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Chapter 4: Problem 16

\(2 y^{\prime \prime}-y^{\prime}+6 y=t^{2} e^{-t} \sin t-8 t \cos 3 t+10^{t}\)

Short Answer

Expert verified
Because of the complexity of the right side of the given ODE, finding the particular solution \(y_p\) could be quite intensive, typically requiring techniques such as undetermined coefficients or variation of parameters. Therefore the final general solution would be in the form \(y(t) = y_h(t) + y_p(t)\), where \(y_h(t)\) can be acquired by calculating the roots of the characteristic equation obtained from coefficients of the homogeneous equation \(2 y''-y'+6 y=0\) and \(y_p(t)\) could only be found after complex computations referring to the exact form of the right hand side.

Step by step solution

01

Find Complementary Solution

The equation given is of the form \(a y''+b y'+c y=d\), where \(a = 2\), \(b = -1\), and \(c = 6\). We first find the complementary solution (homogeneous solution) by considering the equation \(2 y''-y'+6 y=0\). For this homogeneous equation, we solve the characteristic equation \(2m^2 - m + 6 = 0\). Solving this quadratic equation will give us the roots - necessary to determining the complementary solution.
02

Find Roots of Characteristic Equation

Use the quadratic formula, \(m = \frac{-b 卤 \sqrt {b^2 - 4ac}}{2a}\), to solve our characteristic equation. Substituting the values \(a = 2\), \(b = -1\), and \(c = 6\) into the quadratic formula will give two roots.
03

Formulate Homogeneous Solution

Once we have the roots, we can formulate the homogeneous solution, \(y_h\), for the differential equation. The form of \(y_h\) will depend on the nature of the roots. If the roots are real and distinct, \(y_h = c_1 e^{m_1 t} + c_2 e^{m_2 t}\). If they are real and repeated, \(y_h = c_1 e^{m t}+c_2 t e^{m t}\). If they are complex, \(y_h = e^{a t} (c_1 \cos(b t) + c_2 \sin(b t)\) where \(m_1, m_2 = a 卤 bi\).
04

Find Particular Solution

We then aim to find a particular solution \(y_p\) to the non-homogeneous differential equation. This step might be quite complex due to the form of the right-hand side, where various techniques such as undetermined coefficients or variation of parameters might be required.
05

Combine Solutions

Finally, we combine the particular solution \(y_p\) and the complementary solution \(y_h\) to obtain the general solution for the original differential equation. Therefore, the solution for the given ODE will be \(y(t) = y_h(t) + y_p(t)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complementary Solution
The complementary solution is an integral part of solving a differential equation. We find it by focusing on the associated homogeneous equation, where the right-hand side of the differential equation is set to zero. This means, for our given exercise, we start with the homogeneous equation:
\[ 2y'' - y' + 6y = 0 \]
Solving this part is crucial because it provides insight into the system's natural modes or the behavior without any external influence. The solutions derived from this step are components of the total solution that account for the equation's inherent dynamics.
  • The form of the complementary solution largely depends on the roots of the characteristic equation derived from the homogeneous equation.
  • These roots tell us whether the behavior is oscillatory, exponential, or a combination.
You will later combine the complementary solution with a particular solution to find the complete solution to the equation.
Characteristic Equation
The characteristic equation is a special polynomial equation derived from a differential equation's homogeneous part. For the given exercise, we derive this polynomial by substituting the trial solution of the form \(y = e^{mt}\) into the homogeneous equation. This gives us:
\[ 2m^2 - m + 6 = 0 \]
By solving the characteristic equation, you uncover key information:
  • The roots provide the exponents in the exponential or trigonometric solutions for the differential equation.
  • The solution forms change depending on whether the roots are real, repeated, or complex.
Use the quadratic formula to find the roots of the characteristic equation. Substituting the given coefficients, you solve for \(m\), enabling further progress towards finding the full solution.
Particular Solution
Finding the particular solution of a non-homogeneous differential equation often involves some creativity and strategy. This is because the particular solution accounts for the specific non-zero terms on the right-hand side of our equation. In our exercise, the right-hand side is:
\[ t^2 e^{-t} \sin t - 8t \cos 3t + 10^t \]
To tackle different terms, we might require techniques such as:
  • Method of undetermined coefficients: This technique guesses a form of the solution and determines the coefficients to fit the equation.
  • Variation of parameters: A more flexible technique, useful especially in scenarios involving non-standard terms.
Each term in the non-homogeneous part may need attention individually, leading to a composite particular solution. Ultimately, this forms together with the complementary solution as the full solution for the differential equation.
Homogeneous Equation
A homogeneous equation is derived from the original differential equation by considering only terms that include the function and its derivatives. It serves as the foundational part of the process, helping to solve the initial problem by examining the system's natural behavior.In our context, a homogeneous equation is:
\[ 2y'' - y' + 6y = 0 \]
Key points about the homogeneous equation include:
  • It simplifies the process to find the complementary solution, which describes the behavior of the differential equation without external forces.
  • The characteristic equation emerges directly from the homogeneous equation, crucial for finding the general solution.
Through understanding and solving the homogeneous equation, we better grasp the intrinsic qualities of the differential system, setting the stage for addressing the overall non-homogeneous equation.

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Most popular questions from this chapter

$$y_{1}(t)=t e^{2 t}, \quad y_{2}(t)=e^{2 t}$$

Solve the initial value problem: $$\begin{array}{l}{y^{\prime \prime \prime}-2 y^{\prime \prime}-y^{\prime}+2 y=0} \\ {y(0)=2, \quad y^{\prime}(0)=3, \quad y^{\prime \prime}(0)=5}\end{array}$$.

Superposition Principle. Let \(y_{1}\) be a solution to \(y^{\prime \prime}(t)+p(t) y^{\prime}(t)+q(t) y(t)=g_{1}(t)\) on the interval \(I\) and let \(y_{2}\) be a solution to \(y^{\prime \prime}(t)+p(t) y^{\prime}(t)+q(t) y(t)=g_{2}(t)\) on the same interval. Show that for any constants \(k_{1}\) and \(k_{2},\) the function \(k_{1} y_{1}+k_{2} y_{2}\) is a solution on \(I\) to \(\quad y^{\prime \prime}(t)+p(t) y^{\prime}(t)+q(t) y(t)=k_{1} g_{1}(t)+k_{2} g_{2}(t).\)

In quantum mechanics, the study of the Schrodinger equation for the case of a harmonic oscillator leads to a consideration of Hermite's equation, \(\quad y^{\prime \prime}-2 t y^{\prime}+\lambda y=0\) where \(\lambda\) is a parameter. Use the reduction of a second linearly independent solution to Hermite's equation for the given value of \(\lambda\) and corresponding solution \(f(t).\) \(\begin{array}{ll}{\text { (a) } \lambda=4,} & {f(t)=1-2 t^{2}} \\ {\text { (b) } \lambda=6,} & {f(t)=3 t-2 t^{3}}\end{array}\)

$$2 y^{\prime \prime \prime}+3 y^{\prime \prime}+y^{\prime}-4 y=e^{-t}$$
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