Question

Solve differential equation \frac{dy}{dx}+\frac{y}{4x}= 3x, \ y(2)=3

First order differential equations
Solve differential equation $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{\frac{{{y}}}{{{4}{x}}}}={3}{x},\ {y}{\left({2}\right)}={3}$$

2021-02-26

$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{P}{y}={Q}$$
$$\displaystyle{P}={\frac{{{1}}}{{{4}{x}}}},\ {Q}={3}{x}$$
$$\displaystyle{I}.{F}.={e}^{{\int{P}{\left.{d}{x}\right.}}}$$
$$\displaystyle={e}^{{\int{\frac{{{1}}}{{{4}{x}}}}{\left.{d}{x}\right.}}}$$
$$\displaystyle={e}^{{{\frac{{{1}}}{{{4}}}}\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}}}$$
$$\displaystyle={e}^{{{\frac{{{1}}}{{{4}}}}{\log{{x}}}}}$$
$$\displaystyle={e}^{{{\log{\sqrt{{{4}}}}}{\left\lbrace{x}\right\rbrace}}}=\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}$$
$$\displaystyle{y}{\left({I}.{F}.\right)}=\int{\left({I}.{F}.\right)}{Q}{\left.{d}{x}\right.}$$
$$\displaystyle\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}{\left({y}\right)}=\int\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}{\left({3}{x}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle\Rightarrow\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}{\left({y}\right)}=\int{x}^{{{\frac{{{1}}}{{{4}}}}}}{\left({3}{x}\right)}{\left.{d}{x}\right.}=\int{3}{x}^{{{\frac{{{1}}}{{{4}}}}+{1}}}{\left.{d}{x}\right.}$$
$$= 3 \int x^{\frac{5}{4}}dx=4 \Bigg[\frac{x^{\frac{5}{4}+1}}{\frac{5}{4}+1}\Bigg]+c= 3 \Bigg[\frac{x^{\frac{9}{4}}}{\frac{9}{4}}\Bigg]+c =3*\frac{4}{9} \bigg[ x^{\frac{9}{4}}\bigg]+c= \frac{4}{3}x^{\frac{9}{4}}+c$$
$$\displaystyle\Rightarrow\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}{\left({y}\right)}={\frac{{{4}}}{{{3}}}}{x}^{{{\frac{{{9}}}{{{4}}}}}}+{c}$$
$$\displaystyle\Rightarrow{y}{\left({x}\right)}={\frac{{{\frac{{{4}}}{{{3}}}}{x}^{{{\frac{{{9}}}{{{4}}}}}}+{c}}}{{\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}}}}$$ (1)
Using initial condition $$\displaystyle{x}={2}\Rightarrow{y}{\left({2}\right)}={3}$$
$$\displaystyle\Rightarrow{3}={\frac{{{\frac{{{4}}}{{{3}}}}{\left({2}\right)}^{{{\frac{{{9}}}{{{4}}}}}}+{c}}}{{\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}}}}$$
$$\displaystyle\Rightarrow{3}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}={\frac{{{4}}}{{{3}}}}{\left({2}\right)}^{{{\frac{{{9}}}{{{4}}}}}}+{c}$$ $$\displaystyle\Rightarrow{c}={3}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}-{\frac{{{4}}}{{{3}}}}{\left({2}\right)}^{{{\frac{{{9}}}{{{4}}}}}}$$
put the value of c in equation (1)
$$\displaystyle\Rightarrow{y}{\left({x}\right)}={\frac{{{\frac{{{4}}}{{{3}}}}{x}^{{{\frac{{{9}}}{{{4}}}}}}+{3}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}-{\frac{{{4}}}{{{3}}}}{\left({2}\right)}^{{{\frac{{{9}}}{{{4}}}}}}}}{{\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}}}}$$
$$\displaystyle={\frac{{{\frac{{{4}}}{{{3}}}}{x}^{{{\frac{{{9}}}{{{4}}}}}}+{3}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}-{\frac{{{4}}}{{{3}}}}{\left({4}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}\right)}}}{{\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}}}}$$
$$\displaystyle={\frac{{{\frac{{{4}}}{{{3}}}}{x}^{{{\frac{{{9}}}{{{4}}}}}}+{\frac{{{9}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}-{16}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}}}{{{3}}}}}}{{\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}}}}$$
$$\displaystyle={\frac{{{\frac{{{4}}}{{{3}}}}{x}^{{{\frac{{{9}}}{{{4}}}}}}-{\frac{{{7}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}}}{{{3}}}}}}{{\sqrt{{{4}}}{\left\lbrace{x}\right\rbrace}}}}$$
$$\displaystyle={\frac{{{4}}}{{{3}}}}{x}^{{{\frac{{{9}-{1}}}{{{4}}}}}}-{\frac{{{7}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}}}{{{3}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}}}}$$
$$\displaystyle={\frac{{{4}}}{{{3}}}}{x}^{{{\frac{{{8}}}{{{4}}}}}}-{\frac{{{7}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}}}{{{3}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}}}}$$
$$\displaystyle{y}{\left({x}\right)}={\frac{{{4}}}{{{3}}}}{x}^{{{\frac{{{8}}}{{{4}}}}}}-{\frac{{{7}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}}}{{{3}\sqrt{{{4}}}{\left\lbrace{2}\right\rbrace}}}}$$