Question

Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots .\) to denote arbitrary constants. $$y^{\prime \prime}(t)=60 t^{4}-4+12 t^{-3}$$

Short answer

Question: Find the general solution of the second-order differential equation: \(y^{\prime \prime}(t)=60 t^{4}-4+12 t^{-3}\). Answer: The general solution of the given differential equation is \(y(t) = 2t^6-2t^2+\frac{C_1}{3}t^3+Ct+C_2\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Identify the given differential equation

The given differential equation is: $$y^{\prime \prime}(t)=60 t^{4}-4+12 t^{-3}$$

Integrate the first time

Integrate the differential equation with respect to \(t\): $$ \int y^{\prime \prime}(t) \, dt= \int (60 t^{4}-4+12 t^{-3}) \, dt$$ $$y'(t) = \int 60t^4 \, dt - \int 4 \, dt + \int 12t^{-3} \, dt$$ Now integrate each term separately: $$y'(t) = 12t^5 - 4t + C_1t^2 + C$$ Where \(C_1\) and \(C\) are constants.

Integrate the second time

Integrate \(y'(t)\) with respect to \(t\): $$ y(t) = \int y'(t) \, dt = \int (12t^5 - 4t + C_1t^2 + C) \, dt$$ Now integrate each term separately: $$ y(t) = \frac{12}{6}t^6 - \frac{4}{2}t^2 + \frac{C_1}{3}t^3+ Ct + C_2$$ Where \(C_2\) is an arbitrary constant.

Write down the general solution

The general solution of the given differential equation is: $$y(t) = 2t^6-2t^2+\frac{C_1}{3}t^3+Ct+C_2$$

Key concepts

Integration

When solving differential equations, integration is a key mathematical tool. It allows us to reverse the process of differentiation and find the original function given its derivative. In the exercise, we had the second derivative of a function, denoted as \(y''(t)\). To find the original function \(y(t)\), we performed integration twice. First, we integrated \(y''(t)\) to get \(y'(t)\).This involved integrating each term separately, as shown below:

• The term \(60t^4\) became \(12t^5\).
• The constant \(-4\) became \(-4t\) since the integral of a constant \(a\) with respect to \(t\) is \(at\).
• The term \(12t^{-3}\) became \(-6t^{-2}\) since the integral of \(t^n\) is \(\frac{t^{n+1}}{n+1}\), assuming \(n eq -1\).

Once we found \(y'(t)\), we integrated it again to find \(y(t)\), the general solution. Each term was integrated independently:

• \(12t^5\) became \(2t^6\).
• \(-4t\) became \(-2t^2\).
• \(C_1t^2\) involved a simple power rule integration.
• \(C\) became \(Ct\).

Arbitrary Constants

Arbitrary constants play a crucial role when dealing with differential equations. They arise from the integration processes. Every time you integrate, you introduce a new arbitrary constant because integration is the inverse operation of differentiation, which destroys specific constant values.In the original equation, upon the first integration of \(y''(t)\), we introduce an arbitrary constant \(C\). Upon the second integration of \(y'(t)\), we introduce another arbitrary constant \(C_2\). These constants are named arbitrarily as \(C_1, C_2, \ldots\), and they reflect unknown initial conditions or particular solutions that will be determined if additional information is provided, such as initial values or boundary conditions.

General Solution

The general solution of a differential equation incorporates all possible solutions, represented with arbitrary constants. This gives the solution flexibility to fit different conditions.For example, in the exercise, the general solution is expressed as:\[ y(t) = 2t^6 - 2t^2 + \frac{C_1}{3}t^3 + Ct + C_2 \]Here, \(C_1\), \(C\), and \(C_2\) are arbitrary constants. The structure of the solution means it can be tailored to meet specific criteria or initial conditions by determining particular values for these constants.

• It reflects the capability to fit various scenarios by achieving different particular solutions.
• If we had specific initial conditions like \(y(0) = 5\), we would substitute \(t = 0\) in the general solution, set it equal to 5, and solve for one of the arbitrary constants.

This flexibility makes the general solution extremely valuable in both theoretical and practical scenarios, wherever modeling of real-world situations is necessary.