Solve th second order linear equations: y''+4y=\cos2x

danrussekme

danrussekme

Answered question

2021-11-23

Solve th second order linear equations:
y+4y=cos2x

Answer & Explanation

Margaret Plemons

Margaret Plemons

Beginner2021-11-24Added 13 answers

We need to use method of variation of parameters to solve
y+4y=cos(2x)
General form of second order differential equation is y+P(x)y+Q(x)y=R(x)
Here,
P(x)=0
Q(x)=4
R(x)=cos(2x)
Complementary function: (yc(x))
For finding yc(x) we consider the homogenous equation , y+4y=0
This is of the form , (D2+4)y=0, D=ddx
Now, let D=m , and finding the eigen values , we get , m2+4=0
m2+4=0
m2=4
m=±4=±2i
We get the eigen values ±2i
If the eigen values are of the form a±ib then the complementary function is of the form ,
yc(x)=eax(C1cos(bx)+C2sin(bx))
Here, the eigen values are ±2i, therefore,
yc=e0x(C1cos(2x)+C2sin(2x))
=C1cos(2x)+C2sin(2x)
Hence, the complementary function is yc(x)=C1cos(2x)+C2sin(2x)
Now, we find the particular integral.
For that we consider the complementary function yc(x)=C1u(x)+C2v(x)
Since, the complementary function is yc(x)=C1cos(2x)+C2sin(2x) therefore,
u(x)=cos(2x), v(x)=sin(2x)
And get the particular integral as yp(x)=Au(x)+Bv(x)
Where
Unpled

Unpled

Beginner2021-11-25Added 23 answers

I am trying to solve and it does not work, if you can, then please help

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