Question

Solve the initial value problem. (a) \(y^{\prime}=-x e^{x}, \quad y(0)=1\) (b) \(y^{\prime}=x \sin x^{2}, \quad y\left(\sqrt{\frac{\pi}{2}}\right)=1\) (c) \(y^{\prime}=\tan x, \quad y(\pi / 4)=3\) (d) \(y^{\prime \prime}=x^{4}, \quad y(2)=-1, \quad y^{\prime}(2)=-1\) (e) \(y^{\prime \prime}=x e^{2 x}, \quad y(0)=7, \quad y^{\prime}(0)=1\) (f) \(y^{\prime \prime}=-x \sin x, \quad y(0)=1, \quad y^{\prime}(0)=-3\) (g) \(y^{\prime \prime \prime}=x^{2} e^{x}, \quad y(0)=1, \quad y^{\prime}(0)=-2, \quad y^{\prime \prime}(0)=3\) (h) \(y^{\prime \prime \prime}=2+\sin 2 x_{1} \quad y(0)=1, \quad y^{\prime}(0)=-6, \quad y^{\prime \prime}(0)=3\) (i) \(y^{\prime \prime \prime}=2 x+1, \quad y(2)=1, \quad y^{\prime}(2)=-4, \quad y^{\prime \prime}(2)=7\)

Short answer

Answer: The solution to the initial value problem (a) is \(y(x) = -xe^x + e^x\). 2. What is the solution to the initial value problem (b)? Answer: The solution to the initial value problem (b) is \(y(x) = -\frac{1}{2}\cos{(x^2)} + 1\). 3. What is the solution to the initial value problem (c)? Answer: The solution to the initial value problem (c) is \(y(x) = -\ln{|\cos{x}|} + 3 + \ln{\frac{1}{\sqrt{2}}}\). 4. What is the solution to the initial value problem (i)? Answer: The solution to the initial value problem (i) is \(y(x) = \frac{1}{12}x^4 + \frac{1}{6}x^3 + \frac{3}{2}x^2 - 11x\).

Step-by-step solution

Problem (a)

We are given a first order differential equation: \(y^{\prime} = -xe^x\) and an initial condition: \(y(0) = 1\). Step 1: Integrate the differential equation \(\int y^{\prime} dx = \int -xe^x dx\) To integrate the right side, we can use integration by parts: \(u = -x, dv = e^x dx\) Then, \(du = -dx\) and \(v = e^x\). The integration by parts formula is: \(\int u dv = uv - \int v du\) Thus, \(\int -xe^x dx = -xe^x - \int -e^x dx = -xe^x + e^x + C\), where C is the constant of integration. Step 2: Apply the initial condition \(y(0) = 1\) \(1 = -0 + e^0 + C \implies C = 0\) Finally, the solution to the initial value problem is: \(y(x) = -xe^x + e^x\).

Problem (b)

We are given a first order differential equation: \(y^{\prime} = x\sin{(x^2)}\) and an initial condition: \(y\left(\sqrt{\frac{\pi}{2}}\right) = 1\). Step 1: Integrate the differential equation \(\int y^{\prime} dx = \int x\sin{(x^2)} dx\) Let \(u = x^2\), then \(du = 2x\, dx\), so the integral becomes: \(\frac{1}{2}\int \sin{(u)}\, du = -\frac{1}{2}\cos{(u)} + C = -\frac{1}{2}\cos{(x^2)} + C\) Step 2: Apply the initial condition \(y\left(\sqrt{\frac{\pi}{2}}\right) = 1\) \(1 = -\frac{1}{2}\cos{\left(\frac{\pi}{2}\right)} + C \implies C = 1\) Finally, the solution to the initial value problem is: \(y(x) = -\frac{1}{2}\cos{(x^2)} + 1\).

Problem (c)

We are given a first order differential equation: \(y^{\prime} = \tan{x}\) and an initial condition: \(y\left(\frac{\pi}{4}\right) = 3\). Step 1: Integrate the differential equation \(\int y^{\prime} dx = \int \tan{x} dx\) We know that \(\int \tan{x} dx = -\ln{|\cos{x}|} + C\). So we get: \(y(x) = -\ln{|\cos{x}|} + C\) Step 2: Apply the initial condition \(y\left(\frac{\pi}{4}\right) = 3\) \(3 = -\ln{|\cos{\left(\frac{\pi}{4}\right)}|} + C \implies C = 3 + \ln{\frac{1}{\sqrt{2}}}\) Finally, the solution to the initial value problem is: \(y(x) = -\ln{|\cos{x}|} + 3 + \ln{\frac{1}{\sqrt{2}}}\).

Problem (i)

We are given a third order differential equation: \(y^{\prime\prime\prime} = 2x + 1\) and initial conditions: \(y(2) = 1, y^{\prime}(2) = -4, y^{\prime\prime}(2) = 7\). Step 1: Integrate the differential equation 1st integration: \(\int y^{\prime\prime\prime} dx = \int (2x + 1) dx\) \(y^{\prime\prime} (x) = x^2 + x + C_1\) 2nd integration: \(\int y^{\prime\prime} dx = \int (x^2 + x + C_1) dx\) \(y^{\prime}(x) = \frac{1}{3}x^3 + \frac{1}{2}x^2 + C_1x + C_2\) 3rd integration: \(\int y^{\prime} dx = \int \left(\frac{1}{3}x^3 + \frac{1}{2}x^2 + C_1x + C_2\right) dx\) \(y(x) = \frac{1}{12}x^4 + \frac{1}{6}x^3 + \frac{1}{2}C_1x^2 + C_2x + C_3\) Step 2: Apply the initial conditions 1. \(y(2) = 1\) \(\frac{1}{12}(2)^4 + \frac{1}{6}(2)^3 + \frac{1}{2}C_1(2)^2 + C_2(2) + C_3 = 1\) 2. \(y^{\prime}(2) = -4\) \(\frac{1}{3}(2)^3 + \frac{1}{2}(2)^2 + C_1(2) + C_2 = -4\) 3. \(y^{\prime\prime}(2) = 7\) \((2)^2 + (2) + C_1 = 7\) Solving the system of equations, we get \(C_1 = 3, C_2 = -11, C_3 = 0\). Finally, the solution to the initial value problem is: \(y(x) = \frac{1}{12}x^4 + \frac{1}{6}x^3 + \frac{3}{2}x^2 - 11x\).

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving a function and its derivatives. These types of equations often describe how a quantity changes but involve only one independent variable, such as time or space. In ODEs, you might encounter first-order equations, which involve the first derivative of the function, or higher-order ones involving second or third derivatives.

Solving an ODE often involves finding a function that satisfies the equation. This requires methods like integration or separation of variables to solve them. Once the general solution is found, specific solutions can be determined by applying initial or boundary conditions specific to the problem.

ODEs are applied in various fields, such as physics, engineering, biology, and economics, modeling diverse phenomena from electrical circuits to population dynamics.

Integration by Parts

Integration by parts is a powerful technique used to solve integrals involving products of functions. It comes from the product rule of differentiation and is particularly useful when dealing with ODEs where simple integration won't work.

The formula for integration by parts is:
\[\int u\, dv = uv - \int v\, du\]
where \( u \) and \( dv \) are chosen from the product of functions in the integrand. The process involves differentiating \( u \) to get \( du \) and integrating \( dv \) to find \( v \).

An example is the initial value problem with \( y' = -x e^x \). Integration by parts simplifies \( \int -x e^x \, dx \) by letting \( u = -x \) and \( dv = e^x \, dx \), resulting in:
\[-xe^x + e^x + C\]
Properly applying integration by parts can greatly ease solving initial value problems involving differential equations.

Boundary Conditions

Boundary conditions are essential in determining specific solutions to differential equations. They are additional constraints that the solution must satisfy.

In initial value problems, boundary conditions specify the value of the function and sometimes its derivatives at a particular point. For example, a condition like \( y(0) = 1 \) sets the value of the function at \( x = 0 \).

Importantly, boundary conditions allow the determination of integration constants that may appear after solving an ODE. Without these conditions, solutions remain general forms, unable to describe specific physical situations or phenomena.

When applying boundary conditions, they are plugged into the solved differential equation to find the unknown constants, ensuring the solution satisfies both the ODE and the conditions given.

Higher-Order Differential Equations

Higher-order differential equations involve derivatives of the dependent variable beyond the first derivative. Commonly, these include second-order or third-order derivatives.

Solving higher-order differential equations often involves several steps of integration, requiring multiple constants of integration. Consider the differential equation \( y''' = 2x + 1 \). This third-order equation requires three integrations to find \( y(x) \), each introducing a new integration constant.

Again, boundary conditions play a crucial role. With conditions like \( y(2) = 1 \), \( y'(2) = -4 \), and \( y''(2) = 7 \), a system of equations is formed to solve for unknown constants.

• Higher-order equations are widespread in physics and engineering.
• They can describe systems with more complex interactions, such as mechanical vibrations.

Understanding and working through these equations are pivotal in many advanced scientific and mathematical applications.