Given the ODE $x'=3x^{2/3}$ with the initial condition $x(0)=0$, I tried to use an integration multiplicator method, but my solution is incorrect, I assume it's related to equation not being linear. As there are no $t$ argument, I cannot transform equation into $t(x)$ form or can I? Or maybe there is altogether different method how to solve this problem?
2026-03-25 06:50:35.1774421435
How to solve Cauchy problem if equation is not linear and there is no $t$ argument?
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Assuming you mean $x'-3x^{2/3} = 0 \iff x' = 3x^{2/3}$, in which case you integrate both sides: $$ \frac{dx}{dt} = 3x^{2/3} \iff \int dt = \int \frac{dx}{3x^{2/3}} $$ can you finish?