The problem continues! Original differential equation: $y^{\text{'}}2xy=x{e}^{-x^2}$

Copy over your results so far: $,p(x)=2x,f(x)=xe,y_1=e^{-x}$

$2/2$

$($e$)$ We're looking for solutions of the form $y=u{y}_1$ where $u$ is some function of $x.$

Starting with the general form $y^{\text{'}}p(x)y=f(x),$ we can substitute $y=u{y}_1$ to get:

$y^{\text{'}}p(x)y=f(x)$

$(u{y}_1{)}^{\text{'}}p(x)(u{y}_1)=f(x)$

applying the product rule $(u^{\text{'}}{y}_{1}u{y}_1^{\text{'}})up(x)*y_1=f(x)$

$u^{\text{'}}{y}_{1}u{y}_1^{\text{'}}up(x)y_1=f(x)$

factoring out $u,u^{\text{'}}{y}_{1}u(y_1^{\text{'}}p(x)y_1)=f(x)$

remember $y_1^{\text{'}}p(x)y_1=0,u^{\text{'}}{y}_{1}u*0=f(x)$

$u^{\text{'}}{y}_1=f(x)$

Write an equation that represents the condition $u^{\text{'}}{y}_1=f(x)$ for this problem.

$($f$)$ Solve for $u^{\text{'}},$ then integrate both sides to find $u.$ Your answer will involve an unknown constant $c.$

$u^{\text{'}}=$

$u=$

$($g$)$ Write your general solution $y=u{y}_1$ here.

$y=$

$($h$)$ Check that your solution $y$ satisfies the original differential equation.