Question

Use the method of reduction of order to find a second solution of the given differential equation. \(t^{2} y^{\prime \prime}-4 t y^{\prime}+6 y=0, \quad t>0 ; \quad y_{1}(t)=t^{2}\)

Short answer

Question: Find the second solution to the given differential equation using the method of reduction of order: \(t^{2} y^{\prime \prime}-4 t y^{\prime}+6 y=0\), where \(y_1(t) = t^2\). Answer: The second solution to the given differential equation is \(y_{2}(t) = \left(c_1 e^{\sqrt{2}t} + c_2 e^{-\sqrt{2}t}\right) t^2\).

Step-by-step solution

Write down the given equation and known solution

We have the following differential equation and known solution: $$ t^{2} y^{\prime \prime}-4 t y^{\prime}+6 y=0, \quad t>0 $$ $$ y_{1}(t) = t^{2} $$

Apply the method of reduction of order

In the method of reduction of order, we assume a new solution has the form \(y_{2}(t) = v(t) \cdot y_{1}(t)\), where \(y_1(t)\) is the known solution. In our case: $$ y_{2}(t) = v(t) \cdot t^{2} $$

Calculate the derivatives of \(y_{2}(t)\)

Let's find \(y_2'(t)\) and \(y_2''(t)\): $$ y_{2}^{\prime} (t) = \frac{d}{dt}(v(t) \cdot t^{2}) = v^{\prime}(t) \cdot t^{2} + 2t v(t) $$ $$ y_{2}^{\prime \prime} (t) = \frac{d^2}{dt^2}(v(t) \cdot t^{2}) = v^{\prime \prime}(t) \cdot t^{2} + 2v^{\prime}(t) \cdot t + 2v^{\prime}(t) \cdot t + 2v(t) = v^{\prime \prime}(t) \cdot t^{2} + 4t v^{\prime}(t) + 2v(t) $$

Substitute \(y_{2}(t)\), \(y_{2}^{\prime} (t)\), and \(y_{2}^{\prime \prime} (t)\) into the given equation

Now, we'll plug our \(y_{2}(t)\) and its derivatives back into the original differential equation: $$ t^{2}(v^{\prime \prime}(t) \cdot t^{2} + 4t v^{\prime}(t) + 2v(t))- 4t (v^{\prime}(t) \cdot t^{2} + 2t v(t)) + 6(v(t) \cdot t^2) = 0 $$

Simplify the equation

Next, we will eliminate terms and simplify the equation: $$ t^{4}v^{\prime \prime}(t) + 4t^{3}v^{\prime}(t) + 2t^{2}v(t)- 4t^{3}v^{\prime}(t) - 8t^{2}v(t) + 6t^{2}v(t) = 0 $$ $$ t^{4}v^{\prime \prime}(t) - 2t^{2}v(t) = 0 $$

Solve for \(v(t)\)

Now we need to solve for \(v(t)\). Divide the equation by \(t^2\) and separate variables: $$ t^{2}v^{\prime \prime}(t) - 2v(t) = 0 $$ This is now a second-order homogeneous linear differential equation with constant coefficients. The characteristic equation is: $$ r^2 - 2 = 0 $$ The roots are \(r_1 = \sqrt{2}\) and \(r_2 = -\sqrt{2}\). Thus, the general solution for \(v(t)\) is: $$ v(t) = c_1 e^{\sqrt{2}t} + c_2 e^{-\sqrt{2}t} $$

Find \(y_2(t)\)

Finally, we substitute the solution for \(v(t)\) back into the expression for \(y_{2}(t)\): $$ y_{2}(t) = \left(c_1 e^{\sqrt{2}t} + c_2 e^{-\sqrt{2}t}\right) t^2 $$ This is our second solution to the original differential equation.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are a powerful tool for modeling and solving real-world problems where there is a relationship between changing quantities. The equation given in the exercise is \[t^{2} y^{\prime \prime}-4 t y^{\prime}+6 y=0\] This is a second-order differential equation, as it involves the second derivative of the function \(y\), which can describe various phenomena such as motion, growth, or decay in both natural and engineered systems. Typically, the goal is to find a function \( y(t) \) that satisfies the equation over a given range of \( t \). Solving differential equations can involve several different methods, with reduction of order being a useful approach when a solution is already known.

Homogeneous Linear Equations

A homogeneous linear equation is a special type of differential equation where the sum of terms, each consisting of a function and its derivatives, equals zero. In other words, all the terms are dependent on the function and its derivatives, without any independent constant terms.

The given equation, \[t^{2} y^{\prime \prime}-4 t y^{\prime}+6 y=0\] is an example of a homogeneous linear differential equation. The key feature here is that all the terms involve the function \( y \) and its derivatives multiplied by some coefficient, and they add up neatly to zero. For such equations, if \( y_1(t) \) is a particular solution, any linear combination of solutions can also form a solution, which means solving one part can simplify finding others!

Characteristic Equation

When solving linear differential equations, the characteristic equation is a helpful algebraic tool. It allows us to identify the type of solutions we can expect based on the differential equation's coefficients. In this exercise, once the equation for \( v(t) \) is simplified to \[t^{2}v^{\prime \prime}(t) - 2v(t) = 0\] we can reduce it further to a more standard form, leading us to a characteristic equation:

\[r^2 - 2 = 0\]
The roots of this algebraic equation, \(r_1 = \sqrt{2}\) and \(r_2 = -\sqrt{2}\), tell us about the general nature of the solutions for \(v(t)\), which are exponential functions determined by these roots. Recognizing and solving characteristic equations is a crucial step in dealing with linear homogeneous equations, providing insights into the solution structure.

Second Solution

Finding a second, linearly independent solution is essential when solving differential equations, as it allows for constructing the general solution of the equation. Given that one solution \( y_1(t) = t^2 \) is already known, we employ the method of reduction of order to find another one.

We start by assuming a solution of the form \( y_2(t) = v(t) \cdot t^2 \). Substituting this form and finding derivatives leads us to a simplified equation for \( v(t) \):\[t^{2}v^{\prime \prime}(t) - 2v(t) = 0\]This equation is solved using exponential functions based on the roots from the characteristic equation.

The second solution becomes:\[y_{2}(t) = \left(c_1 e^{\sqrt{2}t} + c_2 e^{-\sqrt{2}t}\right) t^2\]This solution complements the first and provides the full range of behaviors consistent with the original differential equation, offering a broader understanding.