Ordinary and Partial Differential Equations

Working Rules For Finding Complementary Function of Linear Differential Equation

2009-04-05T08:20:00.000-07:00

Case 1: -

If the roots are unequal (m = m₁, m₂, m₃) then the complementary function is

C.F = c₁e^m1x + c₂e^m2x + c₃e^m3x

Case 2: -

If the roots are equal (m = m₁, m₁, m₁) then the complementary function is

C.F = (c₁ + c₂x + c₃x²) e^m1x

Case 3: -

If the roots are complex (m = a ± ib) then the complementary function is

C.F = e^ax (c₁cos bx + c₂ sin bx), c₁e^ax cos(bx + c₂) or, c₁e^ax sin (bx + c₂)

And if the two equal part of complex roots (m = a ± ib, a ± ib) then the complementary function is

C.F = e^ax {(c₁ + c₂x) cos bx + (c₃ + c₄x) sin bx}

Case 4: -

If the roots are “a ± √b” then the complementary function is

C.F = e^ax (c₁cos x√b + c₂ sin x√b), c₁e^ax cosh (x√b + c₂) or, c₁e^ax sinh (x√b + c₂)

Rule for solving homogeneous equation

2009-02-21T02:09:00.000-08:00

Rule for solving homogeneous equation

Let the given equation be homogeneous. Then, by definition, the given equation can be put in the form

dy/dx = ƒ(y/x) ---------- (1)

Solution Method: -

To solve (1), let y/x = v

or, y = vx

Differentiating equation (1) with respect to x, we get

dy/dx = v + x (dv/dx)----------- (2)

So the equation becomes

v + x (dv/dx) = ƒ (v)

or, x (dv/dx) = ƒ (v) – v

Separating the variables x and v, we have

dx/x = dv/{ƒ (v) – v}--------------(3)

Integrating equation (3)

After integration, replace v by y/x and finally we will find the required solution.

Example:

Solve: (x³ + 3xy²)dx + (y³ + 3x²y)dy = 0

Given that, (x³ + 3xy²)dx + (y³ + 3x²y)dy = 0

or, dy/dx = - {(x³ + 3xy²)/ (y³ + 3x²y)}

or, dy/dx = - [{1 + 3(y/x)²}/{(y/x)³ + 3(y/x)}] -------(1)

Take, y/x = v

or, y = vx

Differentiating with respect to x, we get

dy/dx = v + x (dv/dx)----------- (2)

From equation (1) and (2)

v + x (dv/dx) = - (1 + 3v²)/(v³ + 3v)

or, x (dv/dx) = - (1 + 3v²)/(v³ + 3v) – v

or, x (dv/dx) = - (v⁴ + 6v² + 1)/(v³ + 3v)

or 4(dx/x) = - {(4v³ + 12v)dv}/(v⁴ + 6v² + 1)

Integrating,

4 lnx = - ln(v⁴+ 6v² + 1) + lnc

or, lnx⁴= ln{c/(v⁴ + 6v² + 1)}

or, x⁴(v⁴ + 6v² + 1) = c

or, y⁴ + 6x²y² + x⁴ = c [as y/x = v]

Answer:

Example:

Solve: dy/dx = y/x + tan(y/x)

Given that, dy/dx = y/x + tan(y/x) --------(1)

Let y/x = v

or, y = vx

Differentiating with respect to x, we get

dy/dx = v + x (dv/dx)----------- (2)

From equation (1) and (2)

v + x (dv/dx) = v + tanv

or, dx/x = cotv dv

Integrating,

ln x = ln sinv + ln c

or, x = c sin(y/x) [as y/x = v]

Answer:

Transformation of some equation in the form in which variables are separable.

2009-02-20T23:35:00.000-08:00

Transformation of some equation in the form in which variables are separable. Equation of the form

Dy/dx = ƒ (ax + by + c)

or, dy/dx = ƒ (ax + by)

can be reduced to an equation in which variables can be separated. For this purpose we use the substitution ax + by + c = v or, ax + by = v

Example:

Solve: dy/dx = (4x + y +1)²

Given that, dy/dx = (4x + y +1)² ----------- (1)

Let, 4x + y +1 = v -------------- (2)

Differentiating equation (1) with respect to x, we get

4 + dy/dx = dv/dx

or, dy/dx = dv/dx – 4 --------------- (3)

We get from equation (2) & (3),

dv/dx – 4 = v²

or, dv/dx = 4 + v²

or, dx = dv/(4 + v²)

Integrating,

x + c = (1/2) tan ^-1 (v/2)

or, 2x + 2c = tan ^-1 (v/2)

or, v = 2 tan (2x + 2c)

or, 4x + y +1 = 2 tan (2x + 2c)

Answer:

Example:

Solve: dy/dx = sin (x + y) + cos (x + y)

Given that, dy/dx = sin (x + y) + cos (x + y) ------------- (1)

Let, x + y = v ------------(2)

Differentiating equation (1) with respect to x, we get

1 + dy/dx = dv/dx

or, dy/dx = dv/dx – 1 --------------- (3)

We get from equation (2) & (3),

dv/dx – 1 = sinv + cosv

or, dv/dx = 1 + sinv + cosv -------------(4)

but, 1 + sinv + cosv = 1 + 2 sin(v/2)cos(v/2) + 2 cos²(v/2) – 1

or, 1 + sinv + cosv = 2 cos²(v/2) [ 1 + tan v/2 ]

So from equation (3)

dx = dv/[2 cos²(v/2) { 1 + tan v/2 }]

or, dx = {(1/2)sec² (v/2)}/ {1 + tan (v/2)}

Integrating,

x + c = ln {1 + tan (v/2)}

or, x + c = ln [1 + tan{(x + y)/2}]

Answer:

First order and First Degree Differential Equation (Separation of variables)

2009-02-18T08:43:00.000-08:00

Separation of variables

If a differential equation of the first order and first degree is of the form

ƒ₁ (x)dx = ƒ₂ (y)dy

Solve: y –x dy/dx = a(y² + dy/dx)

Solution: -

Given that,
y –x dy/dx = a(y² + dy/dx)

=> y – ay² = dy/dx (a + x)

=> y (1 - ay)/dy = (a + x)/dx

=> dx/(a + x) = dy/y (1 - ay)

=> dx/(a + x) = [{a/(1 - ay)} + 1/y] dy

Integrating,

ln(a + x) = - ln(1 - ay) + lny + lnc

=> ln(a + x) = ln{cy/(1 - ay)}

=> a + x = cy/(1 - ay)

Answer: -

Solve: dy/dx = x (2 lnx + 1)/siny +y cosy

Solution: -

Given that,
dy/dx = x (2 lnx + 1)/siny +y cosy

=>siny +y cosy)dy = (2xlnx + x)dx

Integrating,

ſ sinydy + ſ y cosydy = 2 ſ x lnxdx + ſxdx

=> - cosy + y ſ cosydy - ſ (dy/dy ſ cosydy)dy = 2 lnx ſ xdx - 2 ſ(d lnx/dx ſ xdx)dx + x²/2

=> - cosy +y siny - ſ sinydy = x² lnx - ſ xdx + x²/2

=> - cosy + y siny + cosy = x² lnx - x²/2+ x²/2 + c

=> y siny = x² lnx + c

Answer: -

Problem: Differential Equation of a Solution

2009-02-18T08:12:00.000-08:00

a)Find the differential equation of the family of curves y = e^x (A cosx + B sinx), where A and B are arbitrary constants.

Solution: -

Given that, y = e^x (A cosx + B sinx) ------------ (1)

Differentiating (1) with respect to x, we get

y ‘ = e^x (-A sinx + B cosx) + e^x (A cosx + B sinx)

or, y ‘ = e^x (-A sinx + B cosx) + y {using (1)} ------ (2)

Differentiating (2) with respect to x, we get

y “ = -e^x (A cosx + B sinx) + e^x (-A sinx + B cosx) + y’ ---- (3)

Now From Equation (1)

e^x (-A sinx + B cosx) = y’ – y -----------(4)

Hence eliminating A and B from equation (1), (3) and (4), we get

y” = - y + y’ – y + y’

or, y” – 2y’ +2y = 0 ---------- (5)

Equation (5) is the required differential equation.

b)By eliminating the constants a and b obtain the differential equation for which xy = ae^x + be^-x + x² is a solution.

Solution: -

Given that, xy = ae^x + be^-x + x² ---------- (1)

Differentiating (1) with respect to x, we get

xy’ + y = ae^x– be^-x + 2x ------ (2)

Differentiating (2) with respect to x, we get

xy” + y’ + y’ = ae^x + be^-x + 2

or, xy” + 2y’ = xy – x² + 2

or, xy” + 2y’ - xy + x² – 2 = 0 ------- (3)

Equation (3) is the required differential equation.

Definition of Different Kind Differential Equations

2009-02-15T09:23:00.000-08:00

Differential Equation:-
An equation involving derivatives on different of one or more dependent variables with respect to one or more independent variables is called Differential Equation.

Example:
dy = ( x + sinx ) dx