nitraiddQ

2021-02-25

Solve differential equation

Derrick

$\frac{dy}{dx}+Py=Q$

$I.F.={e}^{\int Pdx}$
$={e}^{\int \frac{1}{4x}dx}$
$={e}^{\frac{1}{4}\int \frac{1}{x}dx}$
$={e}^{\frac{1}{4}\mathrm{log}x}$
$={e}^{\mathrm{log}\sqrt{4}\left\{x\right\}}=\sqrt{4}\left\{x\right\}$
$y\left(I.F.\right)=\int \left(I.F.\right)Qdx$
$\sqrt{4}\left\{x\right\}\left(y\right)=\int \sqrt{4}\left\{x\right\}\left(3x\right)dx$
$⇒\sqrt{4}\left\{x\right\}\left(y\right)=\int {x}^{\frac{1}{4}}\left(3x\right)dx=\int 3{x}^{\frac{1}{4}+1}dx$
$=3\int {x}^{\frac{5}{4}}dx=4\left[\frac{{x}^{\frac{5}{4}+1}}{\frac{5}{4}+1}\right]+c=3\left[\frac{{x}^{\frac{9}{4}}}{\frac{9}{4}}\right]+c=3\ast \frac{4}{9}\left[{x}^{\frac{9}{4}}\right]+c=\frac{4}{3}{x}^{\frac{9}{4}}+c$
$⇒\sqrt{4}\left\{x\right\}\left(y\right)=\frac{4}{3}{x}^{\frac{9}{4}}+c$
$⇒y\left(x\right)=\frac{\frac{4}{3}{x}^{\frac{9}{4}}+c}{\sqrt{4}\left\{x\right\}}$ (1)
Using initial condition $x=2⇒y\left(2\right)=3$
$⇒3=\frac{\frac{4}{3}{\left(2\right)}^{\frac{9}{4}}+c}{\sqrt{4}\left\{2\right\}}$
$⇒3\sqrt{4}\left\{2\right\}=\frac{4}{3}{\left(2\right)}^{\frac{9}{4}}+c$

Jeffrey Jordon