Question

For the given differential equation, $$ y^{\prime \prime}-4 y=4 t^{2} $$

Short answer

Question: Find the general solution to the given differential equation: \(y^{\prime \prime} - 4y = 4t^2\). Answer: The general solution to the given differential equation is \(y(t) = C_1 e^{2t} + C_2 e^{-2t} - t^2 - \frac{1}{2}\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Identify the type of differential equation

The given differential equation is $$ y^{\prime \prime} - 4y = 4t^2. $$ It is a second-order linear differential equation with constant coefficients.

Solve the homogeneous equation

We will first solve the homogeneous equation, which is obtained by setting the right-hand side to 0: $$ y^{\prime \prime} - 4y = 0. $$ To solve this, we will find the characteristic equation: $$ r^2 - 4 = 0. $$ Solving for \(r\), we get: $$ r=\pm 2. $$ So, the homogeneous solution is given by: $$ y_h(t) = C_1 e^{2t} + C_2 e^{-2t}. $$ Where \(C_1\) and \(C_2\) are arbitrary constants.

Determine the particular solution

Next, we have to find the particular solution \(y_p(t)\) by substituting a trial function into the differential equation and solving for its coefficients. Given the right-hand side is \(4t^2\), our trial solution takes the form: $$ y_p(t) = At^2+Bt+C. $$ Now, we will calculate the first and second derivatives of our trial function: $$ y_p^{\prime}(t) = 2At + B, $$ $$ y_p^{\prime \prime}(t) = 2A. $$ Substituting our trial function and its derivatives back into the original differential equation, we get: $$ 2A - 4(At^2 + Bt + C) = 4t^2. $$ This implies $$ - 4At^2 - 4Bt - 4C + 2A = 4t^2. $$ So, $$ - 4A = 4 \Rightarrow A = -1, $$ $$ - 4B = 0 \Rightarrow B = 0, $$ $$ - 4C + 2A = 0 \Rightarrow C = -\frac{1}{2}. $$ Thus the particular solution \(y_p(t)\) is: $$ y_p(t) = -t^2 - \frac{1}{2}. $$

Combine homogeneous and particular solutions

Now, we will combine the homogeneous and particular solutions to find the general solution of the given differential equation. Adding \(y_h(t)\) and \(y_p(t)\), we get: $$ y(t) = y_h(t) + y_p(t) = C_1 e^{2t} + C_2 e^{-2t} - t^2 - \frac{1}{2}. $$ This is the general solution to the given differential equation.

Key concepts

Characteristic Equation

When dealing with second-order linear differential equations such as \( y^{\prime \prime} - 4y = 0 \), one of the pivotal steps is solving the characteristic equation. The characteristic equation arises by assuming solutions of the form \( e^{rt} \), which simplifies the differential equation.
To find the characteristic equation, replace the second derivative with \( r^2 \), resulting in \( r^2 - 4 = 0 \).
This forms a quadratic equation that can be solved using algebraic methods. In our case, solving for \( r \), we get \( r = \pm 2 \).
These roots \( r \) tell us about the nature of the solution to the homogeneous equation. Since our roots here are real and distinct, the solutions to the differential equation will be exponential functions, giving us insight into the behavior and form of the general solution.

Homogeneous Solution

Once we have our roots from the characteristic equation, we can write down the homogeneous solution. In the example of the equation \( y^{\prime \prime} - 4y = 0 \), the roots \( r = 2 \) and \( r = -2 \) lead to the homogeneous solution:
\[ y_h(t) = C_1 e^{2t} + C_2 e^{-2t} \]
where \( C_1 \) and \( C_2 \) are constants determined by initial or boundary conditions if provided.

• These exponential terms \( e^{2t} \) and \( e^{-2t} \) reflect the influence of the characteristic roots.
• The terms indicate how quickly the solutions grow if \( t \) is positive or how they decay if \( t \) is negative.

This part of the solution captures the natural response of the system without any external forcing applied to it, representing what the system would do on its own.

Particular Solution

For inhomogeneous differential equations, we seek a particular solution to account for the non-zero right-hand side. In our example, the differential equation is \( y^{\prime \prime} - 4y = 4t^2 \).
The particular solution \( y_p(t) \) aims to match this external 'forcing'. By selecting a trial function based on the form of the right side, in this case \( At^2 + Bt + C \), we can determine a specific solution:

• Derive this trial function and substitute it back into the differential equation.
• Adjust coefficients \( A \), \( B \), and \( C \) to satisfy the equation.

For \( y_p(t) = At^2 + Bt + C \), we find through matching coefficients that \( A = -1 \), \( B = 0 \), and \( C = -\frac{1}{2} \).
Therefore, the particular solution is:
\[y_p(t) = -t^2 - \frac{1}{2}\]
This reflects the particular response of the system due to the external polynomial input.

Constant Coefficients

The term 'constant coefficients' refers to the fact that the coefficients of the derivatives in the differential equation remain unchanged.
In the equation \( y^{\prime\prime} - 4y = 4t^2 \), the coefficient of \( y^{\prime\prime} \) is 1 and that of \( y \) is -4.

• This simplifies the solution process significantly as it allows the use of standard methods like characteristic equations.
• With constant coefficients, solutions will often involve exponential functions or their combinations, as these naturally solve the homogeneous part.

This consistency in the coefficients means that the characteristic equation itself will have constant terms leading to predictable solutions, such as those involving real, complex, or repeated roots.
This allows for simplifications and uniform strategies to solve such differential equations.