Let's generalize a formula $to$ solve the first order linear differential equa$-$

tion:

$y^{\text{'}}+P(x)y=Q(x)$

$By$ getting the integrating factor: $r(x)=e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx},$ then multiply each

term $by$ integrating factor:

$e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}{y}^{\text{'}}+e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}P(x)y=e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}Q(x)$

$(e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}y{)}^{\text{'}}=e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}Q(x)$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}y{)}^{\text{'}}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}Q(x)dx$

$e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}y={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}Q(x)dx$

$y=\frac{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}Q(x)dx}{e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}}$

Then $we$ could use the general solution with formula

$y=\frac{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}Q(x)dx}{e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}P(x)dx}}P(x),Q(x),$ you don't need $to$ solve $it.y^{\text{'}}+x^{5}y=e^x$

Let $P(x)=x^5,Q(x)=e^x,$ then

$y=\frac{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^{5}dx}{e}^x(d)x}{e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{x}^{5}dx}}P(x),Q(x),$ you don't need $to$ solve $it.y^{\text{'}}+x^{4}y=sin(2x)$