 # Solve the linear equations by considering y as a function of x, that is, y = y(x) 3xy'+y=12x Amari Flowers 2021-02-12 Answered
Solve the linear equations by considering y as a function of x, that is,
$y=y\left(x\right)$
$3x{y}^{\prime }+y=12x$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it SchepperJ

Variation of parameters
First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives
$3x{y}^{\prime }=-y⇔\frac{dy}{dx}=-\frac{y}{3x}⇔\frac{dy}{y}=-\frac{dx}{3x}$
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
$\int \frac{dy}{y}=-\frac{1}{3}\int \frac{dx}{x}$
Which is
$\mathrm{ln}|y|=-\frac{1}{3}\mathrm{ln}|x|+c$
By taking exponents, we obtain
$|y|={e}^{-\frac{1}{3}\mathrm{ln}|x|+c}={|x|}^{-\frac{1}{3}}\cdot {e}^{c}$
Hence,we obtain
$y=C{x}^{-\frac{1}{3}}$
where $C\phantom{\rule{0.222em}{0ex}}=±{e}^{c}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{y}_{c}={x}^{-\frac{1}{3}}$ 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 make this work.
Let's assume that $u{y}_{c}$ is a solution of the given equation. Hence, it satisfies the given equation.
Write the equation in the standard form
${y}^{\prime }+\frac{y}{3x}=12$
Substituting $u{y}_{c}$, and its derivative in the equation gives
${\left(u{y}_{c}\right)}^{\prime }+\frac{u{y}_{c}}{3x}=4$
${u}^{\prime }{y}_{c}+u{y}_{c}^{\prime }+\frac{u{y}_{c}}{3x}=4$

since ${y}_{c}$ is a solution
Therefore,
${u}^{\prime }{y}_{c}=4⇒{u}^{\prime }=\frac{4}{{y}_{c}}$
which gives
$u=\int \frac{4}{{y}_{c}}dx$
Now, we can find the function u:
$u=\int \frac{4}{{x}^{-\frac{1}{3}}}dx$
$=4\int {x}^{-\frac{1}{3}}dx$
$=4\cdot \frac{{x}^{\frac{1}{3}+1}}{\frac{1}{3}+1}+c$
$=4\cdot \frac{{x}^{\frac{4}{3}}}{\frac{4}{3}}+c$
$=4\cdot \frac{3}{4}{x}^{\frac{4}{3}}+c$