7.4. Illustrative Examples

Solve:

y′′−3y′+2y =ex

Step 1: Solve the homogeneous part Auxiliary equation: r2−3r+2=0⇒(r−1)(r−2)=0⇒r =1,2 So, y =C e^x+C e^(2x) c 1 2

Step 2: Guess particular solution Since e^x is already part of y , we guess: c y =Ax e^x

Step 3: Substitute into original equation Compute: • y =A e^x+Ax e^x p′ • y =A e^x+A e^x+Ax e^x =2A e^x+Ax e^x p′′ Substitute into LHS: (2A e^x+Ax e^x)−3(A e^x+Ax e^x)+2Ax e^x =e^x Simplify: 2A e^x+Ax e^x−3A e^x−3Ax e^x+2Ax e^x =e^x ⇒(−A e^x)=e^x ⇒A=−1 So, y =−x e^x p

Final Solution: y(x)=C e^x+C e^(2x)−x e^x 1 2

In this example, we are solving a differential equation with an exponential forcing function. The first step involves finding the homogeneous solution by solving the auxiliary equation. Here, we find two roots which allow us to write the complementary function. Next, we assume a form for the particular solution, knowing that e^x is already part of the complementary function. By modifying the guess to Ax e^x, we substitute back into the original equation to find the constants. After simplification, we determine A = -1, giving us the particular solution. Finally, we combine this with the complementary function to arrive at the complete solution.

Imagine a bridge that is swaying due to the wind (exponential forces). The natural vibrations of the bridge represent the complementary function, which is part of its inherent structure. However, the additional sway caused by the wind requires us to adjust our calculations to account for the added force. Just as we adjust our equations to include this effect, civil engineers adjust models to account for real-world forces acting on structures.

Solve:

y′′+y =x2

Step 1: Homogeneous part: Auxiliary equation: r2+1=0⇒r =±i y =C cos(x)+C sin(x) c 1 2

Step 2: Guess particular solution Since RHS is x2, try: y =Ax2+Bx+C p Compute derivatives: • y =2Ax+B p′ • y =2A p′′ Substitute: 2A+(Ax2+Bx+C)=x2 ⇒Ax2+Bx+(2A+C)=x2 Comparing both sides: • A=1 • B =0 • 2A+C =0⇒2+C =0⇒C =−2 So, y =x2−2 p

Final Solution: y(x)=C cos(x)+C sin(x)+x2−2

In this case, we solve a polynomial force applied to a dynamic system. We start by determining the homogeneous solution using the auxiliary equation, which yields complex roots. This leads us to the complementary function involving sine and cosine. For the particular solution, we guess a polynomial form due to the nature of the forcing function. After substituting our guess back into the differential equation, we equate the coefficients of similar powers of x to solve for A, B, and C. Combining these results gives us the final solution.

Consider a child riding a bike down a gentle hill and occasionally pushing off the ground with their feet. The natural motion of the bike represents the system's inherent behavior, while the pushes from the child act like the polynomial forces applied to the bike’s motion. By calculating both the pedal (complementary response) and the child’s pushes (particular integral), we can accurately predict the bike's overall movement.

Solve:

y′′+4y =cos(2x)

Step 1: Homogeneous part: Auxiliary equation: r2+4=0⇒r =±2i y =C cos(2x)+C sin(2x) c 1 2

Since cos(2x) is part of the complementary function, try: y =x(Acos(2x)+Bsin(2x)) p Calculate derivatives and substitute into the equation, then compare coefficients to find A,B.

In this example, we are dealing with a trigonometric function as the forcing term. First, we find the homogeneous solution using the auxiliary equation, leading to a complementary function formed by sine and cosine with a frequency of 2. Since one of our terms is already part of the complementary function, we modify our guess for the particular solution by multiplying by x to avoid repetition. By substituting this new form into the differential equation and comparing coefficients, we can solve for the unknown constants A and B, completing the solution process.

Think of waves in the ocean interacting with a buoy. The natural oscillations of the buoy represent its inherent response to the water (the homogeneous solution). When additional waves hit the buoy, similar to the forcing function, we need to adjust our model to account for these new forces. Just as we analyze both the buoy's natural movements and the added oceanic waves, engineers analyze both complementary and particular responses in their calculations.

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