Physix & Maths

Multiplying $f(x)$ and $g(x)$ to differential equations $f'(x)=\frac{-f(x)}{\sqrt{f^{2}(x)+g^{2}(x)}}$ and $g(x)=1-\frac{g(x)}{\sqrt{f^{2}(x)+g^{2}(x)}}$ respectively.

Now adding both the equations we get $f(x)f'(x)+g(x)g'(x)=g(x)-\frac{f^{2}(x)+g^{2}(x)}{\sqrt{f^{2}(x)+g^{2}(x)}}$

Simplifying it we get $f(x)f'(x)+g(x)g'(x)=g(x)-\sqrt{f^{2}(x)+g^{2}(x)}$

Dividing both side by $\sqrt{f^{2}(x)+g^{2}(x)}$ we get $\frac{f(x)f'(x)+g(x)g'(x)}{\sqrt{f^{2}(x)+g^2(x)}}=-(1-\frac{g(x)}{\sqrt{f^{2}(x)+g^{2}(x)}})$

Substituting $g'(x)$ in the above equation and integrating both sides we get $\sqrt{f^{2}(x)+g^{2}(x)}=c-g(x)$ where $c$ is a constant.

Now putting $x=0$ in the above equation we get the value of $c$ . Then putting $(\psi\rightarrow ∞)$ in the above equation we get the value of $g(\psi)$ which is equal to $\boxed{500}$

$\frac{g'(x)}{f'(x)}=\frac{g(x)-\sqrt{f^2(x) +g^2(x)}}{f(x)}$

$\frac{dg(x)}{df(x)}=\frac{g(x)-\sqrt{f^2(x) +g^2(x)}}{f(x)}$

using $u=\frac{g(x)}{f(x)}$ so

$dg(x)=udf(x)+f(x)du$

when you separate yields:

$\frac{df(x)}{f(x)}+\frac{du}{\sqrt{u^2 +1}}=0$

by integrating we yield that:

$ln|f(x)| +ln|u+\sqrt{u^2 +1}|=C_1$

so $g(x)+\sqrt{f^2(x) +g^2(x)} =e^C_1$

while $g(0)=0$ and $f(0)=1000$ hence $e^C_1 =1000$

when $s\rightarrow ∞$

$g(s)+\sqrt{f^2(s) +g^2(s)} =1000$

while $f(s)=0$ when $S\rightarrow ∞$

$g(s)+\sqrt{0+g^2(s)}=1000 \rightarrow g(s)=\frac{1000}{2} =500$ .