8.7. Advanced Example

Solve the differential equation:

π y′′+y =tanx, 0<x<2

The section begins with a differential equation that we need to solve. This equation is a non-homogeneous second-order linear differential equation, where the left side contains the second derivative of y and a first-power term of y, while the right side is a tangent function. The domain specified here is between 0 and π/2, indicating a range within which we are looking for solutions.

Think of this as a problem in physics where you want to understand how a pendulum swings under the influence of a varying external force (representing the tan(x) function). The values of x are limited to a particular range, just like how you might only observe the pendulum's motion within a specific timeframe.

Step 1: Solve the homogeneous equation:

y′′+y =0⇒r2+1=0⇒r =±i

So, the general solution to the homogeneous equation is: y (x)=C cosx+C sinx

Thus, y (x)=cosx, y (x)=sinx

In this step, we first solve the homogeneous part of the equation, which is y'' + y = 0. We find the characteristic equation, which gives us complex roots (±i). This indicates that the solutions will be oscillatory. Therefore, the general solution to the homogeneous part is expressed using cosine and sine functions, implying that the system's natural response is sinusoidal in nature.

Imagine you are watching a swinging pendulum. The natural motion of the pendulum corresponds to the cosine and sine functions, which describe how it swings back and forth. This represents the system's response when there are no external forces acting upon it.

Step 2: Compute the Wronskian:

W(x)=(cid:12) (cid:12) (cid:12) (cid:12)−cos

sin x

x (cid:12) (cid:12)cos

x

sin x (cid:12)(cid:12)

=cos2x+sin2x=1

The Wronskian is a determinant used to check the linear independence of the solutions we found earlier (cos x and sin x). In this step, we calculate the Wronskian and find that it equals 1. A non-zero Wronskian indicates that the solutions are linearly independent, confirming we can use them to construct the general solution to the non-homogeneous equation.

Think of the Wronskian as a tool that helps us determine if two paths (the solutions) are unique and can coexist without overlapping. If you have two rivers that flow differently without merging, they are analogous to the independent solutions we derived.

Step 3: Compute u (x) and u (x):

1′ 2′

y g sin2x

u =− 2 =−sinx·tanx=−

1′ W cosx

y g sinx

u = 1 =cosx·tanx=

2′ W 1

Next, we compute the functions u1(x) and u2(x) that will help in constructing the particular solution. These functions are derived from the non-homogeneous term (tan x) and the Wronskian we calculated. By substituting the required components into the formulas for u1 and u2, we obtain expressions that account for how the input function (tan x) influences our overall solution.

Imagine trying to fit different pieces into a puzzle. Here, u1 and u2 are like special pieces that adjust the shape of our overall solution to perfectly match the contour defined by the external force represented by tan x.

Now integrate:

Z sin2x Z (cid:18)1−cos2x(cid:19) Z Z

tud =− dx=− dx=− secxdx+ cosxdx

u =− dx=− ln|secx+tanx|+sinx

Z u = sinxdx=−cosx

After deriving u1 and u2, we proceed to integrate these functions. The integration is necessary to find the specific functions u1(x) and u2(x) which will be utilized in constructing the particular solution. It involves using standard integration techniques to find these functions, which can sometimes get complicated depending on the expressions we have.

Think of integration as baking a cake. Each layer (integral) contributes to the final flavor of your cake (the complete solution). If you can’t mix the right ingredients (function components) properly, your cake will not turn out as expected!

Step 4: Construct the particular solution:

y =u y +u y =(sinx−ln|secx+tanx|)cosx−cosx·sinx p 1 1 2 2

With u1 and u2 now integrated, we substitute them back into the assumed form of the particular solution. This involves multiplying the functions we computed by the homogeneous solutions and combining them to form a complete expression that satisfies the original differential equation.

Imagine combining different colors of paint to create a new shade. Each u function we integrated reflects a unique color, and together they create a solution—a new shade—that accurately reflects the effects of the external forces acting on our system.

Final solution: y(x)=C cosx+C sinx−ln|secx+tanx|cosx

The final step is to express the complete solution to the differential equation by combining the homogeneous solution with the particular solution we derived. The result illustrates how the system behaves under the influence of both its natural tendencies and the external forces described by the tan function.

The final equation is like a complete story of a character's journey. The first part features the natural characteristics of the character (homogeneous solution), while the latter part describes how external events (the force of tan x) influence their journey and change their behavior.

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