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{\bf Question}

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Find the solution of the initial value problem $$x^2 t
\frac{dx}{dt} = x^3 + t^3$$

subject to $x(1)=0$

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{\bf Answer}

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$\displaystyle x^2 t \frac{dx}{dt} = x^3 + t^3 \Rightarrow
\frac{dx}{dt} = \frac{x}{t} + \left(\frac{t}{x}\right)^2 =
f\left(\frac{x}{t}\right)$

This is of the form $y = \frac{x}{t} \Rightarrow \frac{dx}{dt} = t
\frac{dy}{dt} + y = y + \frac{1}{y^2} \Rightarrow t\frac{dy}{dt} =
\frac{1}{y^2}$

Thus $\displaystyle \int y^2 dy = \int\frac{dt}{t} \Rightarrow
\frac{1}{3}y^3 = \ln |t| + \mathrm{constant}$

Now apply the initial condition $x(1) = 0 $ for $t>1$ so $|t| = t$

$y(1) = \frac{x(1)}{t = 1} = x(1)$ and $\ln 1 = 0 \Rightarrow
\mathrm{constant} = 0$

Therefore $$x^3 = 3 \ln t \Rightarrow x = \sqrt[3]{3 \ln t}$$



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