 logp} =e^logp2 =p² On integrating, we get

x.(I.F.) = Z

(I.F.)Qdp x.p² =

Z

p².(−npⁿ⁻²) dp x.p² = −

Z

npⁿdp xp² = − n

n+ 1pⁿ⁺¹+c

⇒ x = c

p² − n

n+ 1pⁿ⁻¹ (60)

Substituting this value ofx in the given equation, we get y = 2px+pⁿ

y = 2p c

p² − n n+ 1pⁿ⁻¹

+pⁿ y = 2c

p − 2n

n+ 1pⁿ+pⁿ y = 2c

p − (n−1)

(n+ 1)pⁿ (61)

8

Equation (60) and (61) together give the required solution. Hence, x= c

p² − n

n+ 1pⁿ⁻¹, y= 2c

p −(n−1) (n+ 1)pⁿ

Home Assignment 1. (Ex 7/ p-48) Solve y=x(1 +p) +p²

[Ans : x=2(1−p) +ce^−p, y={2(1−p) +ce^−p}(1+p) +p²] 2. (M Ex.8/ p-48) Solve y=xp²+p

[Ans : x= (logp−p+c)(p−1)⁻², y=xp²+p]

3. (M Ex.10/ p-48) Solvey = ³₂xp+e^p [Ans : x= _p^c3 −2e^p

1

p − _p²2 + _p²3

, y= _2p^3c2 −2e^p

1− _p³ + _p³2

]

The Singular Solution of a First order ODE

(A topic of Unit I)

Consider the ordinary differential equation

f(x, y, p) = 0, where p= dy

dx. (62)

In Lecture 1, we have already discussed the two types of solution of an ordinary differential equation namely: general solution and particular solution.

In case of differential equation (62), its general solution will contain an arbitrary constantc, and therefore the general solution of differential equation (62) is of the form

g(x, y, c) = 0. (63)

Recall that, any particular solution of equation (62) can be obtained from (63) by substituting the suitable values of c.

Besides general solution and particular solution, there is another type of solution of equation (62), which is called Singular solution.

Definition 1. A singular solution of equation (62) is a solution, which is free from arbitrary constants but not a particular solution of the equation.

1The Content of this lecture is based on the reference book

”Frank Ayres, Jr: Theory and Problems of Differential Equations (Schaum‘s Outline Series)”

1