Solve the correct answer linear equations by considering y as a function of x, that is, \displaystyle{y}={y}{\left({x}\right)}.\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{y}= \cos{{x}}

sibuzwaW 2021-03-07 Answered

Solve the correct answer linear equations by considering y as a function of x, that is, \(\displaystyle{y}={y}{\left({x}\right)}.\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{y}= \cos{{x}}\)

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Expert Answer

svartmaleJ
Answered 2021-03-08 Author has 17464 answers

Variation of parameters
First, solve the linear homogeneous equation by separating variables.
Rearranging terms in the equation gives
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=-{y}\Leftrightarrow\frac{{\left.{d}{y}\right.}}{{y}}=-{\left.{d}{x}\right.}\)
Now, the variables are separated, x appears only on the right side, and y only on the left.
Integrate the left side in relation to y, and the right side in relation to x
\(\displaystyle\int\frac{{\left.{d}{y}\right.}}{{y}}=-\int{\left.{d}{x}\right.}\)
which is
\(\displaystyle{\ln}{\left|{y}\right|}=-{x}+{c}\)
By taking exponents, we obtain
\(\displaystyle{\ln}{\left|{y}\right|}={e}^{{-{x}+{c}}}={e}^{{-{x}}}\cdot{e}^{c}\)
Hence,we obtain
\(\displaystyle{y}={C}{e}^{{-{x}}}\)
where \(\displaystyle{C}\:=\pm{e}^{c}{\quad\text{and}\quad}{y}_{{c}}={e}^{{-{x}}}\) is the complementary solution.
Next, we need to find the particular solution \(\displaystyle{y}_{{p}}\).
Therefore, we consider \(\displaystyle{u}{y}_{{c}}\) and try to find u, a function of x, that will make this work.
Let’s assume that \(\displaystyle{u}{y}_{{c}}\) is a solution of the given equation. Hence, it satisfies the given equation. Substituting \(\displaystyle{u}{y}_{{c}}\) and its derivative in the equation gives
\(\displaystyle{\left({u}{y}_{{c}}\right)}'+{u}{y}_{{c}}= \cos{{x}}\)
\(\displaystyle{u}'{y}_{{c}}+{u}{y}'_{{c}}+{u}{y}_{{c}}= \cos{{x}}\)
\(\displaystyle\underbrace{{{u}'{y}_{{c}}+{u}{\left({y}'{c}+{y}_{{e}}\right)}= \cos{{x}}}}_{{={0}\ \text{since}\ {y}_{{c}}\text{is a solution}}}\)
Therefore, \(\displaystyle{u}'{y}_{{c}}= \cos{{x}}{x}\Rightarrow{u}'=\frac{{ \cos{{x}}}}{{y}_{{c}}}\)
which gives
\(\displaystyle{u}=\int\frac{{ \cos{{x}}}}{{y}_{{c}}}{\left.{d}{x}\right.}\)
Now, we can find the function u:
\(\displaystyle{u}=\int\frac{{ \cos{{x}}}}{{{e}^{ -{{x}}}}}{\left.{d}{x}\right.}=\overbrace{{\int{e}^{x} \cos{{x}}}}^{{\text{Integration by parts}\int{u}{d}{v}={u}{v}-\int{v}{d}{u}}}\)
\(\displaystyle={\left|\begin{matrix}{u}= \cos{{x}}\Rightarrow{d}{u}=- \sin{{x}}{\left.{d}{x}\right.}\\{d}{v}={e}^{x}\Rightarrow{v}=\int{d}{v}={e}^{x}\end{matrix}\right|}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\left\lbrace{e}\right\rbrace}^{{{x}}}\cdot{\cos{{\left\lbrace{\left\lbrace{x}\right\rbrace}\right\rbrace}}}-\int{\left\lbrace{e}\right\rbrace}^{{{x}}}{\left\lbrace{\left(-{\sin{{\left\lbrace{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\right)}\right\rbrace}\)
\(\displaystyle={e}^{x}\cdot \cos{{x}}-\int{e}^{x}{\left(- \sin{{x}}{\left.{d}{x}\right.}\right)}\)
\(\displaystyle{e}^{x}\cdot \cos{{x}}+\overbrace{{\int{e}^{x} \sin{{x}}}}_{{\text{Integration by parts}\int{u}{d}{v}={u}{v}-\int{v}{d}{u}}}\)
\(\displaystyle={\left|\begin{matrix}{u}= \sin{{x}}\Rightarrow{d}{u}= \cos{{x}}{\left.{d}{x}\right.}\\{d}{v}={e}^{x}\Rightarrow{v}=\int{d}{v}={e}^{x}\end{matrix}\right|}\)
\(\displaystyle={e}^{x}\cdot \cos{{x}}+{e}^{x} \sin{{x}}-\underbrace{{\int{e}^{x} \cos{{x}}{\left.{d}{x}\right.}}}_{{\text{This is}\ {I}}}\)

\(\displaystyle\Rightarrow{I}={e}^{x}\cdot \cos{{x}}+{e}^{x} \sin{{x}}-{I}\)
Now, let`s solve for I.
\(\displaystyle{2}{I}={e}^{x}\cdot \cos{{x}}+{e}^{x} \sin{{x}}+{c}\)
\(\displaystyle{I}=\frac{1}{{2}}{e}^{x}{\left( \cos{{x}}+ \sin{{x}}\right)}+{c}\)
Since we need to find only one function that will make this work, we don’t need to introduce the constant of integration c. Hence,
\(\displaystyle{u}=\frac{1}{{2}}{e}^{x}{\left( \cos{{x}}+ \sin{{x}}\right)}\)
Recall that \(\displaystyle{y}_{{p}}={u}{y}_{{c}}\).

Therefore,
\(\displaystyle{y}_{{p}}-\frac{1}{{2}}{e}^{x}{\left( \cos{{x}}+ \sin{{x}}\right)}\cdot{e}^{{-{x}}}=\frac{1}{{2}}{\left( \cos{{x}}+ \sin{{x}}\right)}\)
The general solution is
\(\displaystyle{y}={C}{y}_{{c}}+{y}_{{p}}\)
\(\displaystyle={C}{e}^{{-{x}}}+\frac{1}{{2}}{e}^{x}\)
Integrating Factor technique
This equation is linear with \(\displaystyle{P}{\left({x}\right)}={1}{\quad\text{and}\quad}{Q}{\left({x}\right)}= \cos{{x}}\)
Hence,
\(\displaystyle{h}=\int{P}{\left.{d}{x}\right.}=\int{\left.{d}{x}\right.}={x}\)
So, an integrating factor is \(\displaystyle{e}^{h}={e}^{x}\)
and the general solution is
\(\displaystyle{y}{\left({x}\right)}={e}^{{-{h}}}{\left({c}+\int{Q}{e}^{h}{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{{-{x}}}{\left({c}+\int \cos{{x}}\cdot{e}^{x}{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{{-{x}}}{\left({c}+\frac{1}{{2}}{e}^{x}{\left( \cos{{x}}+ \sin{{x}}\right)}\right)}\)
\(\displaystyle={c}{e}^{{-{x}}}+\frac{1}{{2}}{\left( \cos{{x}}+ \sin{{x}}\right)}\)
We get the finally answer is
\(\displaystyle{y}={c}{e}^{{-{x}}}+\frac{1}{{2}}{\left( \cos{{x}}+ \sin{{x}}\right)}\)

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