Differential Equations

DIFFERENTIAL EQUATIONS A differential equation (Abbreviated D.E.) is one which contains within it at least one derivative. EXAMPLES The following are examples of differential equations: (a) dy/dx = 2x + 5

(d) ∂u/∂t = h2(∂2u/∂x2 + ∂2u/∂y2)

(b) d2s/dt2 – k2s + a = 0

(e) (d2w/dx2)3 + xy(dw/dx) + w = k

(c) (x2 + y2)dx – 2xydy = 0

(f) (d3s/dt3)2 + s(ds/dt) + 2st = 0

(g) y’’ + (y’)2 = y

CLASSIFICATIONS OF DIFFERENTIAL EQUATIONS A differential equation may be classified as to the following characteristics: (a) ORDER: The order of a differential equation is the order of the highest – ordered derivative in the equation. (b) DEGREE: The degree of a differential equation is the largest power or exponent of the highest – ordered derivative present in the equation. (c) TYPE: A differential equation may be ordinary or partial as to the type of derivatives or differentials appearing in the equation that is if it contains ordinary derivatives, it is ordinary differential equation and if the derivative is partial, the equation is a partial differential equation.

EXAMPLES Consider the examples (a) to (g). Tabulated below are the classifications of these 7 differential equations: D.E. (a) (b) (c) (d) (e) (f) (g) REMARK:

ORDER 1 2 1 2 2 3 2

DEGREE 1 1 1 1 3 2 1

TYPE Ordinary Ordinary Ordinary Partial Ordinary Ordinary Ordinary

In the preceding examples, equation (a) is the only one given in the explicit form, equation (c) is the only equation written in terms of differentials, and equation (d) is the only partial differential equation.

GENERAL FORM OF THE NTH – ORDERED ORDINARY EQUATION The following functional notation is to be adopted as the general form of the nth – ordinary differential equation F(dny/dcn, dn-1/dxn-1 , … , dy/dx, x,y) = 0 REMARK: F(dny/dxn,dn-1/dxn-1,…,dy/dx, x,y) = 0 defines any function relating the dependent variable y, the

independent variable x, and the derivative of y relative to x up to and including the nth derivative. SOLUTION TO AN ORDINARY DIFFERENTIAL EQUATION The solution to an ordinary differential equation is another equation free of any derivatives or differentials expressing the functional relationship between the dependent variables and which, substituted into the equation, reduces it to an identity. 1

EXAMPLE The solution to the second – ordered differential equation y’’ = 6x + 4 (1) is anyone of the following equations: y1 = x3 + 2x2 y2 = x3 + 2x2 – 3x + 5

y3 = x3 + 2x2 + ax y4 = x3 + 2x2 + ax + b

where a and b are constants. REMARKS: 1. y2, y3 and y4 are all solutions to (1) since these are all free of any derivatives or differentials and which, when substituted into this equation, reduce it to the identity. 6x + 4 = 6x + 4 2. (1) can actually have an infinite number of solutions since the constants a and b in a solution y3 and y4 can assume an infinite set of values.

TYPES OF SOLUTION TO A DIFFERENTIAL EQUATION The two types of solution to an ordinary differential equation are listed below:

(a) PARTICULAR SOLUTION: The solution to a differential equation is said to be particular if it does not contain any arbitrary constants. REMARK: Solution y1 and y2 are particular solutions to (1) since these do not contain any arbitrary constants. (b) GENERAL SOLUTION: The solution is said to be general if it contains at least one arbitrary constant. REMARK: Solutions y3 and y4 in (1) are general solutions to (1). Moreover, if the general solution to an nth – ordered differential equation contains n arbitrary constants, it is called as the complete solution. Solution y4 is an example of this since there are two (2) arbitrary constants corresponding to the order of (1) which is two (2).

The general form of the complete solution to the nth – ordered ordinary differential equation will contain n arbitrary elements corresponding to its order n and will take the following functional notation G(x,y, C1, C2, … , Cn) = 0

*

where C1, C2, … Cn are the n arbitrary constants. REMARK: The preceding statement can easily be verified by considering the following particular discussion: Given the simple nth – ordered differential equation dny/dxn = k

(k = constant)

n successive integrations of this will yield a solution which contain the n arbitrary elements specified in *.

DETERMINATION OF THE PARTICULAR SOLUTION Since there are n arbitrary constants in the complete solution to an nth – ordered differential equation as stated in the preceding section, there must necessarily be n given initial or boundary conditions for the determination of these constants. With the knowledge of these constants, which are obtained upon substitution of the given conditions to the general solution, the corresponding particular solution is determined

REMARKS: 1. The term initial condition is used when the independent variable is the time and the conditions given are at zero – time, t = 0. 2. The term boundary is used if the independent variable is anyone of the spatial coordinates x,y, or z and the conditions given are for known values of anyone of these coordinates. 2

EXAMPLE Consider the fourth solution y4 y4 = x3 + 2x2 + ax + b Suppose the boundary condition given are: y = 5 when x = 1 y = 3 when x = 0 determine the particular solution 1. Substitution of the given boundary conditions to the general solution y4 results in the following or

5 = 13 + 2(1)2 + a(1) + b 2=a+b

or

3=0+0+0+b b=3

(1,5) (0,3)

2. The particular solution to y’’ = 6x + 4 under the given conditions then is y = x3 + 2x2 – x + 3. SUPPLEMENTARY PROBLEMS 1. Determine the order, degree, and type of the following differential equations: (a) (x + y)dx + (3x2 - 1)dy = 0 (b) x(d2y/dt2) – y(d2x/dt2) = k

(c) ∂2v/∂x2 + ∂2w/∂x2 = 0 (d) x(d2y/dx2) + (dy/dx)4 – y = 0

(e) y’’’ + 4y’ + 3y = x

2. Prove that each equation is a solution of the given differential equation: (a) y = x2 + 4x ; xdy/dx = x2 + y (b) y = Asin5x + Bcos5x ; d2y/dx2 + 25y = 0 (c) y = (x + 1)e-x ; y’ + y – e-x = 0

(d) lny = C1ex + C2e-x ;yy’’ – (y’)2 = y2lny (e) y -3 = x3(3ex + C); xy’ + y + x4y4ex = 0

SEPARATION OF VARIABLES (a) In a solution of an ordinary first-ordered differential equation, the simples is the one which the variables, say x and y, can be separated. By this, it means that the function of x times dx is separable from the function of y times dy.

(b) From F(dny/dcn, dn-1/dxn-1 , … , dy/dx, x,y) = 0, the general form of the first-ordered differential equation may be derived, F(dy/dx, x,y) = 0

REMARK: F(dy/dx, x,y) = 0 may also be written in terms of the differentials dx and dy M(x,y)dx + N(x,y)dy = 0 or simply

Mdx + Ndy = 0

where M and N are both functions of x and y.

(c) For the case where the variables are separable, M(x.y)dx + N(x,y)dy = 0 or Mdx + Ndy = 0 is written as f(x)dx + g(y)dy = 0 which is the standard form of the first-ordered ordinary differential equation with variables separable.

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(d) It is easily seen that the standard solution to (fx)dx + g(y)dy = 0 is ∫ f(x)dx + ∫ g(y)dy = C REMARK: For the differential equation in three (3) variables Mdx + Ndy + Pdz = 0 where M, N, and P are all functions of x, y, and z, the method of separation of variables may also be applied if it is reducible to the form f(x)dx + g(y)dy + h(z)dz = 0 the solution to which is once more obtained by integration.

TEST ON SEPARATION OF VARIABLES (a) To determine whether the variables in M(x,y)dx + N(x,y)dy = 0 are separable or not, see if M(x,y) is factorable into h(x)•Ø(y) N(x,y) is factorable into g(x)•v(y). (b) If the function M and N are thus factorable, the variables are separable otherwise x and y can not be possibly separated. REMARK: With both M and N factorable as given above, M(x,y)dx + N(x,y)dy = 0 may now be written as h(x)•Ø(y) + g(x)•v(y).= 0 or h(x)/g(x) + v(y)/Ø(y) = 0 which is the standard for defined by f(x)dx + g(y)dy = 0 and whose integration leads to the solution given by ∫ f(x)dx + ∫ g(y)dy = C EXAMPLE PROBLEMS PROBLEM 1

Find the complete solution of the differential equation y/dx = (2x + 2xy2)/(y + 2x2y)

SOLUTION (a) The given D.E. may be written as

2x(1 + y2)dx – y(1 + 2x2)dy = 0 or

(2x)dx/(1 + 2x2) – (y)dy/(1 + y2) = 0

(b) Integration of each term leads to (1/2)ln(1 + 2x2) – (1/2)ln(1 + y2) = C sqr[(1 + 2x2)/ (1 + y2)] = C1 and 1 + 2x2 = C2(1 + y2)

or

ln sqr[(1 + 2x2)/ (1 + y2)] = C

(C1 = eC) (C2 = C1)2

PROBLEM 2 Find the complete solution of LdI + RIdt = 0 SOLUTION (a) Write the given D.E. as dI/I + Rdt/L = 0 (b) Integrate ln I + Rt/L = C ln I = C – kt (k = R/L) and I = C1e -kt (C1 = eC)

(L and R are constants)

PROBLEM 3 Determine the particular solution of exeydx – e-2ydy = 0 when x = 0. SOLUTION (a) From the given D.E. exdx – e-2ydy = 0 (b) Integration gives ex + e-3y/3 = C (c) When x = 0, y = 0 and C = 4/3 So the particular solution is ex + e-3y/3 = 4/3 or 3ex + e-3y/3 = 4

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PROBLEM 4

x cos2ydx + tanydy = 0

Find the complete solution of

(a) From the given equation xdx + tanydy/cos2y = 0 (b) Integrate each term

x2/2 + (cosy)-2/2 = C

or

xdx + sinydy(cosy)-3 = 0

or

xx2 + sec2y = C1

(C1 = 2C)

PROBLEM 5 Find the general solution of dy/dx = (2 – y)2/2(1 + x)1/2 SOLUTION (a) From the given equation (2 – y)2dx – 2(1 + x)1/2dy = 0 or 1dx/(1 + x)1/2 – 2dy(2 – y)2 = 0 (b) Integrate

∫(1 + x)-1/2dx – 2(2 – y)-2dy = 0 2(1 + x)1/2 – 2(2 – y)-1 = C (1 + x)1/2 – 1/(2 – y) = C1

and (C1 = C/2)

PROBLEM 6 Find the particular solution of y’ = y/x(x – x3); y = -2 when x = 2 SOLUTION (a) Separating the variables, we have dy/y – dx/x2(1-x2) = 0 (b) Integrate each term, the second term with the use of reduction to partial fractions method, lny + 1/x + ln sqr(x – 1)/(x + 1) = C or 1/x + lny sqr(x – 1)/(x + 1) = C (c) When x = 2, y = -2 and C = -½ + ln (-2)/sqr(3) So the particular solution is PROBLEM 7 Find the particular solution of SOLUTION (a) From the given equation, separate the variables (b) Integration gives (c) When y = 1, x = 4 and C = (1/2)tan-12 The particular solution is

1/x + lny sqr(x – 1)/(x + 1) + ln (-2)/sqr(3) ydx – 4dy = x2dy; x = 4 when y =1 dx/(x2 + 4) – dy/y = 0 ½[tan-1(x /2)] – lny = C (1/2)tan-1(x/2) – lny = (1/2)tan-12

PROBLEM 8 Find the complete solution of lnx dx/dy = 2/xy3 SOLUTION (a) The equation is written as xlnxdx – 2y-3dy = 0 (b) Integrate each term, the first by the method of integration by parts (x2/2)lnx – x2/4 + 1/y2 = C SUPPLEMENTARY PROBLEMS Find the complete solutions of the following differential equations: 1. dy/dx = (4x + xy2)/(y – x2y) 6. xeydy + (x2 + 1)dx/y = 0 2 2 ans. (4 + y )(1 – x ) = C ans. (y – 1)ey + x2/2 + lnx = C 2. y’ – 2y = y2; y = 3 when x = 0 7. xy3dx + (y + 1)e-xdy = 0 ans. 5y = 3(y + 2)e2x ans. ex(x – 1) = 1/y + 1/2y2 + C 3. dr = b(cosӨdr + rsinӨdӨ) (b is a constant) 8. 2xyy’ = 1 + y2; y = 3 when x = 2 ans. r = C(1 – bcosӨ) ans. y2 = 5x – 1 4. ylnx lnydx + dy = 0 9. (4z + x2z)dz + (1 + z2)dx – (yz2 + y)(4 + x2)dy = 0 ans. xlnx + ln(lny) = x+ C ans. ln(1 + z2) + tan-1(x/2) – y2 = C 2 2 2 5. (y’) = (1 – y )/(1 – x ); y = ½ when x = 1 10. dz – (4x3 – 4x3z + x3z2)dx + (4y – 1)(2 – z)2dy = 0 ans. sin-1x – sin-1y = π/3 ans. 1/(2 – z) – x4/4 + 2y2 – y = C

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Joe Randy B. Tabungao 12-06-06