91Ó°ÊÓ

A particular solution, \(y_{p}\), is a specific answer to the non-homogeneous differential equation. When we have a differential equation like \(y'' - 4y = \sinh x\), it's essential to cater to the entire equation, not just its homogeneous part like \(y'' - 4y = 0\). This means we must find a solution that includes the non-homogeneous part \(\sinh x\).
To find the particular solution, observe the non-homogeneous function, \(\sinh x\), which can be expressed as \(\frac{e^x - e^{-x}}{2}\). Therefore, we guess that the particular solution is of the form \(y_{p} = A e^{x} + B e^{-x}\). This type of assumption is important since it helps build a solution that matches the non-homogeneous term.
By substituting \(y_{p}\) and its derivatives into the original differential equation, we can solve for the coefficients \(A\) and \(B\), ensuring the left-hand side matches the right-hand side. In this case, we find:

• \( A = -\frac{1}{6} \)
• \( B = \frac{1}{6} \)

Replacing these values in our guessed form, the particular solution is \( y_{p} = -\frac{1}{6} e^{x} + \frac{1}{6} e^{-x} \), completing our task for the non-homogeneous part.

The complementary solution, \(y_{c}\), deals with the homogeneous portion of the differential equation. For the equation \(y'' - 4y = \sinh x \), we strip away the non-homogeneous term \(\sinh x\) and solve for the homogeneous equation \( y'' - 4y = 0 \).
To proceed, use the characteristic equation. By setting up \(y'' - 4y = 0 \), we're led to form the characteristic equation: \( r^2 - 4 = 0 \). Solving this yields the roots \( r = 2 \) and \( r = -2 \).
Considering these roots, the solutions are exponential functions:

• \( e^{2x} \)
• \( e^{-2x} \)

Thus, the complementary solution is expressed as \( y_{c} = C_{1} e^{2x} + C_{2} e^{-2x} \). Here, \(C_{1}\) and \(C_{2}\) are constants determined by initial conditions or additional constraints from the problem. This part of the solution represents all possible solutions to the homogeneous equation, filling in the structure for solving the entire differential equation.

The characteristic equation is central to solving linear homogeneous differential equations. It's a simple algebraic equation derived from the differential equation's highest derivatives. For the equation \(y'' - 4y = 0 \), assume a solution in the form of \(y = e^{rx}\), where \(r\) is a constant.
Substituting this form into the differential equation turns it from a differential equation into a simple polynomial equation, termed as the characteristic equation: \(r^2 - 4 = 0\). Solving for \(r\), we find two solutions: \(r = 2\) and \(r = -2\).
These roots of the characteristic equation guide us in crafting the complementary solution. The characteristic equation allows us to transform complex differential equations into more manageable algebraic problems. It's a bridge between the world of differential calculus and algebra, paving the path to solving homogeneous differential equations efficiently.