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Question

(A) Given the differential equation: $t^2 dx + (x^2 - tx + t^2) dt = 0$. Rearranging the equation: $t^2 dx = -(x^2 - tx + t^2) dt$, which gives $\frac

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$x = Vt$

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(A) Given the differential equation: $t^2 dx + (x^2 - tx + t^2) dt = 0$. Rearranging the equation: $t^2 dx = -(x^2 - tx + t^2) dt$, which gives $\frac{dx}{dt} = -\frac{x^2 - tx + t^2}{t^2}$. This can be written as $\frac{dx}{dt} = -(\frac{x}{t})^2 + \frac{x}{t} - 1$. This is a homogeneous differential equation of the form $\frac{dx}{dt} = f(\frac{x}{t})$. To solve a homogeneous differential equation, we use the substitution $x = Vt$, where $V$ is a function of $t$.

The substitution required to reduce the differential equation $t^2 dx + (x^2 - tx + t^2) dt = 0$ to a differential equation which can be solved by the variables separable method is

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