Hello,

I have solved this differential equation, but I want to know whether all the steps are correct.

y'(x)=(4x+3y-1)^2

Let u=4x+3y-1 then du/dx =4+3y'(x)

This implies y'(x)=(1/3)u'(x)-4/3

Now

(1/3)u'(x)-4/3=u^2

Hence,u'(x)=3u^2+4

du/(3u^2+4)=dx

Integrating both the sides,we get

1/2sqrt(3) arctan(sqrt(3)u)/2=x+c

So my ans is sqrt(3)*(4x+3y-1)= 2tan(2sqrt(3)*x)+c

If I am wrong, reply where I am wrong.