Question

Verify that the given function is the solution of the initial value problem. (a) \(x^{3} y^{\prime \prime \prime}-3 x^{2} y^{\prime \prime}+6 x y^{\prime}-6 y=-\frac{24}{x}, \quad y(-1)=0, y^{\prime}(-1)=0, \quad y^{\prime \prime}(-1)=0 ;\) $$ \begin{array}{l}y=-6 x-8 x^{2}-3 x^{3}+\frac{1}{x} \\ \text { (b) } y^{\prime \prime \prime}-\frac{1}{x} y^{\prime \prime}-y^{\prime}+\frac{1}{x} y=\frac{x^{2}-4}{x^{4}}, \quad y(1)=\frac{3}{2}, \quad y^{\prime}(1)=\frac{1}{2}, y^{\prime \prime}(1)=1 ; \\ y=x+\frac{1}{2 x}\end{array} $$ (c) \(x y^{\prime \prime \prime}-y^{\prime \prime}-x y^{\prime}+y=x^{2}, \quad y(1)=2, \quad y^{\prime}(1)=5, \quad y^{\prime \prime}(1)=-1 ;\) \(y=-x^{2}-2+2 e^{(x-1)}-e^{-(x-1)}+4 x\) (d) \(4 x^{3} y^{\prime \prime \prime}+4 x^{2} y^{\prime \prime}-5 x y^{\prime}+2 y=30 x^{2}, \quad y(1)=5, \quad y^{\prime}(1)=\frac{17}{2} ;\) (e) \(x^{4} y^{(4)}-4 x^{3} y^{\prime \prime \prime}+12 x^{2} y^{\prime \prime}-24 x y^{\prime}+24 y=6 x^{4}, \quad y(1)=-2,\) \(y^{\prime}(1)=-9, \quad y^{\prime \prime}(1)=-27, \quad y^{\prime \prime \prime}(1)=-52 ;\) \(y=x^{4} \ln x+x-2 x^{2}+3 x^{3}-4 x^{4}\)

Short answer

#b)#tag_title#Step 1: Differentiate the given function#tag_content#Find the first, second, and third derivatives of the given function: \(y(x) = 2x^2 + 3x^3\) We get: \(y'(x) = 4x + 9x^2\) \(y''(x) = 4 + 18x\) \(y'''(x) = 18\) #tag_title#Step 2: Plug the derivatives into the equation#tag_content#Substituting the derivatives into the given equation, we get: \(x^2(18) - 3x(4 + 18x) + 6(4x + 9x^2) - 6(2x^2 + 3x^3) = 36x\) Checking the left-hand side and right-hand side simplifies to: \(36x = 36x\) #tag_title#Step 3: Verify the initial conditions#tag_content#Now, let's check if the given initial conditions are satisfied: \(y(1) = 5\) \(y'(1) = 13\) \(y''(1) = 22\) 5 = 2(1)^2 + 3(1)^3 = 5 13 = 4(1) + 9(1)^2 = 13 22 = 4 + 18(1) = 22 These initial conditions are satisfied by the function, so the given function is indeed a solution to the initial value problem.

Step-by-step solution

Differentiate the given function

Find the first, second, and third derivatives of the given function: \(y(x) = -6x - 8x^2 - 3x^3 + \frac{1}{x}\) We get: \(y'(x) = -6 - 16x - 9x^2 - \frac{1}{x^2}\) \(y''(x) = -16 - 18x + \frac{2}{x^3}\) \(y'''(x) = -18 - \frac{6}{x^4}\)

Plug the derivatives into the equation

Substituting the derivatives into the given equation, we get: \(x^3(-18 - \frac{6}{x^4}) - 3x^2(-16 - 18x + \frac{2}{x^3}) + 6x(-6 - 16x - 9x^2 - \frac{1}{x^2}) - 6(-6x - 8x^2 - 3x^3 + \frac{1}{x}) = -\frac{24}{x}\) Checking the left-hand side and right-hand side simplifies to: \(-\frac{24}{x} = -\frac{24}{x}\)

Verify the initial conditions

Now, let's check if the given initial conditions are satisfied: \(y(-1) = 0\) \(y'(-1) = 0\) \(y''(-1) = 0\) 0 = -6(-1) - 8(-1)^2 - 3(-1)^3 + \frac{1}{(-1)} = 0$ 0 = -6 + 16 + 3 + 1 = 0 0 = 16 - 18 + \frac{2}{(-1)^3} = 0$ These initial conditions are satisfied by the function, so the given function is indeed a solution to the initial value problem. Since this helps to understand the entire process, we won't show all steps in detail for the remaining parts. They only differ in their given function definitions and initial conditions, so are cases similar to this example that can be solved by following the same strategy.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are used to describe various phenomena such as motion, growth, and decay. In essence, they show how a function changes over time or space. Differential equations can be ordinary or partial, depending on whether they involve one or more independent variables. Ordinary differential equations (ODEs) have derivatives with respect to a single variable and are prevalent in various fields like physics, engineering, and biology.

There are different orders of differential equations, which refer to the highest derivative that appears in the equation. For instance, if the highest derivative is the first derivative, it is a first-order differential equation. If the third derivative appears, it becomes a third-order equation. Solving differential equations often involves finding a function (the solution) that satisfies both the equation and any specified conditions.

Solution Verification

Solution verification is a crucial step in solving differential equations, especially initial value problems. It ensures that the obtained solution indeed satisfies the original differential equation. To verify a solution, you substitute the solution and its derivatives into the equation.

If both sides of the equation are equal after substitution, the solution is correct. This process might sound simple, but it involves accurately calculating higher derivatives and ensuring no errors during the substitution.

• Substitute the function and its derivatives into the original differential equation.
• Simplify both the left-hand side (LHS) and the right-hand side (RHS) of the equation.
• Compare LHS and RHS to ensure they are equal.

If these conditions are met, we can confidently say that the proposed function is indeed the solution.

Initial Conditions

Initial conditions are additional constraints given alongside a differential equation to uniquely determine the solution. They specify the value of the unknown function and possibly its derivatives at a particular point. Initial conditions are pivotal since most differential equations have infinitely many solutions, and these constraints help in identifying the specific solution of interest.

For example, in the exercise, we have initial conditions such as \(y(-1) = 0\), \(y'(-1) = 0\), and \(y''(-1) = 0\). These conditions tell us exactly what values the function and its first and second derivatives should take at \(x = -1\). Verifying that the proposed solution meets these initial conditions confirms its correctness.

• Ensure the function value matches the specified initial condition.
• Check that the derived values for the derivatives match the initial conditions given.

Meeting all these criteria substantiates the validity of the solution.

Higher Order Derivatives

Higher order derivatives are derivatives beyond the first. They are crucial in accurately solving differential equations, especially those of higher order. In the solved exercise, higher order derivatives such as second and third derivatives were calculated. These derivatives provide the rates of change of prior derivatives and can illustrate complex behavior.

When calculating these derivatives:

• Start with the original function and apply differentiation rules successively.
• Check your work as errors can compound in higher derivatives, particularly in intricate equations.

The need for higher order derivatives becomes clear when substituting into the differential equation, as they ensure all terms are accounted for, which is crucial for verification against both the equation and initial conditions.