UkusakazaL

2021-02-20

Find the general solution to the equation $x\left(\frac{dy}{dx}\right)+3\left(y+{x}^{2}\right)=\frac{\mathrm{sin}x}{x}$

Lacey-May Snyder

Write this equation as follows:
$\frac{dy}{dx}+3\frac{y}{x}+\frac{\mathrm{sin}x}{{x}^{2}}-3x\left(1\right)$
We first solve the differential equation
$\frac{dy}{dx}+3\frac{y}{x}=0⇒\frac{dy}{y}=-3\frac{dx}{x}⇒\int \frac{dy}{y}=-3\int \frac{dx}{x}$
Thus,
$\mathrm{ln}|y|=-3\mathrm{ln}|x|+C=\frac{\mathrm{ln}1}{|{x}^{3}|}+C$
Here we used that $\alpha \mathrm{ln}\beta =In{\beta }^{\alpha }$. Now use the exponential function:
$|y|=\frac{1}{|{x}^{3}|}\cdot {e}^{C}$
Recall that $|\alpha |=±\alpha$, so we can write
$y=±{e}^{C}\cdot \frac{1}{{x}^{3}}$
Defining a new constant $D=±{e}^{C}$ yields
$y=\frac{D}{{x}^{3}}$
This is the solution of the homogeneous equation. To obtain the solution of the initial equation, we regard D as a function of x, so
$\frac{dy}{dx}=\frac{{D}^{\prime }{x}^{3}-3D{x}^{2}}{{x}^{6}}=\frac{{D}^{\prime }}{{x}^{3}}-3\frac{D}{{x}^{4}}$
Plugging this and (2) into (1) yields
$\frac{{D}^{\prime }}{{x}^{3}}-3\frac{D}{{x}^{4}}+3\frac{D}{{x}^{4}}=\frac{\mathrm{sin}x}{{x}^{2}}-3x⇒\frac{{D}^{\prime }}{{x}^{3}}=\frac{\mathrm{sin}x}{{x}^{2}}-3x$
Thus,
${D}^{\prime }=x\mathrm{sin}x-3{x}^{4}⇒D=\int \left(x\mathrm{sin}x-3{x}^{4}\right)dx=\int x\mathrm{sin}xdx-3\int {x}^{4}dx$
Now,
$\int x\mathrm{sin}xdx=\left\{\left(u=x⇒du=dx\right),\left(dv=\mathrm{sin}xdx⇒v=-\mathrm{cos}x\right)\right\}$
$=-x\mathrm{cos}x+\int \mathrm{cos}xdx$
$=-x\mathrm{cos}x+\mathrm{sin}x+C$
Therefore,

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