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##\frac{dy}{dx}=4y^2-y##

##\frac{dx}{dy}=\frac{1}{4y^2-y}=\frac{1}{y\left(4y-1\right)}=\frac{4}{4y-1}-\frac{1}{y}##

##\int _{ }^{ }\frac{dx}{dy}dy=\int _{ }^{ }\frac{4}{4y-1}dy-\int _{ }^{ }\frac{1}{y}dy##

##x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)+c##

Substituting in ##y=1## and ##x=0##, we can find ##c##:

##x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)+c##

##0=\ln \left(3\right)+c,\ c=-\ln \left(3\right)##

##x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)-\ln \left(3\right)##

##x=\ln \left(\frac{\left|4y-1\right|}{3\left|y\right|}\right)##

##3e^x=\left|4-\frac{1}{y}\right|##

##4-\frac{1}{y}=-3e^x,4-\frac{1}{y}=3e^x##

Because we know that ##y## has to equal postive ##1## when ##x## equals ##0##;

##4-1=-3e^0,\ 4-1=3e^0##

##3=-3,\ 3=3##

As the first statement is false, the second one must be true.

##∴4-\frac{1}{y}=3e^x##

##\frac{1}{y}=4-3e^x##

##y=\frac{1}{4-3e^x}##.

So as you can see it's relatively simple to solve a differential equation of the form ##\frac{dy}{dx}=f\left(y\right)##, but I'm wondering if it would be possible to solve one in the form of ##\frac{d^2y}{dx^2}=f\left(y\right)##. I've tried looking it up online but I haven't been able to find anything useful.