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. *

  • 2
    $\begingroup$ Hint: Try a substitution, maybe $v = 2x + 3 y$. Find he derivative, substitute and solve. $\endgroup$ – Moo Mar 3 '17 at 3:08

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.