Solve ODE $y'' + (y')^2 + y = \ln(x)$
Solution 1:
Answer
Take $$u(x)=ln(x)-y$$ so $$u'=1/x-y'$$ and $$u''=-1/x^2-y''$$
We have: $$u''+(2/x)u'-(u')^2+u'=0$$ Now, we employ the substitution $$p(x)=u'$$ It is immediate that $$p'+(2/x)p=p^2-p$$
$$p(x)= 1/(x*(c1*x*exp(x)+x*Ei(-x)+1))$$
Now continue...
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