Question

Using the method of variation of parameters, solve
+ 4y = cot 2x​

Answer 1

Answer:

keep things simple, we are only going to look at the case:

d2ydx2 + p dydx + qy = f(x)

where p and q are constants and f(x) is a non-zero function of x.

The complete solution to such an equation can be found by combining two types of solution:

The general solution of the homogeneous equation d2ydx2 + p dydx + qy = 0

Particular solutions of the non-homogeneous equation d2ydx2 + p dydx + qy = f(x)

Note that f(x) could be a single function or a sum of two or more functions.

Once we have found the general solution and all the particular solutions, then the final complete solution is found by adding all the solutions together.

This method relies on integration.

The problem with this method is that, although it may yield a solution, in some cases the solution has to be left as an integral.

Start with the General Solution

On Introduction to Second Order Differential Equations we learn how to find the general solution.

Basically we take the equation

d2ydx2 + p dydx + qy = 0

and reduce it to the "characteristic equation":

r2 + pr + q = 0

Which is a quadratic equation that has three possible solution types depending on the discriminant p2 − 4q. When p2 − 4q is

positive we get two real roots, and the solution is

y = Aer1x + Ber2x

zero we get one real root, and the solution is

y = Aerx + Bxerx

negative we get two complex roots r1 = v + wi and r2 = v − wi, and the solution is

y = evx ( Ccos(wx) + iDsin(wx) )

The Fundamental Solutions of The Equation

In all three cases above "y" is made of two parts:

y = Aer1x + Ber2x is made of y1 = Aer1x and y2 = Ber2x

y = Aerx + Bxerx is made of y1 = Aerx and y2 = Bxerx

y = evx ( Ccos(wx) + iDsin(wx) ) is made of y1 = evxCcos(wx) and y2 = evxiDsin(wx)

y1 and y2 are known as the fundamental solutions of the equation

And y1 and y2 are said to be linearly independent because neither function is a constant multiple of the other.

The Wronskian

When y1 and y2 are the two fundamental solutions of the homogeneous equation

d2ydx2 + p dydx + qy = 0

then the Wronskian W(y1, y2) is the determinant of the matrix

matrix for the Wronskian

So

W(y1, y2) = y1y2' − y2y1'

The Wronskian is named after the Polish mathematician and philosopher Józef Hoene-Wronski (1776−1853).

Since y1 and y2 are linearly independent, the value of the Wronskian cannot equal zero.

The Particular Solution

Using the Wronskian we can now find the particular solution of the differential equation

d2ydx2 + p dydx + qy = f(x)

using the formula:

yp(x) = −y1(x)∫ y2(x)f(x)W(y1, y2) dx + y2(x)∫ y1(x)f(x)W(y1, y2) dx

Example 1: Solve d2ydx2 − 3 dydx + 2y = e3x

1. Find the general solution of d2ydx2 − 3 dydx + 2y = 0

The characteristic equation is: r2 − 3r + 2 = 0

Factor: (r − 1)(r − 2) = 0

r = 1 or 2

So the general solution of the differential equation is y = Aex+Be2x

So in this case the fundamental solutions and their derivatives are:

y1(x) = ex

y1'(x) = ex

y2(x) = e2x

y2'(x) = 2e2x

2. Find the Wronskian:

W(y1, y2) = y1y2' − y2y1' = 2e3x − e3x = e3x

3. Find the particular solution using the formula:

yp(x) = −y1(x)∫ y2(x)f(x)W(y1, y2) dx + y2(x)∫ y1(x)f(x)W(y1, y2) dx

4. First we solve the integrals:

∫ y2(x)f(x)W(y1, y2) dx

= ∫ e2xe3xe3x dx

= ∫e2xdx

= 12 e2x

So:

−y1(x)∫ y2(x)f(x)W(y1, y2) dx = −(ex)( 12 e2x) = − 12 e3x

And also:

∫ y1(x)f(x)W(y1, y2) dx

= ∫ exe3xe3x dx

= ∫exdx

= ex

So:

y2(x)∫ y1(x)f(x)W(y1, y2) dx = (e2x)( ex) = e3x

Finally:

yp(x) = −y1(x)∫ y2(x)f(x)W(y1, y2) dx + y2(x)∫ y1(x)f(x)W(y1, y2) dx

= − 12 e3x + e3x

= 12 e3x

and the complete solution of the differential equation d2ydx2 − 3 dydx + 2y = e3x is

y = Aex + Be2x + 12 e3x

Which looks like this (example values of A and B):

Aex + Be2x + 12e3x

Step-by-step explanation:

hope it helps you

Answer 2

Complete question

Using the method of variation of parameters solve ($(D^{2} +4)y = cot2x$

Answer:

The complete solution is $y=(Acos2x +bsin2x) - \frac{1}{4} sin2xlog(cosec2x +cot2x)$

Step-by-step explanation:

Given: $(D^{2} +4)y = cot2x$

To find: solve the above equation using the method of variation of parameters.

Solution:

The equation is $m^{2} +4 = 0$

Where, m ± 2i

The complementary function = $e^{ox} [Acos2x +Bsin2x]$

$f_{1} = cos2x$

Differentiate with respect to x

$f_{1} ' = - 2sin2x$

$f_{2} = sin2x$

Differentiate with respect to x

$f_{2} ' = 2cos2x$

$f_{1} f_{2} ' - f_{1}' f_{2} = 2(cos^{2} 2x + sin^{2} 2x)$

                 = 2

P.I = $Pf_{1} + Qf_{2}$

P = -∫ $\frac{f_{2}X }{f_{1} f_{2} ' - f_{1}' f_{2} } dx$

   = - ∫ $\frac{sin2xcot2x}{2} dx$

P = $- \frac{1}{2}$∫cos2x dx

    =  - $\frac{1}{4} sin2x$

Q = ∫$\frac{f_{2}X }{f_{1} f_{2} ' - f_{1}' f_{2} } dx$

 =  ∫ $\frac{cos2xcot2x}{2} dx$

 =  $\frac{1}{2}$  ∫ $\frac{cos^{2}2x }{sin2x} dx$

  = $\frac{1}{2}$  ∫$\frac{1 - sin^{2}2x }{sin2x} dx$

  =  $\frac{1}{2}$ ∫ cosec2x - sin2x)dx

  =$\frac{1}{2}${ ${ -\frac{1}{2} log(cosec2x + cot2x)+\frac{1}{2} cos2x$}

P.I = $Pf_{1} + Qf_{2}$

= $\frac{1}{4} sin2x[cos2x - log(cosec2x +cot2x)] - \frac{1}{4} cos2xsin2x$

= $- \frac{1}{4} sin2xlog(cosec2x + cot2x)$

$y=(Acos2x +bsin2x) - \frac{1}{4} sin2xlog(cosec2x +cot2x)$

Final answer:

The complete solution is $y=(Acos2x +bsin2x) - \frac{1}{4} sin2xlog(cosec2x +cot2x)$

SPJ3