First Order Differential Equations

Course No. : MATH GC241 Course Title : MATHEMATICS-III Instructor-In-Charge: Dr. P. Dhanumjaya

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.1/97

Instructors Dr. Ranadev Datta Dr. Prashanth Das Mr. Harish V. N. Mrs. Jessica Pereira

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.2/97

Text Book Simmons G. F., Differential Equations with Applications and Historical Notes, MH, 2nd Edn., 1991.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.3/97

Reference Books Erwin Kreyszig, Advanced Engineering Mathematics, John Wiley & Sons, 8th Edn. William E Boyce and R C Diprima, Elementary Differential Equations and Boundary Value Problems, John Wiley & Sons, 7th Edn, 1991. E A Coddington, An introduction to Ordinary Differential Equations, Prentice Hall, 1961.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.4/97

Contents Introduction to first order Differential Equations and their Solution Techniques Introduction to Second order Differential Equations and their Solution Techniques Power Series solutions to second order DE’s with variable coefficients Introduction to System of Differential Equations

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.5/97

Introduction to Laplace Transform Introduction to Fourier Series Introduction to Partial Differential Equations (i) Hyperbolic Equation (Wave Equation) (ii) Parabolic Equation (Heat Equation) (iii) Elliptic Equation (Laplace Equation)

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.6/97

Evaluation Scheme

______________________________________ E.C Duration Weightage(%) ______________________________________ Test-1 1 hour 25 ______________________________________ Test-2 1 hour 25 ______________________________________ Tut Test/Quiz ***** 10 ______________________________________ Comp. Exam 3 hour 40 ______________________________________

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.7/97

Differential Eqns What is Differential Equation?. A differential equation is an equation that contains a derivative (or derivatives) of an unknown function.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.8/97

Differential Eqns DE’S

ODE

PDE

Elliptic

Hyperbolic Parabolic

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.9/97

ODE A differential equation is said to be ordinary differential equation if all the derivatives are with respect to a single independent variable. Example: dy + t2 y = sin t dt dy d3 y + 5t + 10y = 0. 3 dt dt

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.10/97

PDE A differential equation is said to be partial differential equation if there are derivatives with respect to two or more independent variables. Example: ∂y ∂y +c = 0 ∂t ∂x ∂ 2u ∂ 2u + 2 = 0. 2 ∂x ∂y

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.11/97

Order of a Diff. Eqn The order of a differential equation is the highest order derivative involving in the differential equation. Example: 2

dy dy + 2 dt dt The order is 2.

!3

+ y = 0.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.12/97

Degree of a Diff. Eqn The degree of a differential equation is the highest power of highest order derivative involving in the differential equation. Example: 2

dy dy + 2 dt dt The degree is 1.

!3

+ y = 0.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.13/97

Applications Newton’s Law of cooling : The temperature of a object T (t) changes according to the temperature of it’s environment. It is found by experiment that dT = T − Ta , dt where Ta is the ambient temperature of the environment. cold

T=Ta

T = T(t) hot

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.14/97

Applications Electrical Circuits : In a standard RLC circuit the charge on the capacitor at time t is q(t) which is related to the current i(t) by i = dq dt . In the circuit below the voltage drops across the various components becomes dq 1 d2 q L 2 + R + q = E(t). dt dt C R

E

L C

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.15/97

Applications Falling Bodies : The position of a particle is denoted by s(t) with the velocity and acceleration given by dv d2 s ds = 2. v= , a= dt dt dt A body falling under gravity alone will have the equation d2 s = −g 2 dt where g = 9.8 m/s2 .

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.16/97

Applications Springs, Buildings and Cars suspension : Hooke found that the force on a beam (or spring) is proportional to the amount of displacement. This applies to Springs, Building supports, and Car suspensions. If x(t) is the displacement then Hooke’s law states d2 x m 2 = −kx. dt

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.17/97

Linear and nonlinear DE’s : A differential equation written as D(y(t)) = 0 is linear if D(y1 + y2 ) = D(y1 ) + D(y2 ). Note 1. A differential equation is linear if all the derivative terms and the unknown function y are not raised to any power or part of a function.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.18/97

Example 1 d2 y +y =0 2 dt is linear. d2 y 3 + ty = t dt2 is linear. d2 y 2 +y =0 2 dt is nonlinear.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.19/97

Homogeneous and Nonhomogeneous DE’s : A differential equation is said to be homogeneous if D(y(t)) = 0 whereas it is inhomogeneous if D(y(t)) = f (t).

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.20/97

Example 2 d2 y +y =0 2 dt is homogeneous d2 y 3 + y = t dt2 is inhomogeneous dy + xy 2 = sin x dx is inhomogeneous.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.21/97

Number of Arbitrary Constants Consider the differential equation dy = t. dt The solution is t2 y(t) = + C, 2 where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.22/97

Similarly, we take d2 y =t 2 dt then

dy t2 = + C1 , dt 2 and by integrating again, we get t3 y(t) = + C1 t + C2 , 6

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.23/97

where C1 and C2 are two different arbitrary constants. Note 2. The number of arbitrary constants in the solution of a differential equation is the same as the order of the differential equation. For a physical example, the constants are found from the various conditions imposed by the problem.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.24/97

Example 3 A falling body problem where an object was dropped from a height of 10 m will be d2 s ds m 2 = mg − k , dt dt s(0) = 10, s′ (0) = 0, which means that at time t = 0 (initially) the height was 10 m and the velocity was zero. This is called an initial value problem (IVP).

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.25/97

Example 4 The temperature in a long beam, which would have equation d2 T = 0, 2 dx T (0) = 100, T (1) = 0. This corresponds to the temperature at one end x = 0 is 100 degrees and at the other end x = 1 is zero. This is called a boundary value problem (BVP).

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.26/97

General Form The general form of nth order ordinary differential equation is n 2 dy d y d y F x, y, , 2 , · · · , n  = 0, dx dx dx 



(or) using the prime notation for derivatives, we rewrite as F x, y, y ′ , y ′′ , · · · , y (n) = 0. 



Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.27/97

First order ODE The general first order ODE can be written as dy F x, y, = 0. dx !

We normally expect that the above first order ODE will have a solution and solution contain one arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.28/97

Example We consider the following examples: dy dx

!2

+ 1 = 0,

has no real-valued solutions at all. dy dx

!2

+ y 2 = 0,

has only the solution y = 0, which contains no arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.29/97

Situations of this kind raise difficult theoretical questions about the existence and nature of solutions of differential equations.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.30/97

Existence of a Solution We consider the first order ODE (1)

dy = f (x, y). dx

We assume that f (x, y) is a continuous function throughout some rectangle R in the xy-plane. The geometric meaning of a solution of (1) as follows:

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.31/97

If P0 = (x0 , y0 ) is a point in R then the number dy dx

!

= f (x0 , y0 )

P0

determines a direction at P0 . Now let P1 = (x1 , y1 ) be a point near to P0 in this direction, then dy dx

!

= f (x1 , y1 )

P1

to determine a new direction at P1 .

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.32/97

If we continue like this, we obtain a broken line. y

.P0

.P

.P

2

. R

1 x

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.33/97

If we assume that the successive points move closer to one another then the broken line approaches a smooth curve through the initial point P0 . This curve is a solution y = y(x) of equation (1).

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.34/97

Note that the different initial point produce a different curve. Thus, the solution of (1) form a family of curves called integral curves.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.35/97

Separable Equations A differential equation which can be written in the form M (x) dx + N (y) dy = 0, where M is a function of x alone and N is a function of y alone, is said to be separable. The solution is Z

M (x) dx +

Z

N (y) dy = C,

where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.36/97

Example 1 Find the solution of the equation dy y = = ex+y . dx ′

Solution.

We write dy = ex ey . dx

Now separating the variables, we obtain e−y (dy/dx) = ex .

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.37/97

Integrating with respect to x and then replacing (dy/dx) dx by dy yields Z

e−y dy =

Z

ex dx.

Thus, we get the general solution −e−y = ex + C, where C is any arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.38/97

Example Solve dy = ex y 3 − xex y 3 , dx y(0) = −1. The equation is separable and the solution is Solution.

1 −2 − y = (2 − x)ex + C. 2 Substituting the initial condition, we get C = − 25 .

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.39/97

Therefore, the solution of the given problem is 1 y = . x 5 − 2(2 − x)e 2

This gives 1 y = −q . 5 − 2(2 − x)ex

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.40/97

Example We consider the differential equation dy 1 . = y dt 1+e This equation can be written in separable form as Solution.

dy (1 + e ) = 1. dt y

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.41/97

The general solution for the given differential equation is y + ey = t + C, where C is an arbitrary constant. Note that, we have obtained implicit solution for a given differential equation.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.42/97

Orthogonal Trajectories Two families of curves are said to be mutually orthogonal if the product of the slopes is −1 at every point of intersection between curves of each family. Orthogonal trajectories can be determined from differential equations.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.43/97

Example We consider the family of curves y = Cx. Differentiating with respect to x, we get dy = C. dx By eleminating C, we obtain dy y = . dx x

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.44/97

Curves orthogonal to these curves must satisfy the differential equation x dy =− . dx y Solving, we get x2 + y 2 = K. Thus, the family of concentric circles centered at the origin is orthogonal to the family of stright lines through the origin.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.45/97

y

−3

−2

−1

1

2

3

x

Orthogonal Trajectories

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.46/97

How to find the orthogonal trajectories of a given family of curves

First find the differential equation of the family dy Replace dx by − dx dy to obtain the differential equation of the orthogonal trajectories

Solve the new differential equation

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.47/97

Homogeneous Equations An equation of the form dy = f (x, y) dx is said to be homogeneous whenever the function f does not depend on x and y separately, but only on their ratio xy or xy . Thus, homogeneous equations are of the form dy y =f . dx x !

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.48/97

The function f (x, y) is called homogeneous of degree n if f (tx, ty) = tn f (x, y). Example: x x+y 1 √ x + 2y , sin , ln , x + y + xy, y x x+y 2

2

!

are homogeneous of degree 2, 0, 0, 1 and -1, respectively.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.49/97

A differential equation (2)

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

is called homogeneous if M (x, y) and N (x, y) are homogeneous functions of the same degree. Writing the equation (2) in the form dy M (x, y) y =− =F . dx N (x, y) x !

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.50/97

We simplify the homogeneous equation by introducing a new dependent variable v as a function of x to represent ratio of y to x. Thus (3)

y = xv.

The given differential equation becomes dy = F (v). dx Differentiating (3) with respect to x, we obtain

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.51/97

dv dy =v+x . dx dx Using as

dy dx

= F (v), we rewrite the above equation dv F (v) − v = . dx x

(or) dv 1 1 = . (F (v) − v) dx x

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.52/97

Integrating on both sides with respect to x and replacing (dv/dx) dx by dv, we get the solution for v. Finally, we replace v by solution y.

y x

to get the required

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.53/97

Example Find the general solution of x

2 dy

dx

= y 2 + xy + x2 .

We note that the given differential equation is homogeneous differential equation. Solution.

We now substitute y = vx and simplifying, we arrive 1 1 dv = . 2 v + 1 dx x

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.54/97

Integrating on both sides, we obtain tan−1 (v) = ln |x| + C, where C is an arbitrary constant. We replace v = xy to get the required solution for the given differential equation tan

−1

y = ln |x| + C, x !

where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.55/97

Reduction to Separable Form We consider the differential equation of the form dy a1 x + b1 y + c1 =f . dx a2 x + b2 y + c2 !

We made the above equation homogeneous by change of variables. We put x = x1 + h, y = y1 + k.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.56/97

Substituting x and y in the given differential equation and we choose h and k so that (4)

a1 h + b1 k + c1 = 0, a2 h + b2 k + c2 = 0,

we obtain dy1 a1 x1 + b1 y1 =f , dx1 a2 x1 + b2 y1 !

which is homogeneous.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.57/97

The equation (4) can be solved for h and k provided D = a1 b2 − a2 b1 6= 0.

We note that by translating the origin to the point (h, k) we have removed the constant terms in the linear expressions.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.58/97

If D = 0, the first degree terms of numerator and denominator are proportional and the equation falls into the form dy = f (ax + by + c). dx We put z = ax + by + c then dz = a + bf (z). dx We can solve the above equation by using variable seperable method.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.59/97

Example Find the general solution of dy (1 + x − 2y) + (4x − 3y − 6) = 0. dx

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.60/97

Example Find the general solution of (x + y + 1) dx + (2x + 2y − 1) dy = 0. Solution.

Here D = a1 b2 − a2 b1 = 0. We put z =x+y

so that dz dy =1+ . dx dx

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.61/97

Thus dz z+1 =1− dx 2z − 1

(or)

Solving, we get

dz z−2 = . dx 2z − 1

2z + 3 ln |z − 2| = x + C, where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.62/97

Substituting z = x + y, we obtain 2(x + y) + 3 ln |x + y − 2| = x + C, where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.63/97

Exact equations A differential equation M (x, y) dx + N (x, y) dy = 0 is said to be exact if there exists a function, denoted by U (x, y) such that ∂U ∂U dU = dx + dy = M dx + N dy. ∂x ∂y

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.64/97

That is, the equation is the exact differential of a function of x and y then the differential equation takes the form dU = 0, and its solutions are defined implicitly by U (x, y) = C, where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.65/97

Example Solve the differential equation 1 x dx − 2 dy = 0. y y We note that the given equation is exact, since it may be written Solution.

x d =0 y !

and its solution is x = C. y

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.66/97

Note We note that the differential equation 1 x dx − 2 dy = 0, y y can be rewritten in the form y dx − x dy = 0 is not exact.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.67/97

A differential equation M (x, y) dx + N (x, y) dy = 0 is said to be exact if and only if ∂M ∂N = . ∂y ∂x The general solution is U (x, y) = C.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.68/97

Example Test the equation ey dx + (xey + 2y) dy = 0 for exactness and solve it, if it is exact. Solution.

Here M = ey and N = xey + 2y.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.69/97

Then ∂M ∂N y = e , and = ey . ∂y ∂x We have ∂N ∂M = . ∂y ∂x Therefore, the given equation is exact.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.70/97

We have ∂U ∂U y = e and = xey + 2y. ∂x ∂y Integrating, we get U =

Z

ey dx + g(y)

= xey + g(y).

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.71/97

Now differentiating with respect to y, we get ∂U = xey + g ′ (y). ∂y This gives 2

g(y) = y . The solution is xey + y 2 = C where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.72/97

Integrating Factors We note that the equation y dx − x dy = 0 is not exact. But if we multiply by equation

1 y2 ,

we have an exact

1 x x dx − 2 dy = d = 0, y y y !

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.73/97

(or) by

x y3 , 2



2



x x 1x   dx − 3 dy = d = 0, 2 2 y y 2y (or) by

1 xy ,

dx dy − = d (ln |x| − ln |y|) = 0. x y 1 The multipliers y12 , yx3 , xy are called integrating factors of the given equation.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.74/97

Definition A nonzero function µ(x, y) is called integrating factor of the equation M (x, y) dx + N (x, y) dy = 0, if the equation µ(x, y) (M (x, y) dx + N (x, y) dy) = 0 is exact.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.75/97

Note The differential equation M (x, y) dx + N (x, y) dy = 0 has an infinite number of integrating factors. In general case, it is very difficult to find an integrating factor.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.76/97

Integrating Factors We now diccuss some simple cases: (i) If, in the equation M (x, y) dx + N (x, y) dy = 0, we have 1 ∂M ∂N − N ∂y ∂x

!

= f (x)

a function of x alone.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.77/97

Then µ(x) = e is an integrating factor. If

R

f (x) dx

1 ∂N ∂M − M ∂x ∂y

!

= f (y)

is a function of y alone. Then µ(y) = e

R

f (y) dy

is an integrating factor.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.78/97

Example Solve the differential equation 2 sin(y 2 ) dx + xy cos(y 2 ) dy = 0. Solution.

The general solution is x4 sin(y 2 ) = C, U (x, y) = 2

where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.79/97

Example Solve the differential equation (2xy 2 − y) dx + (y 2 + x + y) dy = 0. Solution.

The general solution is x U (x, y) = x − + y + ln |y| = C, y 2

where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.80/97

Linear Equations The general form of a first order linear differential equation is dy A(x) + B(x)y + C(x) = 0. dx On division by the first coefficient, it can be put into the form (5)

dy + P (x)y = Q(x). dx

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.81/97

We note that d e dx

 R

P (x) dx



y =e

We multiply (5) by e

R

R

P (x) dx

P (x) dx

dy + Py . dx !

, we get

R d  R P (x) dx  y = Q e P (x) dx . e dx

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.82/97

Integrating now yields e

R

P (x) dx

y=

Z

Q(x)e

R

P (x) dx

+ C,

(or) y = e−

R

P (x) dx

Z

Q(x)e

R

P (x) dx

+C



is the general solution of the equation (5).

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.83/97

Example Solve the differential equation dy 1 + y = x2 . dx x Solution.

Here P (x) = x1 , Q(x) = x2 .

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.84/97

The integrating factor is e

R

P dx

=e

R

1 x

dx

= x.

The general solution is yx =

Z

x2 (x) dx + C,

or

x3 C y= + , 4 x where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.85/97

Example Find the solution of the initial value problem (IVP) ′

y − 2xy = x, y(0) = 1. Solution.

The solution is 1 3 x2 y=− + e . 2 2

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.86/97

Bernoulli’s Equation The nonlinear differential equation (6)

dy + P (x)y = Q(x)y n , dx

where n is a constant but not necessary an integer, known as Bernoulli’s equation. If n = 0, n = 1, the equation is linear and we know the solution methods.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.87/97

For n ≥ 2, we use the substitution v = y 1−n to reduce Bernoulli’s equation to linear equation. The above equation (6), we rewrite as y

−n dy

dx

+ P (x)y 1−n = Q(x).

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.88/97

Taking v = y 1−n , we get dv −n dy = (1 − n)y dx dx (or) y

−n dy

1 dv = . dx (1 − n) dx

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.89/97

Finally, we get the linear equation of the form 1 dv + P (x)v = Q(x). (1 − n) dx We solve the linear equation for v and we replace v = y 1−n to get the required general solution for the equation (6).

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.90/97

Example Solve the differential equation x2 y ′ + 2xy − y 3 = 0. Solution.

The general solution is y

−2

2 −1 = x + Cx4 , 5

where C is an arbitrary constant.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.91/97

Reduction of Order The general second order differential equation has the form (7)

F (x, y, y ′ , y ′′ ) = 0.

We consider two special types of second order equations that can be solved by first order methods.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.92/97

Dependent Variable missing If y is not explictly present, then equation (7) can be written as F (x, y ′ , y ′′ ) = 0.

(8)

In this case, we introduce a new dependent variable p, by putting dp y = p and y = . dx ′

′′

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.93/97

Then (8) transforms into the first order equation dp F x, p, = 0. dx !

We solve for p and replace p by for y.

dy dx

and we solve

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.94/97

Example Solve the differential equation xy ′′ − y ′ = 3x2 . Solution.

The required general solution is 2 x y = x3 + C1 + C2 , 2

where C1 and C2 are arbitrary constants.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.95/97

Independent Variable missing If x is not explicitly present, then the second order equation can be written F (y, y ′ , y ′′ ) = 0. Let y ′ = p and dp dy dp dp y = = =p . dx dy dx dy ′′

Then dp F y, p, p = 0. dy !

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.96/97

Example Solve y ′′ + k 2 y = 0. Solution.

The general solution is sin

−1

y = ±kx + C2 , C1 !

where C1 and C2 are arbitrary constants.

Course No. : MATH GC241 Course Title : MATHEMATICS-III – p.97/97