Problem 40 - Advanced Engineering Mathematics

Question

Solve the given boundary-value problem. \(y^{\prime \prime}+3 y=6 x, y(0)+y^{\prime}(0)=0, y(1)=0\)

Step-by-Step Solution

Answer

The solution is: \(y(x) = -2\cos(\sqrt{3}x) + \frac{2(\cos(\sqrt{3})-1)}{\sin(\sqrt{3})}\sin(\sqrt{3}x) + 2x\).

Step 1: Solve the Complementary Equation

First, we solve the homogeneous equation related to the given boundary-value problem: \[ y'' + 3y = 0 \] The characteristic equation is \[ r^2 + 3 = 0 \]Solving for \(r\), we find \(r = ext{plus or minus } irac{ ext{sqrt}(3)}{ ext{sqrt}(3)}\). This indicates that the complementary solution is of the form:\[ y_c(x) = C_1 ext{cos}( ext{sqrt}(3)x) + C_2 ext{sin}( ext{sqrt}(3)x) \]

Step 2: Find the Particular Solution

Next, we find a particular solution to the non-homogeneous equation \[ y'' + 3y = 6x \]We use the method of undetermined coefficients. Assume a particular solution of the form \[ y_p(x) = Ax + B \]The second derivative \(y_p''(x) = 0\). Substituting into the equation:\[ 3(Ax + B) = 6x \]This gives us \[ 3Ax + 3B = 6x \] Comparing coefficients, we get \(A = 2\) and \(B = 0\). Hence the particular solution is:\[ y_p(x) = 2x \]

Step 3: Form the General Solution

The general solution to the differential equation is the sum of the complementary and particular solutions:\[ y(x) = y_c(x) + y_p(x) = C_1 ext{cos}( ext{sqrt}(3)x) + C_2 ext{sin}( ext{sqrt}(3)x) + 2x \]

Step 4: Apply Boundary Conditions

Apply the boundary conditions to find the constants \(C_1\) and \(C_2\).First condition: \[ y(0) + y'(0) = 0 \]Substituting \(x = 0\):\[ C_1 + 2 = 0 \]This gives \(C_1 = -2\).Second condition: \[ y(1) = 0 \]Substituting \(x = 1\):\(-2 ext{cos}( ext{sqrt}(3)) + C_2 ext{sin}( ext{sqrt}(3)) + 2 = 0\).

Step 5: Solve for Constants

From the equation obtained in Step 4:\[ -2 ext{cos}( ext{sqrt}(3)) + C_2 ext{sin}( ext{sqrt}(3)) + 2 = 0 \]Isolate \(C_2\) and solve:\[ C_2 ext{sin}( ext{sqrt}(3)) = 2 ext{cos}( ext{sqrt}(3)) - 2 \]Therefore, \[ C_2 = \frac{2( ext{cos}( ext{sqrt}(3)) - 1)}{ ext{sin}( ext{sqrt}(3))} \]

Step 6: Write the Solution with Constants

With \(C_1 = -2\) and \(C_2\) calculated, the solution can be expressed as:\[ y(x) = -2 ext{cos}( ext{sqrt}(3)x) + \left(\frac{2( ext{cos}( ext{sqrt}(3)) - 1)}{ ext{sin}( ext{sqrt}(3))}\right) ext{sin}( ext{sqrt}(3)x) + 2x \]

Key Concepts

Complementary Solution

Understanding the complementary solution is key in solving differential equations. It represents the general solution to the homogeneous part of a differential equation, like our equation: \[ y'' + 3y = 0 \]This is the equation derived by setting the non-homogeneous part (in our case, 6x) to zero.
To find the complementary solution, we determine the roots of the characteristic equation linked to the differential equation. Here, the characteristic equation:\[ r^2 + 3 = 0 \]reveals roots that are imaginary numbers, which tells us that our complementary solution involves sinusoidal functions.
These roots \( r = \pm i \sqrt{3} \)led us to the complementary solution:\[ y_c(x) = C_1 \cos( \sqrt{3}x ) + C_2 \sin( \sqrt{3}x ) \]where \(C_1\) and \(C_2\) are arbitrary constants. These express any function that satisfies the homogeneous part of the differential equation.

Particular Solution

The particular solution is specific to the non-homogeneous part of the differential equation. In our example, this is the term 6x in the equation \(y'' + 3y = 6x\).
To find it, one effective method is using undetermined coefficients. This approach involves assuming a form for the particular solution based on the non-homogeneous term and then finding coefficients that satisfy the equation.
For our problem, we assumed \(y_p(x) = Ax + B\),with a second derivative \(y_p''(x) = 0\).When plugged into our original equation, we got:\[3(Ax + B) = 6x\]
By equating the coefficients from both sides, we discovered:

• \(A = 2\)
• \(B = 0\)

This results in the particular solution:\[y_p(x) = 2x\]which only satisfies the specific form of the original differential equation’s non-homogeneous part.

Undetermined Coefficients

The method of undetermined coefficients is a popular technique to find a particular solution to linear differential equations. It is especially used when the non-homogeneous term is a simple function like polynomials, exponentials, or trigonometric functions.
In this method, you guess the form of the particular solution based on the non-homogeneous term.
Let's break this down:

• Identify the type of non-homogeneous term (e.g., polynomial \(6x\)).
• Make an educated guess regarding the form (e.g. \(y_p(x) = Ax + B\)).
• Substitute back into the differential equation to determine the coefficients (A and B in our case).

This approach allows you to find a solution part that compensates the non-homogeneous component of the equation, which complements the solution to the homogeneous part. Thus, together they form the complete solution to the differential equation.