Question

First find the general solution (involving a constant \(\mathrm{C}\) ) for the given differential equation. Then find the particular solution that satisfies the indicated condition. $$ \frac{d z}{d t}=t^{2} z^{2} ; z=1 / 3 \text { at } t=1 $$

Short answer

The particular solution is \(z = -\frac{27}{9t^3 - 30}\).

Step-by-step solution

Separate Variables

The given differential equation is \( \frac{dz}{dt} = t^2 z^2 \). To solve it by separation of variables, divide both sides by \( z^2 \) and multiply by \( dt \) to get \( \frac{dz}{z^2} = t^2 dt \). This allows us to integrate each side separately.

Integrate Both Sides

Integrate \( \int \frac{dz}{z^2} \) on the left and \( \int t^2 dt \) on the right. The integral of \( \frac{dz}{z^2} \) is \( -\frac{1}{z} + C_1 \), and the integral of \( t^2 \) is \( \frac{t^3}{3} + C_2 \). Thus, the integrated form is \(-\frac{1}{z} = \frac{t^3}{3} + C\), where \(C\) is the constant of integration combining \(C_1\) and \(C_2\).

Solve for the General Solution

Rearrange the equation \(-\frac{1}{z} = \frac{t^3}{3} + C\) to solve for \(z\). Taking the reciprocal gives \(z = -\frac{1}{\frac{t^3}{3} + C}\), which is the general solution involving the constant \(C\).

Substitute Initial Condition

Use the initial condition \(z = \frac{1}{3}\) at \(t = 1\) to find the constant \(C\). Substitute these values into the general solution: \(\frac{1}{3} = -\frac{1}{\frac{1^3}{3} + C}\). Simplifying gives \(\frac{1}{3} = -\frac{1}{\frac{1}{3} + C}\).

Solve for Constant C

Cross-multiply to solve for \(C\): \(\frac{1}{3} (\frac{1}{3} + C) = -1\). This simplifies to \(\frac{1}{9} + \frac{C}{3} = -1\). Further simplify to find \(C\): \(\frac{C + 1}{3} = -1\), thus \(C = -1 - \frac{1}{9}\), or \(C = -\frac{10}{9}\).

Write the Particular Solution

Substitute \(C = -\frac{10}{9}\) back into the general solution to get the particular solution: \(z = -\frac{1}{\frac{t^3}{3} - \frac{10}{9}}\). Simplifying gives \(z = -\frac{1}{\frac{9t^3}{27} - \frac{30}{27}}\) or \(z = -\frac{27}{9t^3 - 30}\).

Key concepts

Separation of Variables

Separation of Variables is a powerful technique used to solve certain types of differential equations. It involves rearranging the equation so that each variable appears on a different side. This allows us to integrate both sides separately. In the provided exercise, we start with the differential equation \( \frac{dz}{dt} = t^2 z^2 \).
By dividing both sides by \( z^2 \) and multiplying by \( dt \), we effectively isolate \( z \) terms on one side and \( t \) terms on the other, giving us \( \frac{dz}{z^2} = t^2 dt \). This separation is crucial as it sets the stage for the next process, integration.

Integration

Once the variables are separated, the next step is to integrate both sides of the resulting equation. Integration is a fundamental operation in calculus which can often reverse the process of differentiation.
In our problem, we integrate \( \int \frac{dz}{z^2} \) on the left side and \( \int t^2 dt \) on the right side. This yields the antiderivatives \( -\frac{1}{z} + C_1 \) and \( \frac{t^3}{3} + C_2 \), respectively.
Combining constants, the integral equation becomes \(-\frac{1}{z} = \frac{t^3}{3} + C\), where \( C \) encompasses both \( C_1 \) and \( C_2 \). Here, integration helps us transition towards finding the general form of the solution.

Initial Conditions

Initial conditions are values provided in a differential equation problem that allow us to find a unique solution, known as the particular solution. They are crucial because, without them, a differential equation with its general solution often contains arbitrary constants.
For this exercise, the initial condition given is \( z = \frac{1}{3} \) when \( t = 1 \). By substituting these values into the general solution, \( -\frac{1}{\frac{1}{3} + C} = \frac{1}{3} \), we can solve for the constant \( C \). These conditions ensure that the solution not only satisfies the differential equation but also aligns with known data points.

General Solution

The general solution to a differential equation encompasses all possible solutions, typically containing one or more arbitrary constants. These constants represent the infinite number of possible solutions before applying any additional conditions.
In the context of the exercise, after integrating and rearranging, we find the general solution: \( z = -\frac{1}{\frac{t^3}{3} + C} \).
This solution holds for any value of \( C \), and illustrates the breadth of solutions covered by the given differential equation. The presence of \( C \) is essential, as it allows the general solution to be adapted to specific cases through the use of additional conditions.

Particular Solution

A particular solution arises from applying initial conditions to the general solution, thus eliminating the arbitrary constants and leading to a specific answer tailored to the problem's conditions.
In this case, substituting the initial condition \( z = \frac{1}{3} \) at \( t = 1 \) into the general solution \( z = -\frac{1}{\frac{t^3}{3} + C} \), we simplify to find \( C = -\frac{10}{9} \).
Reinserting this back into the general solution gives the particular solution: \( z = -\frac{27}{9t^3 - 30} \).
This particular solution not only satisfies the original differential equation but also honors the specific initial conditions provided.