Kymani Shepherd

2022-04-21

Verify $y=x\mathrm{tan}x$ is solution to ODE: $x{y}^{\prime }=y+{x}^{2}+{y}^{2}$
Verify $Sy=x\mathrm{tan}\left\{x\right\}$
is a solution to $x{y}^{\prime }=y+{x}^{2}+{y}^{2}$

jeffster830gyz

Step 1
$y=x\mathrm{tan}\left(x\right)$.
Verify that $x{y}^{\text{'}}=y+{x}^{2}+{y}^{2}$.
Step 2
Alternate approach:
$\frac{d}{dx}\mathrm{tan}\left(x\right)={\mathrm{sec}}^{2}\left(x\right).$
${\mathrm{tan}}^{2}\left(x\right)+1={\mathrm{sec}}^{2}\left(x\right).$
LHS equals:
$x\left[x{\mathrm{sec}}^{2}\left(x\right)+\mathrm{tan}\left(x\right)\right]={x}^{2}\left[{\mathrm{tan}}^{2}\left(x\right)+1\right]+y$
$={x}^{2}{\mathrm{tan}}^{2}\left(x\right)+{x}^{2}+y={y}^{2}+{x}^{2}+y.$

elvis0217t2x

$\begin{array}{rl}{y}^{\text{'}}& =\mathrm{tan}x+x\frac{1}{{\mathrm{cos}}^{2}x}\\ {y}^{\text{'}}& =\mathrm{tan}x+x\frac{{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}\\ {y}^{\text{'}}& =\mathrm{tan}x+x+x{\mathrm{tan}}^{2}x\\ {y}^{\text{'}}& =y/x+x+{y}^{2}/x\\ x{y}^{\text{'}}& =y+{x}^{2}+{y}^{2}\end{array}$