Question

Find the general solution to the differential equations. $$ y^{\prime}=2^{x}+\cos ^{-1} x $$

Short answer

The general solution is \( y = \frac{2^x}{\ln(2)} + x \cos^{-1}(x) + \sqrt{1-x^2} + C \).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \( y' = 2^x + \cos^{-1}(x) \). This is a first-order differential equation because it involves the first derivative of \( y \), and it is non-homogeneous due to the presence of \( 2^x + \cos^{-1}(x) \).

Integrate Both Sides

To find the general solution, we integrate both sides of the equation with respect to \( x \). Start by integrating the left-hand side, which is \( y' \,dx = dy \). So integrating \( dy \) simply yields \( y \). For the right-hand side, integrate \( 2^x + \cos^{-1}(x) \) with respect to \( x \).

Integrate \( 2^x \)

The integral of \( 2^x \) with respect to \( x \) is \( \frac{2^x}{\ln(2)} \). This is because the derivative of \( \log(2) \times 2^x \) with respect to \( x \) gives \( 2^x \).

Integrate \( \cos^{-1}(x) \)

The integral of \( \cos^{-1}(x) \) with respect to \( x \) is more complex. It is typically calculated using integration by parts, yielding \( x \cos^{-1}(x) + \sqrt{1-x^2} + C_1 \), where \( C_1 \) is an integration constant.

Write the General Solution

Combine the results from Steps 3 and 4 to write the general solution. The solution is:\[ y = \frac{2^x}{\ln(2)} + x \cos^{-1}(x) + \sqrt{1-x^2} + C \]where \( C \) is the constant of integration accounting for all arbitrary constants.

Key concepts

First-Order Differential Equation

A first-order differential equation is an equation that involves the first derivative of a function. Derivatives indicate how a function changes as its input changes. In the context of this exercise, we have the equation\( y' = 2^x + \cos^{-1}(x) \). This is a clear example of a first-order differential equation because the highest derivative is \( y' \), representing the first derivative of \( y \) with respect to \( x \).
First-order differential equations can usually be solved through straightforward integration. They are often the starting point in learning about differential equations because they are simpler to handle than higher-order ones. In practical applications, they can model a variety of dynamics such as population growth or cooling processes.

Non-Homogeneous Differential Equation

A non-homogeneous differential equation is one that has terms that are not all related to the function and its derivatives. In our case, the right-hand side of the equation \( y' = 2^x + \cos^{-1}(x) \) is not zero, indicating a non-homogeneous differential equation.
Non-homogeneous equations include additional functions that remove them from being purely related to the derivatives. This makes them more complex as they model systems influenced by external factors. For instance, in physics, they represent systems where external forces, like friction or an applied force, are acting.

• They differ from homogeneous equations, which have zero on one side and only involve terms with the function and its derivatives.
• Solving such equations usually requires finding a specific solution to the non-homogeneous part and adding it to the solution of the associated homogeneous equation.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. This is necessary when straightforward integration isn’t possible or becomes cumbersome. The formula for integration by parts is a reworking of the product rule for derivatives and is written as\( \int u \, dv = uv - \int v \, du \). In the context of this exercise, integration by parts is used to find the integral of \( \cos^{-1}(x)\).
Here's how integration by parts works:

• First, identify two parts of the function to be integrated: one part to differentiate (\( u\)) and one part to integrate (\( dv\)).
• Differentiating \( u\) gives \( du\), and integrating \( dv\) gives \( v\).
• Plug these into the formula \( \int u \, dv = uv - \int v \, du \) to find the solution.

This method enables us to tackle more complex integrals step by step by simplifying the integration process. It is particularly handy for inverse trigonometric functions like \( \cos^{-1}(x) \).

General Solution

The general solution of a differential equation encompasses all possible solutions. For our equation, after integrating each part separately, we synthesize the individual answers into a general solution: \[ y = \frac{2^x}{\ln(2)} + x\cos^{-1}(x) + \sqrt{1-x^2} + C \]Here, \( C \) is the constant of integration, accounting for any constant shift in the solution.
Understanding the general solution is crucial because:

• It provides a complete set of solutions for the differential equation.
• Given specific initial conditions, the exact value of \( C \) can be determined to find a particular solution.

By combining the results from integrating each piece (both \( 2^x \) and \( \cos^{-1}(x) \)), we get the most comprehensive version of the solution.