2021-12-26

The following differential equations appear similar but have very different solutions:
$dydx=x$,
$dydx=y$
Solve both subject to the initial condition $y\left(1\right)=2$

Lindsey Gamble

Given:
Separation of variables

Integrate both sides

Use:
$y=\frac{{x}^{1+1}}{1+1}+C$
$y=\frac{{x}^{2}}{2}+C$
$y\left(1\right)=2$ (Given)
Solve the original equation at $x=1$
$2=\frac{{1}^{2}}{1}+C$
$2=1+C$
$C=1$
The constant is $C=1$ and place in the equation for the final answer
$y=\frac{{x}^{2}}{2}+1$
Given:
Separation of variables

Integrate both sides

Use:
$\mathrm{ln}|y|=x+C$
$y\left(1\right)=2$ (Given)
Solve the original equation at $x=1$
$\mathrm{ln}|2|=1+C$
$C=\mathrm{ln}|2|-1$
The constant is $C=\mathrm{ln}|2|-1$ and place in the equation for the final answer
$\mathrm{ln}|y|=x+\mathrm{ln}|2|-1$
Simplify:
$\mathrm{ln}|y|-\mathrm{ln}|2|=x-1$
Using: $\mathrm{ln}|\frac{y}{x}|=\mathrm{ln}|y|-\mathrm{ln}|x|$
$\mathrm{ln}|\frac{y}{2}|=x-1$
Result: $y=\frac{{x}^{2}}{2}+1\text{and}\mathrm{ln}|\frac{y}{2}|=x-1$

Raymond Foley

1) $\frac{dy}{dx}=x$
$⇒\int dy=\int xdx+c$, c is int const.
$⇒y=\frac{{x}^{2}}{2}+c$
Now, $y\left(7\right)=3$
$\therefore y=3\text{when}x=7$
$\therefore 3=\frac{49}{2}+c⇒c=3-\frac{49}{2}=\frac{6-49}{2}=-\frac{43}{2}$
$\therefore y=\frac{1}{2}{x}^{2}-\frac{43}{2}=\frac{1}{2}\left({x}^{2}-43\right)$
2) $\frac{dy}{dx}=y$
$⇒\int \frac{dy}{y}=\int dx+\mathrm{log}c,\mathrm{log}c$ is int const.
$⇒{\mathrm{log}}_{e}y=x+{\mathrm{log}}_{e}e$
$⇒{\mathrm{log}}_{e}y-\mathrm{log}c=x⇒{\mathrm{log}}_{e}\left(\frac{y}{c}\right)=x$
$⇒\frac{y}{c}={e}^{x}⇒y=c{e}^{x}$
Since
$\therefore 3=c{e}^{7}⇒c=\frac{3}{{e}^{7}}$
$\therefore y=\frac{3}{{e}^{7}}{e}^{x}=3{e}^{x-7}$

karton