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Okay experts, I was trying to solve questions on differentials last night, but got across this question which I have no idea how to solve. $\frac{dx}{dy}=(2x+3y-4)^2$. Can someone please let me know how to solve such kind of differentials?

My try:*I tried to take the right hand side as $t$ and solve it but could not get any further. Please help. *

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    $\begingroup$ Hint: Try a substitution, maybe $v = 2x + 3 y$. Find he derivative, substitute and solve. $\endgroup$ – Moo Mar 3 '17 at 3:08
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Consider $$\frac{dx}{dy}=(2x+3y-4)^2$$ and define $$2x+3y-4=z\implies x=-\frac{3 y}{2}+\frac{z}{2}+2\implies\frac{dx}{dy}=-\frac{3}{2}+\frac{1}{2}\frac{dz}{dy}$$ So, the differential equation becomes $$-\frac{3}{2}+\frac{1}{2}\frac{dz}{dy}=z^2\implies \frac{dz}{dy}-2z^2=3$$ which becomes simple.

I am sure that you can take it from here.

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