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

danrussekme 2021-11-23 Answered
Solve th second order linear equations:
\(\displaystyle{y}{''}+{4}{y}={\cos{{2}}}{x}\)

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Expert Answer

Margaret Plemons
Answered 2021-11-24 Author has 1433 answers
We need to use method of variation of parameters to solve
\(\displaystyle{y}{''}+{4}{y}={\cos{{\left({2}{x}\right)}}}\)
General form of second order differential equation is \(\displaystyle{y}{''}+{P}{\left({x}\right)}{y}'+{Q}{\left({x}\right)}{y}={R}{\left({x}\right)}\)
Here,
\(\displaystyle{P}{\left({x}\right)}={0}\)
\(\displaystyle{Q}{\left({x}\right)}={4}\)
\(\displaystyle{R}{\left({x}\right)}={\cos{{\left({2}{x}\right)}}}\)
Complementary function: \(\displaystyle{\left({y}_{{c}}{\left({x}\right)}\right)}\)
For finding \(\displaystyle{y}_{{c}}{\left({x}\right)}\) we consider the homogenous equation , \(\displaystyle{y}{''}+{4}{y}={0}\)
This is of the form , \(\displaystyle{\left({D}^{{2}}+{4}\right)}{y}={0},\ {D}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}\)
Now, let \(\displaystyle{D}={m}\) , and finding the eigen values , we get , \(\displaystyle{m}^{{2}}+{4}={0}\)
\(\displaystyle\Rightarrow{m}^{{2}}+{4}={0}\)
\(\displaystyle\Rightarrow{m}^{{2}}=-{4}\)
\(\displaystyle\Rightarrow{m}=\pm\sqrt{{-{4}}}=\pm{2}{i}\)
We get the eigen values \(\displaystyle\pm{2}{i}\)
If the eigen values are of the form \(\displaystyle{a}\pm{i}{b}\) then the complementary function is of the form ,
\(\displaystyle{y}_{{c}}{\left({x}\right)}={e}^{{{a}{x}}}{\left({C}_{{1}}{\cos{{\left({b}{x}\right)}}}+{C}_{{2}}{\sin{{\left({b}{x}\right)}}}\right)}\)
Here, the eigen values are \(\displaystyle\pm{2}{i}\), therefore,
\(\displaystyle{y}_{{c}}={e}^{{{0}{x}}}{\left({C}_{{1}}{\cos{{\left({2}{x}\right)}}}+{C}_{{2}}{\sin{{\left({2}{x}\right)}}}\right)}\)
\(\displaystyle={C}_{{1}}{\cos{{\left({2}{x}\right)}}}+{C}_{{2}}{\sin{{\left({2}{x}\right)}}}\)
Hence, the complementary function is \(\displaystyle{y}_{{c}}{\left({x}\right)}={C}_{{1}}{\cos{{\left({2}{x}\right)}}}+{C}_{{2}}{\sin{{\left({2}{x}\right)}}}\)
Now, we find the particular integral.
For that we consider the complementary function \(\displaystyle{y}_{{c}}{\left({x}\right)}={C}_{{1}}{u}{\left({x}\right)}+{C}_{{2}}{v}{\left({x}\right)}\)
Since, the complementary function is \(\displaystyle{y}_{{c}}{\left({x}\right)}={C}_{{1}}{\cos{{\left({2}{x}\right)}}}+{C}_{{2}}{\sin{{\left({2}{x}\right)}}}\) therefore,
\(\displaystyle{u}{\left({x}\right)}={\cos{{\left({2}{x}\right)}}},\ {v}{\left({x}\right)}={\sin{{\left({2}{x}\right)}}}\)
And get the particular integral as \(\displaystyle{y}_{{p}}{\left({x}\right)}={A}{u}{\left({x}\right)}+{B}{v}{\left({x}\right)}\)
Where \(\displaystyle{A}=-\int{\frac{{{v}{\left({x}\right)}{R}{\left({x}\right)}}}{{{W}{\left({x}\right)}}}}{\left.{d}{x}\right.}\) and \(\displaystyle{B}=\int{\frac{{{u}{\left({x}\right)}{R}{\left({x}\right)}}}{{{W}{\left({x}\right)}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{R}{\left({x}\right)}={\cos{{\left({2}{x}\right)}}}\) from (1)
W is the wronskian is u and v, that is, \[W(x)=\begin{vmatrix}u(x) & v(x) \\u'(x) &v'(x) \end{vmatrix}\]
\[\Rightarrow W=\begin{vmatrix}\cos(2x) &\sin(2x) \\-2\sin(2x) &2\cos(2x) \end{vmatrix}\]
\(\displaystyle={2}{{\cos}^{{2}}{\left({2}{x}\right)}}+{2}{{\sin}^{{2}}{\left({2}{x}\right)}}={2}{\left({{\cos}^{{2}}{\left({2}{x}\right)}}+{{\sin}^{{2}}{\left({2}{x}\right)}}\right)}={2}\)
(Since, \(\displaystyle{{\cos}^{{2}}{x}}+{{\sin}^{{2}}{x}}={1}\))
Now, we evaluate A and B.
\(\displaystyle{A}=-\int{\frac{{{v}{\left({x}\right)}{R}{\left({x}\right)}}}{{{W}{\left({x}\right)}}}}{\left.{d}{x}\right.}\)
\(\displaystyle=-\int{\frac{{{\sin{{\left({2}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}}}{{{2}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{-{1}}}{{{4}}}}\int{2}{\sin{{\left({2}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{-{1}}}{{{4}}}}\int{\sin{{\left({4}{x}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{\frac{{{\cos{{\left({4}{x}\right)}}}}}{{{4}}}}={\frac{{{\cos{{\left({4}{x}\right)}}}}}{{{16}}}}\)
Now we find B,
\(\displaystyle{B}=\int{\frac{{{u}{\left({x}\right)}{R}{\left({x}\right)}}}{{{W}{\left({x}\right)}}}}{\left.{d}{x}\right.}\)
\(\displaystyle=\int{\frac{{{\cos{{\left({2}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}}}{{{2}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}\int{{\cos}^{{2}}{\left({2}{x}\right)}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}\int{\frac{{{1}+{\cos{{\left({4}{x}\right)}}}}}{{{2}}}}{\left.{d}{x}\right.}\) using (\(\displaystyle{{\cos}^{{2}}{x}}={\frac{{{1}+{\cos{{2}}}{x}}}{{{2}}}}\))
\(\displaystyle={\frac{{{1}}}{{{4}}}}\int{\left({1}+{\cos{{\left({4}{x}\right)}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{\left({x}+{\frac{{{\sin{{\left({4}{x}\right)}}}}}{{{4}}}}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{\left({x}+{\frac{{{\sin{{\left({4}{x}\right)}}}}}{{{4}}}}\right)}\)
Now, we have A and B as,
\(\displaystyle{A}={\frac{{{\cos{{\left({4}{x}\right)}}}}}{{{16}}}},\ {B}={\frac{{{1}}}{{{4}}}}{\left({x}+{\frac{{{\sin{{\left({4}{x}\right)}}}}}{{{4}}}}\right)}\)
Now , we put these values in particular integral \(\displaystyle{y}_{{p}}{\left({x}\right)}={A}{u}{\left({x}\right)}+{B}{v}{\left({x}\right)}\)
\(\displaystyle\Rightarrow{y}_{{p}}{\left({x}\right)}={A}{u}{\left({x}\right)}+{B}{v}{\left({x}\right)}\)
\(\displaystyle={\frac{{{\cos{{\left({4}{x}\right)}}}}}{{{16}}}}{\cos{{\left({2}{x}\right)}}}+{\frac{{{1}}}{{{4}}}}{\left({x}+{\frac{{{\sin{{\left({4}{x}\right)}}}}}{{{4}}}}\right)}{\sin{{\left({2}{x}\right)}}}\)
\(\displaystyle={\frac{{{x}}}{{{4}}}}{\sin{{\left({2}{x}\right)}}}+{\frac{{{\cos{{\left({4}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}}}{{{16}}}}+{\frac{{{\sin{{\left({4}{x}\right)}}}{\sin{{\left({2}{x}\right)}}}}}{{{16}}}}\)
\(\displaystyle={\frac{{{x}}}{{{4}}}}{\sin{{\left({2}{x}\right)}}}+{\frac{{{1}}}{{{16}}}}{\left({\cos{{\left({4}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}+{\sin{{\left({4}{x}\right)}}}{\sin{{\left({2}{x}\right)}}}\right)}\)
We have the identity , \(\displaystyle{\cos{{\left({x}\right)}}}{\cos{{\left({y}\right)}}}+{\sin{{\left({x}\right)}}}{\sin{{\left({y}\right)}}}={\cos{{\left({x}-{y}\right)}}}\)
\(\displaystyle\Rightarrow{y}_{{p}}{\left({x}\right)}={\frac{{{x}}}{{{4}}}}{\sin{{\left({2}{x}\right)}}}+{\frac{{{1}}}{{{16}}}}{\left({\cos{{\left({4}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}+{\sin{{\left({4}{x}\right)}}}{\sin{{\left({2}{x}\right)}}}\right)}\)
\(\displaystyle={\frac{{{x}}}{{{4}}}}{\sin{{\left({2}{x}\right)}}}+{\frac{{{1}}}{{{16}}}}{\cos{{\left({4}{x}-{2}{x}\right)}}}\)
\(\displaystyle={\frac{{{x}}}{{{4}}}}{\sin{{\left({2}{x}\right)}}}+{\frac{{{1}}}{{{16}}}}{\cos{{\left({2}{x}\right)}}}\)
Now ,we have , the solution \(\displaystyle{y}{\left({x}\right)}={y}_{{c}}{\left({x}\right)}+{y}_{{p}}{\left({x}\right)}\)
\(\displaystyle\Rightarrow{y}{\left({x}\right)}={C}_{{1}}{\cos{{\left({2}{x}\right)}}}+{C}_{{2}}{\sin{{\left({2}{x}\right)}}}+{\frac{{{x}}}{{{4}}}}{\sin{{\left({2}{x}\right)}}}+{\frac{{{1}}}{{{16}}}}{\cos{{\left({2}{x}\right)}}}\)
\(\displaystyle={\left({C}_{{1}}+{\frac{{{1}}}{{{16}}}}\right)}{\cos{{\left({2}{x}\right)}}}+{C}_{{2}}{\sin{{\left({2}{x}\right)}}}+{\frac{{{x}}}{{{4}}}}{\sin{{\left({2}{x}\right)}}}\)
\(\displaystyle={C}_{{3}}{\cos{{\left({2}{x}\right)}}}+{C}_{{2}}{\sin{{\left({2}{x}\right)}}}+{\frac{{x}}{{4}}}{\sin{{\left({2}{x}\right)}}}\)
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Unpled
Answered 2021-11-25 Author has 1349 answers
I am trying to solve and it does not work, if you can, then please help
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