chillywilly12a

2021-02-24

Solve the linear equations by considering y as a function of x, that is,

lobeflepnoumni

Skilled2021-02-25Added 99 answers

Variation of parameters

First, solve the linear homogeneous equation by separating variables.

Rearranging terms in the equation gives

$\frac{dy}{dx}=-\frac{y}{x}\iff \frac{dy}{y}=-\frac{dx}{x}$

Now, the variables are separated, x appears only on the right side, and yonly on the left.

Integrate the left side in relation to y, and the right side in relation to x

$\int \frac{dy}{y}=-\int \frac{dx}{x}$

Which is

$\mathrm{ln}\left|y\right|=-\mathrm{ln}\left|x\right|+c$

By taking exponents, we obtain

$\left|y\right|={e}^{-\mathrm{ln}\left|x\right|+c}=\frac{1}{\left|x\right|}\cdot {e}^{c}$

Hence,we obtain

$y=\frac{1}{C}x$

where$C{\textstyle \phantom{\rule{0.222em}{0ex}}}=\pm {e}^{c}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{y}_{c}=\frac{1}{x}$ is the complementary solution .

Next, we need to find the particular solution${y}_{p}$ .

Therefore, we consider$u{y}_{c}$ , and try to find u, a function of x, that will makethis work.

Let's assume that$u{y}_{c}$ is a solution of the given equation. Hence, it satisfiesthe given equation.

Substituting$u{y}_{c}$ , and its derivative in the equation gives

${\left(u{y}_{c}\right)}^{\prime}+\frac{u{y}_{c}}{x}=x$

${u}^{\prime}{y}_{c}+u{y}_{c}^{\prime}+\frac{u{y}_{c}}{x}=x$

${u}^{\prime}{y}_{c}+\underset{\text{=0}\text{}}{\underset{\u23df}{u({y}_{c}^{\prime}+\frac{{y}_{c}}{x})}}=x$

since${y}_{c}$ is a solution

Therefore,

$u}^{\prime}{y}_{c}=x\Rightarrow {u}^{\prime}=\frac{x}{{y}_{c}$

which gives

$u=\int \frac{x}{{y}_{c}}dx$

Now, we can find the function u:

$u=\int \left(\frac{x}{\frac{1}{x}}\right)dx$

$=\int {x}^{2}dx$

$=\frac{{x}^{3}}{3}+c$

Since we need to find only one function that will male this work, we don’tneed to introduce the constant of integration c. Hence,

$u=\frac{{x}^{3}}{3}$

Recall that${y}_{p}=u{y}_{c}$

Therefore

$y}_{p}=\frac{1}{x}\cdot \frac{{x}^{3}}{3$

$=\frac{{x}^{2}}{3}$

The general solution is

$y=C{y}_{c}+{y}_{p}$

$=\frac{C}{x}+\frac{{x}^{2}}{3}$

Integrating Factor technique

This equation is linear with$P\left(x\right)=\frac{1}{x}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}Q\left(x\right)=x$

Hence,$h=\int \frac{1}{x}=\mathrm{ln}\left|x\right|$

So, an integrating factor is

First, solve the linear homogeneous equation by separating variables.

Rearranging terms in the equation gives

Now, the variables are separated, x appears only on the right side, and yonly on the left.

Integrate the left side in relation to y, and the right side in relation to x

Which is

By taking exponents, we obtain

Hence,we obtain

where

Next, we need to find the particular solution

Therefore, we consider

Let's assume that

Substituting

since

Therefore,

which gives

Now, we can find the function u:

Since we need to find only one function that will male this work, we don’tneed to introduce the constant of integration c. Hence,

Recall that

Therefore

The general solution is

Integrating Factor technique

This equation is linear with

Hence,

So, an integrating factor is