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Sunday, April 5, 2009

Working Rules For Finding Complementary Function of Linear Differential Equation

Case 1: -

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

C.F = c1em1x + c2em2x + c3em3x

Case 2: -

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

C.F = (c1 + c2x + c3x2) em1x

Case 3: -

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

C.F = eax (c1cos bx + c2 sin bx), c1eax cos(bx + c2) or, c1eax sin (bx + c2)

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

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

Case 4: -

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

C.F = eax (c1cos x√b + c2 sin x√b), c1 eax cosh (x√b + c2) or, c1 eax sinh (x√b + c2)

Saturday, February 21, 2009

Rule for solving homogeneous equation

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:

Friday, February 20, 2009

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

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:

Wednesday, February 18, 2009

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

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

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

Solution: -

Given that, y = ex (A cosx + B sinx) ------------ (1)

Differentiating (1) with respect to x, we get

y ‘ = ex (-A sinx + B cosx) + ex (A cosx + B sinx)

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

Differentiating (2) with respect to x, we get

y “ = -ex (A cosx + B sinx) + ex (-A sinx + B cosx) + y’ ---- (3)

Now From Equation (1)

ex (-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 = aex + be-x + x2 is a solution.

Solution: -

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

Differentiating (1) with respect to x, we get

xy’ + y = aex – be-x + 2x ------ (2)

Differentiating (2) with respect to x, we get

xy” + y’ + y’ = aex + be-x + 2

or, xy” + 2y’ = xy – x2 + 2

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

Equation (3) is the required differential equation.

Sunday, February 15, 2009

Definition of Different Kind Differential Equations

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