Question

Pls help me solve the questions in d PDF

Answer 1

I'll take a stab at what I think is the most interesting problem in the attachment.
$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y+x-2}{y-x-4}$

Substitute $y=s-m$ and $x=t-n$, so that $\mathrm dy=\mathrm ds$ and $\mathrm dx=\mathrm dt$. Then the ODE becomes

$\dfrac{\mathrm ds}{\mathrm dt}=\dfrac{s-m+t-n-2}{s-m-(t-n)-4}=\dfrac{s+t-m-n-2}{s-t-m+n-4}$

We would like to have this be a homogeneous equation, which is to say, given

$\dfrac{\mathrm ds}{\mathrm dt}=f(t,s)$

there is some $\alpha$ for which $f(kt,ks)=k^\alpha f(t,s)$. In order for this to be true, we require that the constant terms in the numerator and denominator vanish. That is,

$\begin{cases}-m-n-2=0\\-m+n-4=0\end{cases}$

which is easy to solve; you'd find that $m=-3$ and $n=1$. Now, the ODE is

$\dfrac{\mathrm ds}{\mathrm dt}=\dfrac{s+t}{s-t}$

Substitute $s=tu$, so that $\dfrac{\mathrm ds}{\mathrm dt}=u+t\dfrac{\mathrm du}{\mathrm dt}$, and the ODE is

$u+t\dfrac{\mathrm du}{\mathrm dt}=\dfrac{tu+t}{tu-t}=\dfrac{u+1}{u-1}$
$t\dfrac{\mathrm du}{\mathrm dt}=\dfrac{u+1-u(u-1)}{u-1}$
$t\dfrac{\mathrm du}{\mathrm dt}=\dfrac{1+2u-u^2}{u-1}$

which is separable. You have

$\dfrac{u-1}{1+2u-u^2}\,\mathrm du=\dfrac{\mathrm dt}t$

and integrating both sides is easy. Substituting $v=1+2u-u^2$, so that $\mathrm dv=(-2u+2)\,\mathrm du$, you have

$\displaystyle-\frac12\int\dfrac{\mathrm dv}v\,\mathrm du=\int\dfrac{\mathrm dt}t$
$-\dfrac12\ln|v|=\ln|t|+C$
$-\dfrac12\ln|1+2u-u^2|=\ln|t|+C$
$-\dfrac12\ln\left|1+\dfrac{2s}t-\dfrac{s^2}{t^2}\right|=\ln|t|+C$
$-\dfrac12\ln\left|1+\dfrac{2(y-3)}{x+1}-\dfrac{(y-3)^2}{(x+1)^2}\right|=\ln|x+1|+C$