Question

(a) Verify that \(\left\\{t, t^{2}, t^{4}\right\\}\) is a fundamental set of solutions of the differential equation $$ t^{3} y^{\prime \prime \prime}-4 t^{2} y^{\prime \prime}+8 t y^{\prime}-8 y=0 \text {. } $$ (b) Find the general solution of $$ t^{3} y^{\prime \prime \prime}-4 t^{2} y^{\prime \prime}+8 t y^{\prime}-8 y=2 \sqrt{t}, \quad t>0 . $$

Short answer

What is the general solution of the nonhomogeneous differential equation \(t^3 y^{'''} - 4t^2 y^{''} + 8t y' - 8y = 2\sqrt{t}\)? Answer: Yes, the given set of solutions \(\left\{t, t^{2}, t^{4}\right\}\) is a fundamental set for the homogeneous differential equation. The general solution of the nonhomogeneous differential equation is \(y(t) = c_1t + c_2t^2 + c_3t^4 + \frac{4}{3}t^{\frac{3}{2}}, \quad t > 0\).

Step-by-step solution

Verifying the given set of solutions for the homogeneous equation

To verify if \(\left\{t, t^{2}, t^{4}\right\}\) is a fundamental set of solutions, we will substitute each expression into the given differential equation and see if the equation is satisfied. - First, let \(y = t\): $$t^3 \big(\frac{d^3 t}{dt^3}\big) - 4t^2 \big(\frac{d^2 t}{dt^2}\big) + 8t \big(\frac{dt}{dt}\big) - 8t = 0$$ - Then, let \(y = t^2\): $$t^3\big(\frac{d^3 t^2}{dt^3}\big) - 4t^2\big(\frac{d^2 t^2}{dt^2}\big) + 8t\big(\frac{d t^2}{dt}\big) - 8t^2 = 0$$ - Finally, let \(y = t^4\): $$t^3\big(\frac{d^3 t^4}{dt^3}\big) - 4t^2\big(\frac{d^2 t^4}{dt^2}\big) + 8t\big(\frac{d t^4}{dt}\big) - 8t^4 = 0$$

Solving the differential equation with the given nonhomogeneous term

Now we have to find the general solution of the nonhomogeneous differential equation. We can do this using the method of variation of parameters. Since we already have a fundamental set of solutions, the general solution of the given differential equation is: $$y(t)=c_{1}t+c_{2}t^{2}+c_{3}t^{4}+u(t), \quad t>0$$ Differentiating the equation with respect to \(t\) and substituting it into the nonhomogeneous differential equation, we get the equation for \(u(t)\): $$t^3\big(\frac{d^3 u}{dt^3}\big) - 4t^2\big(\frac{d^2 u}{dt^2}\big) + 8t\big(\frac{du}{dt}\big) - 8u = 2\sqrt{t}, \quad t>0$$ Now we find \(u(t)\) by integrating the given nonhomogeneous part of the differential equation: $$\int 2\sqrt{t}\, dt = \frac{4}{3}t^{\frac{3}{2}} + C$$ This gives us \(u(t) = \frac{4}{3}t^{\frac{3}{2}}\). Plugging this expression back into our general solution, we obtain: $$y(t) = c_1t + c_2t^2 + c_3t^4 + \frac{4}{3}t^{\frac{3}{2}}$$ So, the general solution of the given differential equation is: $$y(t) = c_1t + c_2t^2 + c_3t^4 + \frac{4}{3}t^{\frac{3}{2}}, \quad t > 0$$

Key concepts

Fundamental Set of Solutions

When it comes to differential equations, a 'fundamental set of solutions' refers to a collection of solutions that are both independent from each other and span the entire solution space of a homogeneous linear differential equation. Think of it like finding a set of unique building blocks that can be combined to construct any possible solution for a specific differential equation.

In our exercise, the set \( \{t, t^2, t^4\} \) was tested against the homogeneous differential equation to see if it qualifies. We look for each function in the set to satisfy the equation when plugged in. Once verified, like in our example, this fundamental set can be used to build the general homogeneous solution by multiplying each function by an arbitrary constant and summing them up, thus embodying all possible solutions to the equation.

Nonhomogeneous Differential Equation

A 'nonhomogeneous differential equation' stands as a modification to its homogeneous counterpart, displaying a crucial characteristic: an additional function that is not equal to zero. In the given exercise, the nonhomogeneous term is \( 2\sqrt{t} \). Solving such an equation requires us to find a 'particular solution' that addresses the nonhomogeneous part, coupled with the 'general solution' of the related homogeneous equation.

While the general solution of the homogeneous part represents myriad solutions enveloped by the fundamental set, the particular solution is an individual, specific solution to the full nonhomogeneous equation. The sum of these two solutions gives the 'general solution' to the nonhomogeneous equation, ensuring it covers all potential solutions, both caused by the equation's nature and its particular non-zero term.

Method of Variation of Parameters

To address the challenges posed by nonhomogeneous differential equations, we employ the 'method of variation of parameters'. This ingenious technique involves tweaking the constants in the fundamental set's general solution into functions that shift in a way to accommodate the nonhomogeneous term.

What we're doing is allowing the parameters (constants) to vary and finding their specific forms by substituting back into the original equation. It requires integration, which can be intimidating, yet is straightforward once we absorb the method's logic. The outcome is a particular solution that, when added to the homogeneous solution, grants us the general solution to the entire nonhomogeneous equation. It's like finding the missing piece of a puzzle that perfectly fits.

General Solution of Differential Equation

Wrapping up the concepts is the 'general solution of a differential equation', which embodies the complete set of solutions for either a homogeneous or nonhomogeneous equation. It encompasses all the variations that are possible within the scope of the equation, neatly encapsulated in a single expression.

For our nonhomogeneous example, the general solution incorporates both a combination of the fundamental set of solutions for the homogeneous part and the particular solution for the nonhomogeneous term. This comprehensive answer, represented by a formula, is elegantly customizable with varying constants (or parameters). These constants are essentially what enable the general solution to morph into any specific solution required by an initial condition or other constraints. It's the master key to unlock any particular scenario the differential equation might represent.