Question

(b) find an expression for ( y = f(x) ) by solving the differential equation ( \frac{dy}{dx} = \frac{3x^{2}}{y} ) with the initial condition ( f(2) = - 8 ).

Explanation:

Step1: Separate variables

Separate the variables in the differential equation \(\frac{dy}{dx}=\frac{3x^{2}}{y}\). We get \(y\;dy = 3x^{2}\;dx\).

Step2: Integrate both sides

Integrate both sides of the equation.
For the left - hand side, \(\int y\;dy=\frac{y^{2}}{2}+C_{1}\).
For the right - hand side, \(\int3x^{2}\;dx=x^{3}+C_{2}\).
So, \(\frac{y^{2}}{2}=x^{3}+C\) (where \(C = C_{2}-C_{1}\)).

Step3: Use the initial condition

We are given the initial condition \(f(2)=-8\). Substitute \(x = 2\) and \(y=-8\) into the equation \(\frac{y^{2}}{2}=x^{3}+C\).
\(\frac{(-8)^{2}}{2}=2^{3}+C\).
\(\frac{64}{2}=8 + C\).
\(32=8 + C\), then \(C = 24\).

Step4: Solve for \(y\)

From \(\frac{y^{2}}{2}=x^{3}+24\), we can solve for \(y\).
\(y^{2}=2x^{3}+48\).
Since \(y=-8\) when \(x = 2\), \(y=-\sqrt{2x^{3}+48}\) (we take the negative square root because of the initial condition \(y=-8\)).

Answer:

\(y =-\sqrt{2x^{3}+48}\)