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Best questions on differential eqns

OK ,I tried solving this and what I did was 1)Assumed x-3 =X and y+3=Y ,and then 2)We get expression only in X and Y devoid of constant.. 3)What to do now I can't think..

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    you can insert the solutions a) to d) and lokk if the equation is fulfilled – 2017-01-20
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    But it will be way too complicated... – 2017-01-20

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Hint: Introduce a substitution for $x=u+x_0$ and $y=v+y_0$.

$$\frac{x+2y-3}{2x+y+3}=\frac{(u+x_0)+2(v+y_0)-3}{2(u+x_0)+(v+y_0)+3}=\frac{u+2v+x_0+2y_0-3}{2u+v+2x_0+y_0+3}$$

Now, we have to choose $(x_0,y_0)$ in such a way that:

$$x_0+2y_0-3=0 \qquad \wedge \qquad 2x_0+y_0+3=0.$$

If you have found $x_0$ and $y_0$ this substitution will reduce your problem to:

$v'=(\frac{u+2v}{2u+v})^2=(\frac{1+2v/u}{2+v/u})^2.$

The solution of the linear system is $(x_0,y_0)=(-3,3)$. This ODE can be solved by using $v=u\cdot z(u)$ substitution.

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    But it is not only that there is a square of the expression on RHS that's causing problem.... – 2017-01-20
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    That is not a big problem. The square term will become a function of $z$ only after substitution. BTW I think just trying the solutions will be faster. – 2017-01-20