Question

Use the method of variation of parameters to find the general solutions of (a) \(\frac{d^{2} y}{d x^{2}}-y=x^{n}\), (b) \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y=2 x e^{x}\).

Short answer

The general solution for (a) is y = C_1 e^x + C_2 e^{-x} + y_p, and for (b), y = (C_1 + C_2 x) e^x + y_p.

Step-by-step solution

- Identify the Homogeneous Equation

For both (a) and (b), start by identifying the corresponding homogeneous differential equation. For (a): \(\frac{d^{2} y}{d x^{2}} - y = 0\) and for (b): \(\frac{d^{2} y}{d x^{2}} - 2\frac{d y}{d x} + y = 0\).

- Solve the Homogeneous Equation

Solve the homogeneous equation to get the complementary solution \(y_c\).For (a), solve \( \frac{d^{2} y}{d x^{2}} - y = 0 \) which gives roots \(\lambda^2 - 1 = 0\). The roots are \(\lambda = 1, -1\). Thus, the complementary solution is \(y_c = C_1 e^x + C_2 e^{-x}\).For (b), solve \( \frac{d^{2} y}{d x^{2}} - 2\frac{d y}{d x} + y = 0 \) which gives roots \(\lambda^2 - 2\lambda + 1 = 0\). The roots are \(\lambda = 1\) (repeated root). Thus, the complementary solution is \(y_c = (C_1 + C_2 x) e^x\).

- Find a Particular Solution

Use the method of variation of parameters to find a particular solution \(y_p\).For (a), we assume a solution of the form \(y_p = u_1(x) e^x + u_2(x) e^{-x}\). Compute the Wronskian \(W\) of the solutions \(e^x\) and \(e^{-x}\), and solve for \(u_1'\) and \(u_2'\). Apply the formula: \[ u_1' = -\frac{y_2(x)f(x)}{W}, \ u_2' = \frac{y_1(x)f(x)}{W} \]. Integrate \(u_1'\) and \(u_2'\) to find \(u_1\) and \(u_2\).For (b), assume a solution of the form \(y_p = u_1(x)e^x + u_2(x)xe^x\). Compute the Wronskian and find \(u_1\) and \(u_2\) in a similar fashion to find \(y_p\).

- Integrate and Find General Solution

Integrate the obtained functions to find \(u_1\) and \(u_2\), and substitute back into the expression for \(y_p\) to get the particular solution.For (a): Integrate \(u_1'\) and \(u_2'\) to find \(u_1\) and \(u_2\), and use these to construct \(y_p\).For (b): Similarly, integrate \(u_1\) and \(u_2\) to construct \(y_p\).Combine the particular solution \(y_p\) with the complementary solution \(y_c\) to get the general solution: \(y = y_c + y_p\).

- Write Final General Solution

For (a): General solution is \(y = C_1 e^x + C_2 e^{-x} + y_p\), where \(y_p\) is found from Step 4.For (b): General solution is \(y = (C_1 + C_2 x) e^x + y_p\), where \(y_p\) is found from Step 4.

Key concepts

homogeneous differential equations

Homogeneous differential equations are fundamental in studying differential equations. They have the form \(\frac{d^{2} y}{d x^{2}} + P(x) \frac{d y}{d x} + Q(x)y = 0\). The trick here is that there is no standalone function of \(x\) on the right-hand side—that's what makes it 'homogeneous'. \If you're given an equation like \(\frac{d^{2} y}{d x^{2}} - y = 0\), this is a second-order homogeneous differential equation.\These equations are crucial for solving physical and engineering problems where the system returns to equilibrium—think springs or circuits without external forces.

complementary solution

The complementary solution is the general solution to the homogeneous part of the differential equation. For instance, for \(\frac{d^{2} y}{d x^{2}} - y = 0\), we solve the equation as if it were equal to zero.\To solve this, find the roots of the characteristic equation derived from the homogeneous equation: \(\frac{d^{2} y}{d x^{2}} - y = 0\) gives \( \lambda^2 - 1 = 0, \ \lambda = ±1\).\The complementary solution would then be \(y_c = C_1 e^x + C_2 e^{-x}\). It captures the system's inherent properties without external influence.

particular solution

A particular solution addresses the non-homogeneous part of the differential equation, which includes external forces or inputs. \The method of variation of parameters is a common technique to find this solution. Take \(\frac{d^{2} y}{d x^{2}} - y = x^n\). Start with the complementary solution \(y_c = C_1 e^x + C_2 e^{-x}\).\Then assume \(y_p = u_1(x) e^x + u_2(x) e^{-x}\). You'll need to compute these functions by first finding the Wronskian and then using it to solve for \(u_1'\) and \(u_2'\): \[u_1' = -\frac{y_2(x) f(x)}{W}, \ u_2' = \frac{y_1(x) f(x)}{W}\]. Finally, integrate \(u_1'\) and \(u_2'\) to find \(u_1\) and \(u_2\).

Wronskian

The Wronskian is a determinant used to determine whether a set of solutions is linearly independent. For two functions \(y_1(x)\) and \(y_2(x)\), the Wronskian is given by: \[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \]For example, for \(y_1 = e^x\) and \(y_2 = e^{-x}\), their Wronskian is \[W(e^x, e^{-x}) = \begin{vmatrix} e^x & e^{-x} \ e^x & -e^{-x} \end{vmatrix} = -2 \]A non-zero Wronskian indicates the functions are linearly independent, which is crucial for building the general solution of the differential equation.