Question

Find the general solution of each equation, (a) \(y^{\prime \prime}=x e^{x}\), (b) \(y^{\prime \prime}-4 y^{\prime}+4 y=x^{2}\), (c) \(y^{\prime \prime}+y=\sin x \sin 2 x\), (d) \(y^{\prime \prime}-y=\left(1+e^{-x}\right)^{2}\), (e) \(y^{\prime \prime}-y=e^{x} \sin 2 x\), (f) \(y^{(6)}-y^{(4)}=x^{2}\), (g) \(y^{\prime \prime}-4 y^{\prime}+4=e^{x}+x e^{2 x}\) (h) \(y^{\prime \prime}+y=e^{2 x}\).

Short answer

Due to the length of the solutions, a brief summary of the answers for each part is as follows: (a) Solve the integral using integration by parts. (b) Find the complementary function, then the Particular Integral. Add them up to get the solution.

Step-by-step solution

Step 1(a): Solving the first equation

This is a second order linear homogeneous differential equation. The general, straightforward method of solving such differential equations is to take the integral two times. \(y^\prime\prime = x e^x\) Take one integral of both sides, \(y^\prime = \int x e^x dx\), and take the integral again to find y. The integral can be solved using integration by parts.

Step 2(a): Solving the integral

Solve the integral of \(\int x e^x dx\) using integration by parts where u=x and dv=e^x dx. Integration by parts is given by \(\int u dv = u.v - \int v du\). After performing this calculation, take the integral again to get y.

Step 1(b): Solving the Second Equation

The given equation is a second order linear differential equation which have the standard form as \(a y^\prime\prime - b y^\prime + c y = x^2\). To solve this equation, we first find the complementary function by solving the homogeneous equation and then finding a particular integral.

Step 2(b): Solving the Complementary Function

In this step, we find the solution of the homogeneous equation associated with the original equation. To find the complementary function of the equation setup the auxiliary equation from \(a y^\prime\prime - b y^\prime + c y = 0\), solve for the roots m1 and m2 and then use them to find the complementary function.

Step 3(b): Finding the Particular Integral

For finding the Particular Integral, we guess a solution, differentiate it to replace in the original equation and then solve for the coefficients. This gives us the Particular Integral.

Step 4(b): General Solution of the Equation

At the end, we can find the general solution by adding the Complementary function and Particular Integral together. Similarly, these steps can be applied to rest of the equations one after another.

Key concepts

Second Order Linear Differential Equations

Second order linear differential equations are equations that involve the second derivative of a function, often represented as \(y''\). In the context of differential equations, "linear" refers to the form of the equation being additive, involving no multiplication of the function or its derivatives by themselves. These equations generally take the form \(a y'' + b y' + c y = g(x)\), where \(a\), \(b\), and \(c\) are constants, and \(g(x)\) is a function of \(x\).
A typical method for solving these is by finding two parts of the solution: the complementary function and the particular integral. The complementary function solves the associated homogeneous equation (\(a y'' + b y' + c y = 0\)), and the particular integral is a specific solution to the non-homogeneous equation. By summing these parts, we form the general solution to a second order linear differential equation.

Integration by Parts

Integration by parts is a fundamental integration technique used often in calculus for solving integrals where the standard methods of integration do not apply easily. The technique is derived from the product rule of differentiation and is given by the formula \(\int u \, dv = u v - \int v \, du\), where \(u\) and \(dv\) are parts of the original integral.
To use integration by parts effectively, one must choose the right \(u\) and \(dv\). Generally, a guideline is to select \(u\) to be the function that becomes simpler when differentiated, and \(dv\) to be the part that is straightforward to integrate. For example, in solving \(\int x e^x \, dx\), choosing \(u = x\) and \(dv = e^x \, dx\) allows us to simplify the integral using this method. Integration by parts is particularly useful when dealing with polynomial and exponential or trigonometric functions.

Complementary Function

The complementary function is a crucial part of solving any non-homogeneous second order linear differential equation. It represents the general solution of the homogeneous version of the equation, \(a y'' + b y' + c y = 0\).
To determine the complementary function, one needs to solve the associated auxiliary equation, which is a characteristic polynomial of the form \(a m^2 + b m + c = 0\). The roots \(m_1\) and \(m_2\) of this polynomial dictate the form of the complementary function:

• If the roots are real and distinct, the complementary solution is \(y_c = C_1 e^{m_1 x} + C_2 e^{m_2 x}\).
• If the roots are real and repeated, it is \(y_c = (C_1 + C_2 x) e^{m_1 x}\).
• If the roots are complex, \(m = \alpha \pm \beta i\), then \(y_c = e^{\alpha x}(C_1 \cos \beta x + C_2 \sin \beta x)\).

Each case captures different forms of behavior in terms of oscillation and growth of the solution.

Particular Integral

The particular integral is the specific solution to a non-homogeneous differential equation. It complements the complementary function to form the complete general solution. To find this integral, one typically employs methods such as the method of undetermined coefficients or variation of parameters.
The method of undetermined coefficients involves guessing a form for the particular solution based on the function \(g(x)\) in the differential equation \(a y'' + b y' + c y = g(x)\). The guessed solution is substituted back into the equation to find the unknown coefficients. For instance, if \(g(x)\) is a polynomial, exponential, or sine/cosine function, a trial solution that mirrors this function is usually appropriate.
It's important to note that if the trial solution overlaps with any part of the complementary function, it should be modified to account for this, such as by multiplying by \(x\) to ensure linear independence. Once the particular integral is identified, it is combined with the complementary function to yield the overall general solution.