Initial-value problems

Worked solution

1. 1

   Integrate the derivative.

   $y=\int \left(3x^2-2x\right)dx=x^3-x^2+C$

   Each polynomial term uses the reverse power rule.

2. 2

   Substitute the known point.

   $\begin{array}{rl}4& =1^3-1^2+C\\ C& =4\end{array}$

   The condition supplies a value of the function rather than a value of its derivative.

3. 3

   Write the particular function.

   $y=x^3-x^2+4$

   The computed constant selects one member of the antiderivative family.

4. 4

   Verify both requirements.

   $\begin{array}{rl}{y}^{\prime }& =3x^2-2x\\ y\left(1\right)& =1-1+4=4\end{array}$

   A valid solution must satisfy both the derivative equation and the initial condition.

Worked solution

1. 1

   Integrate the exponential rate.

   $V\left(t\right)=8e^{t/2}+C$

   Differentiating the exponential contributes a factor of one half, so the coefficient doubles from 4 to 8.

2. 2

   Use the initial volume.

   $\begin{array}{rl}10& =8e^0+C\\ 10& =8+C\\ C& =2\end{array}$

   The exponential term already contributes 8 litres at the start.

3. 3

   State the volume function.

   $V\left(t\right)=8e^{t/2}+2 \mathrm{L}$

   The particular constant gives a volume, measured in litres, consistent with the starting amount.

4. 4

   Check the rate and starting value.

   $\begin{array}{rl}{V}^{\prime }\left(t\right)& =4e^{t/2}\\ V\left(0\right)& =8+2=10\end{array}$

   Both the specified flow rate and the specified initial volume are recovered.