Solve the linear equations by considering y as a function of x, that is, y=y(x).xy′+(1+x)y=e−xsin⁡2x

Variation of parameters
First, solve the linear homogeneous equation by separating variables.
Rearranging terms in the equation gives
${\displaystyle x{y}^{\prime }=-(1+x)y\iff \frac{dy}{dx}=-\frac{(1+x)y}{x}\iff \frac{d}{y}=-\frac{(1+x)dx}{x}}$
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
${\displaystyle x{y}^{\prime }=-(1+x)y\iff \frac{dy}{dx}=-\frac{(1+x)y}{x}\iff \frac{d}{y}=-\frac{(1+x)dx}{x}}$
Let’s solve the integral on the right side.
${\displaystyle \int \frac{(1+x)dx}{x}=\int (\frac{1}{x}+\frac{x}{x})dx}$
${\displaystyle =\int (\frac{1}{x}+1)dx}$
${\displaystyle =\int \frac{dx}{x}+\int dx}$
${\displaystyle =\ln \left|x\right|+x+c}$
Therefore,
${\displaystyle \ln \left|y\right|=-\ln \left|x\right|-x+c}$
By taking exponents, we obtain
${\displaystyle \left|y\right|=e^{-\ln \left|x\right|-x+c}=\frac{1}{\left|x\right|e^x}\cdot e\cdot c}$
Hence,we obtain
${\displaystyle y=\frac{C}{x{e}^x}}$
where ${\displaystyle C=\pm e^c\text{and}{y}_c=\frac{1}{x{e}^x}}$ 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.
Write the equation in the standard form (divide it by x 4 0)
${\displaystyle y^{\prime }+\frac{(1+x)y}{x}=\frac{e^{-x}\sin 2x}{x}}$
Substituting ${\displaystyle displaysty\le \left\{u\right\}{\left\{y\right\}}_{\left\{c\right\}}}$ and its derivative in the equation gives
${\displaystyle u{y}_c}$
${\displaystyle {\left(u{y}_c\right)}^{\prime }+\frac{(1+x)u{y}_c}{x}=\frac{e^{-x}\sin 2x}{x}}$
${\displaystyle u^{\prime }{y}_c+u(y^{\prime }c+\underbrace{\frac{(1+x)y_c}{x}})=0\text{since}{y}_c\text{is a solution}}$