Question

# 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}}

Second order linear equations

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

2021-03-08

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
$$\displaystyle{y}={c}{e}^{{-{x}}}+\frac{1}{{2}}{\left( \cos{{x}}+ \sin{{x}}\right)}$$