Solve th second order linear equations: y+4y=cos⁡2x

Name: Solve th second order linear equations: y+4y=cos⁡2x
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve th second order linear equations: y+4y=cos⁡2x

Question

Solve th second order linear equations:  y+4y=cos⁡2x

Margaret Plemons · Accepted Answer

We need to use method of variation of parameters to solve

y^{″} + 4 y = \cos (2 x)

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 (2 x)

Complementary function:

(y_{c} (x))

For finding

y_{c} (x)

we consider the homogenous equation ,

y^{″} + 4 y = 0

This is of the form ,

(D^{2} + 4) y = 0, D = \frac{d}{d x}

Now, let

D = m

, and finding the eigen values , we get ,

m^{2} + 4 = 0

\Rightarrow m^{2} + 4 = 0

\Rightarrow m^{2} = - 4

\Rightarrow m = \pm \sqrt{- 4} = \pm 2 i

We get the eigen values

\pm 2 i

If the eigen values are of the form

a \pm i b

then the complementary function is of the form ,

y_{c} (x) = e^{a x} (C_{1} \cos (b x) + C_{2} \sin (b x))

Here, the eigen values are

\pm 2 i

, therefore,

y_{c} = e^{0 x} (C_{1} \cos (2 x) + C_{2} \sin (2 x))

= C_{1} \cos (2 x) + C_{2} \sin (2 x)

Hence, the complementary function is

y_{c} (x) = C_{1} \cos (2 x) + C_{2} \sin (2 x)

Now, we find the particular integral.
For that we consider the complementary function

y_{c} (x) = C_{1} u (x) + C_{2} v (x)

Since, the complementary function is

y_{c} (x) = C_{1} \cos (2 x) + C_{2} \sin (2 x)

therefore,

u (x) = \cos (2 x), v (x) = \sin (2 x)

And get the particular integral as

y_{p} (x) = A u (x) + B v (x)

Where

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Unpled · Accepted Answer

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

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

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