Question

find the general solution of dy/dx=(x+y)^2 /5,y(0)=1
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Answer 1

We're given the differential equation,

$\small\longrightarrow\dfrac{dy}{dx}=\dfrac{(x+y)^2}{5}\quad\quad\dots(1)$

Substitute,

$\small\longrightarrow u=x+y$

$\small\longrightarrow y=u-x$

$\small\longrightarrow\dfrac{dy}{dx}=\dfrac{du}{dx}-1$

Then (1) becomes,

$\small\longrightarrow\dfrac{du}{dx}-1=\dfrac{u^2}{5}$

$\small\longrightarrow\dfrac{du}{dx}=\dfrac{u^2+5}{5}$

$\small\longrightarrow\dfrac{dx}{du}=\dfrac{5}{u^2+5}$

$\small\displaystyle\longrightarrow x=\int\dfrac{5}{u^2+5}\ du$

$\small\displaystyle\longrightarrow x=5\int\dfrac{1}{u^2+\left(\sqrt5\right)^2}\ du$

We have,

• $\small\displaystyle\int\dfrac{dx}{(ax+b)^2+c^2}=\dfrac{1}{ac}\,\tan^{-1}\left(\dfrac{ax+b}{c}\right)+C$

Thus,

$\small\longrightarrow x=5\cdot\dfrac{1}{1\cdot\sqrt5}\,\tan^{-1}\left(\dfrac{u}{\sqrt5}\right)+C$

$\small\longrightarrow x=\sqrt5\,\tan^{-1}\left(\dfrac{u}{\sqrt5}\right)+C$

Taking $\smallu=x+y,$

$\small\longrightarrow x=\sqrt5\,\tan^{-1}\left(\dfrac{x+y}{\sqrt5}\right)+C\quad\quad\dots(2)$

$\small\longrightarrow x-\sqrt5\,\tan^{-1}\left(\dfrac{x+y}{\sqrt5}\right)=C$

At $\small(x,\ y)=(0,\ 1),$

$\small\longrightarrow -\sqrt5\,\tan^{-1}\left(\dfrac{1}{\sqrt5}\right)=C$

Then (2) becomes,

$\small\longrightarrow x=\sqrt5\,\tan^{-1}\left(\dfrac{x+y}{\sqrt5}\right)-\sqrt5\,\tan^{-1}\left(\dfrac{1}{\sqrt5}\right)$

$\small\longrightarrow x=\sqrt5\left[\tan^{-1}\left(\dfrac{x+y}{\sqrt5}\right)-\tan^{-1}\left(\dfrac{1}{\sqrt5}\right)\right]$

We have,

• $\small\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\dfrac{a-b}{1+ab}\right)$

Then,

$\small\text{ \longrightarrow x=\sqrt5\tan^{-1}\left(\dfrac{\frac{x+y-1}{\sqrt5}}{1+\frac{x+y}{5}}\right) }$

$\small\longrightarrow x=\sqrt5\tan^{-1}\left(\dfrac{\sqrt5(x+y-1)}{x+y+5}\right)$

$\small\text{ \longrightarrow\dfrac{x+y-1}{x+y+5}=\dfrac{\tan\left(\dfrac{x}{\sqrt5}\right)}{\sqrt5} }$

By rule of componendo and dividendo,

$\small\text{ \longrightarrow\dfrac{(x+y+5)+(x+y-1)}{(x+y+5)-(x+y-1)}=\dfrac{\sqrt5+\tan\left(\dfrac{x}{\sqrt5}\right)}{\sqrt5-\tan\left(\dfrac{x}{\sqrt5}\right)} }$

$\small\text{ \longrightarrow\dfrac{x+y+2}{3}=\dfrac{\sqrt5+\tan\left(\dfrac{x}{\sqrt5}\right)}{\sqrt5-\tan\left(\dfrac{x}{\sqrt5}\right)} }$

$\small\text{ \longrightarrow\underline{\underline{y=\dfrac{3\left(\sqrt5+\tan\left(\dfrac{x}{\sqrt5}\right)\right)}{\sqrt5-\tan\left(\dfrac{x}{\sqrt5}\right)}-x-2}} }$