Question

Finding general solutions Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots\) to denote arbitrary constants. $$u^{\prime \prime}(x)=55 x^{9}+36 x^{7}-21 x^{5}+10 x^{-3}$$

Short answer

Question: Determine the general solution of the given second-order differential equation: \(u''(x) = 55x^9 + 36x^7 - 21x^5 + 10x^{-3}\). Answer: The general solution is \(u(x) = \frac{1}{2}x^{11} + \frac{1}{2}x^9 - \frac{1}{2}x^7 - 5x^{-1} + C_1x + C_2\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Integrate the equation once

Integrate both sides of the equation with respect to x: $$\int u''(x) dx = \int (55x^9 + 36x^7 - 21x^5 + 10x^{-3}) dx$$ The integral of the left side will give us the first derivative of u(x), \(u'(x)\). Now, we integrate term-by-term on the right side. $$u'(x) = \int 55x^9 dx + \int 36x^7 dx - \int 21x^5 dx + \int 10x^{-3} dx + C_1$$ $$u'(x) = \frac{55}{10}x^{10} + \frac{36}{8}x^8 - \frac{21}{6}x^6 + 10\int x^{-3} dx + C_1$$

Simplify the equation

Simplify the constants in each term: $$u'(x) = \frac{11}{2}x^{10} + \frac{9}{2}x^8 - \frac{7}{2}x^6 + 10\int x^{-3} dx + C_1$$ Now, integrate the last term: $$u'(x) = \frac{11}{2}x^{10} + \frac{9}{2}x^8 - \frac{7}{2}x^6 - 5x^{-2} + C_1$$

Integrate the equation once more

Integrate both sides of the equation with respect to x: $$\int u'(x) dx = \int\left( \frac{11}{2}x^{10} + \frac{9}{2}x^8 - \frac{7}{2}x^6 - 5x^{-2} + C_1\right)dx$$ The integral of the left side will give us the function u(x). Now, we integrate term-by-term on the right side. $$u(x) = \frac{11}{22}x^{11} + \frac{9}{18}x^9 - \frac{7}{14}x^7 + 5\int x^{-2} dx + C_1x + C_2$$ $$u(x) = \frac{1}{2}x^{11} + \frac{1}{2}x^9 - \frac{1}{2}x^7 + 5\int x^{-2} dx + C_1x + C_2$$

Simplify the equation

Now, integrate the remaining term: $$u(x) = \frac{1}{2}x^{11} + \frac{1}{2}x^9 - \frac{1}{2}x^7 - 5x^{-1} + C_1x + C_2$$ So, the general solution of the given differential equation is: $$u(x) = \frac{1}{2}x^{11} + \frac{1}{2}x^9 - \frac{1}{2}x^7 - 5x^{-1} + C_1x + C_2$$

Key concepts

Integration Techniques

Mastering integration techniques is crucial when solving differential equations, as they require the anti-derivative of a function to find solutions. The process involves reversing differentiation, effectively finding the original function whose derivative is given.

For example, when confronted with a term like \( 55x^9 \) in a differential equation, the integration process involves adding 1 to the exponent and then dividing by the new exponent. Similarly, for terms with negative exponents like \( 10x^{-3} \), the same rule applies, but special attention is needed since integrating \( x^{-1} \) leads to a natural logarithm and not a polynomial term.

Often, an equation may contain multiple terms to integrate. In such cases, each term is integrated individually, term-by-term, as done in step 1 of our example exercise. Differentiation and integration are inverse processes, and hence, the knowledge of differentiation rules directly applies to integration. Handling each term according to its specific rule—be it a polynomial, rational function, or any other type—is a strategy that simplifies the overall integration process and helps achieve the correct solution.

General Solutions of Differential Equations

The general solution of a differential equation represents the entire set of possible solutions that can satisfy the equation. Differential equations often contain derivatives, which denote rates of change, and the solutions to these equations provide a function that models a particular physical scenario.

In the context of second-order differential equations, such as \( u''(x) = 55 x^9 + 36 x^7 - 21 x^5 + 10 x^{-3} \), we seek a function \( u(x) \) whose second derivative equals the given expression. Finding the general solution involves integrating the equation twice, as derivatives are 'undone' by integration.

In our worked example, the general solution was found by integrating the second derivative, yielding the first derivative \( u'(x) \), and then integrating once more to find \( u(x) \). Since integration introduces indefinite integration constants, the final expression includes these arbitrary constants, denoted by \( C_1 \) and \( C_2 \) in this case, which makes it a general solution and not a particular one.

Arbitrary Constants

The concept of arbitrary constants is fundamental to solving differential equations. When integrating a function to find its anti-derivative, the process introduces an indefinite integration constant, typically denoted by \( C \). This constant arises due to the fact that the derivative of a constant is zero and hence the original constant value is lost during differentiation.

For example, when given \( u'(x) \) and asked to find \( u(x) \) by integration, we must add a constant \( C_1 \) to the result, acknowledging all possible vertical shifts of the original function. If we're dealing with a second-order differential equation, as in our example, a second integration introduces another arbitrary constant \( C_2 \) into the solution.

These constants are vital in forming the general solution and can be determined if initial conditions or boundary conditions are provided, turning the general solution into a specific or particular solution. Without these constants, we would not have a complete family of solutions to the differential equation.