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. $$ \frac{d y}{d t}=y^{4} ; y=1 \text { at } t=0 $$

Short answer

The particular solution is \( y = \left( \frac{1}{1 - 3t} \right)^{1/3} \).

Step-by-step solution

Separate Variables

To solve the differential equation \( \frac{dy}{dt} = y^4 \), we start by separating the variables. We rearrange the equation to get \( \frac{dy}{y^4} = dt \). This allows us to integrate each side with respect to its own variable.

Integrate Both Sides

Next, we integrate both sides of the equation. The integral of \( \frac{1}{y^4} \) with respect to \( y \) is \( -\frac{1}{3y^3} + C_1 \), and the integral of \( dt \) is \( t + C_2 \). Setting the constants of integration together gives: \[-\frac{1}{3y^3} = t + C \].

Solve for y

Solve the equation \(-\frac{1}{3y^3} = t + C \) for \( y \):1. Multiply through by -3 to isolate the term with \( y \): \[\frac{1}{y^3} = -3t - 3C\].2. Take the reciprocal to solve for \( y^3 \): \[y^3 = \frac{1}{-3t - 3C}\].3. Finally, solve for \( y \) by taking the cube root: \[y = \left(\frac{1}{-3t - 3C}\right)^{1/3}\].This is the general solution.

Apply Initial Condition

We use the initial condition given, \( y = 1 \) at \( t = 0 \), to find the particular solution. Substitute \( y = 1 \) and \( t = 0 \) into the general solution:\[1 = \left(\frac{1}{0 - 3C}\right)^{1/3}\].Cubing both sides, we have:\[1 = \frac{1}{-3C}\],Solving this gives \(-3C = 1\) or \(C = -\frac{1}{3}\).

Substitute C into General Solution

Substitute \( C = -\frac{1}{3} \) back into the general solution to get the particular solution:\[y = \left(\frac{1}{-3t + 1}\right)^{1/3}\]. This equation describes the particular solution satisfying the condition \( y = 1 \) at \( t = 0 \).

Key concepts

General Solution

The general solution of a differential equation involves integration and typically includes an arbitrary constant, often denoted as \( C \). This constant represents a family of solutions, covering all possible curves or functions that satisfy the given differential equation. For example, in the differential equation \( \frac{dy}{dt} = y^4 \), we begin the process by integrating after variable separation. When we separate variables, we rearrange to have \( \frac{dy}{y^4} = dt \), which allows us to integrate each side. After integration, we obtain \(-\frac{1}{3y^3} = t + C \), representing the general solution. The presence of \( C \) is crucial as it accounts for different initial conditions and scenarios. In this equation, our journey from the differential to the integrated form highlights the flexibility of solutions due to this arbitrary constant.

Particular Solution

A particular solution emerges when we utilize an initial condition to determine the arbitrary constant \( C \), thus pinpointing a specific curve from the family described by the general solution. In the same differential equation \( \frac{dy}{dt} = y^4 \), we start with the general solution \( y = \left(\frac{1}{-3t - 3C}\right)^{1/3} \). With the initial condition \( y = 1 \) when \( t = 0 \), we substitute these values back into the general solution to find \( C \). Setting \( y = 1 \) and \( t = 0 \), we solve \( 1 = \left(\frac{1}{0 - 3C}\right)^{1/3} \), eventually finding \( C = -\frac{1}{3} \). By substituting \( C = -\frac{1}{3} \) into the general solution, we arrive at the particular solution: \( y = \left(\frac{1}{-3t + 1}\right)^{1/3} \). This expression uniquely satisfies the initial condition provided in the problem.

Initial Conditions

Initial conditions are specific values provided in differential equations that allow us to find particular solutions from the general form. They serve as constraints to identify the particular path or solution that fits the real-world problem being modeled. For the differential equation \( \frac{dy}{dt} = y^4 \) with the initial condition \( y = 1 \) when \( t = 0 \), this condition helps anchor our arbitrary constant \( C \) to a specific value. By substituting the initial values into our general solution \( y = \left(\frac{1}{-3t - 3C}\right)^{1/3} \), we compute \( C \) and tailor the broad solution to address this specific scenario precisely. Initial conditions essentially "nail down" the solution so we can apply it practically, ensuring that predictions or analyses are accurate for given starting points.

Variable Separation

Variable separation is a technique used to solve differential equations by rearranging terms so that each variable appears on only one side of the equation. This enables us to integrate both sides individually, helping unravel the relationship between variables. For instance, in the differential equation \( \frac{dy}{dt} = y^4 \), we use variable separation by getting \( \frac{dy}{y^4} = dt \). This step involves moving all terms involving \( y \) to one side and \( t \) to the other. By doing so, we prepare each side for integration. Integrating \( \frac{1}{y^4} \) with respect to \( y \) and \( dt \) with respect to \( t \), we achieve \(-\frac{1}{3y^3} = t + C \). This result forms the basis for determining solutions that follow any given conditions. The elegance of variable separation lies in transforming a potentially complex differential equation into solvable, integrable forms.