Mary Keefe

2022-01-19

I need help
Solve the following first order differential equation:
$3t{x}^{\prime }-2x=\frac{{t}^{3}}{{x}^{2}}$

Dabanka4v

Expert

Given
$3t{x}^{\prime }-2x=\frac{{t}^{3}}{{x}^{2}}$
$3t\frac{dx}{dt}-2x=\frac{{t}^{3}}{{x}^{2}}$
$\frac{dx}{dt}-\frac{2x}{3t}=\frac{{t}^{3}}{{x}^{23}t}$
$\frac{dx}{dt}-\frac{2x}{3t}=\frac{{t}^{2}}{3{x}^{2}}$
${x}^{2}\frac{dx}{dt}-\frac{2{x}^{3}}{3t}={t}^{2}v$
$3{x}^{2}\frac{dx}{dt}-\frac{2{x}^{3}}{t}=3{t}^{2}$
Let ${x}^{3}=u$
$3{x}^{2}\frac{dx}{dt}=\frac{du}{dt}$
$\frac{du}{dt}-\frac{2}{t}u=e{t}^{2}$
$\frac{du}{dt}+Pu=Q$ where

$If={e}^{\int Pdt}={e}^{\int \frac{-2}{t}dt}={e}^{-2\mathrm{ln}t}=\frac{1}{{t}^{2}}$
Solution:
$u\left(If\right)=\int Q\left(If\right)dt+c$
${x}^{3}\frac{1}{{t}^{2}}=\int 3{t}^{2}\frac{1}{{t}^{2}}dt+c=3t+c$
$\frac{{x}^{3}}{{t}^{2}}=3t+c$

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