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$x{y}^{\prime}=y+{x}^{2}+{y}^{2}$

Verify

is a solution to

jeffster830gyz

Beginner2022-04-22Added 21 answers

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:

$x[x{\mathrm{sec}}^{2}\left(x\right)+\mathrm{tan}\left(x\right)]={x}^{2}[{\mathrm{tan}}^{2}\left(x\right)+1]+y$

$={x}^{2}{\mathrm{tan}}^{2}\left(x\right)+{x}^{2}+y={y}^{2}+{x}^{2}+y.$

elvis0217t2x

Beginner2022-04-23Added 13 answers

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

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