Solving the Differential Equation y' - 4y 3sin(2x) Using the Method of Undetermined Coefficients

In this article, we will demonstrate the step-by-step approach to solving the differential equation

The differential equation is given by:

y' - 4y 3sin(2x)

We will use the method of undetermined coefficients to find both the homogeneous and particular solutions. Let's begin!

Step 1: Solve the Homogeneous Equation

First, let's solve the associated homogeneous equation:

y' - 4y 0

By setting up the characteristic equation, we get:

r - 4 0

Solving for r, we find:

r 4

The roots of the characteristic equation yield a general solution for the homogeneous part:

{% math %}y_h C_1 e^{4x}{% endmath %}

However, this form is not suitable for our given non-homogeneous equation. We will use the appropriate form based on the right-hand side, i.e., trigonometric functions.

Step 2: Find a Particular Solution

We need to find a particular solution to the non-homogeneous equation. Given that the right-hand side is in the form of (3sin(2x)), we will use a trial form involving (x) to find the particular solution:

Assume:

{% math %}y_p x(Acos(2x) Bsin(2x)){% endmath %}

Step 3: Compute Derivatives

First derivative:

{% math %} y_p' Acdot2sin(2x) - 2Bcos(2x) (Acos(2x) Bsin(2x)) Ax(-2sin(2x) 2cos(2x)){% endmath %}

Second derivative:

{% math %} y_p'' -4Acos(2x) - 4Bsin(2x) 2(-2Asin(2x) - 2Bcos(2x)) 2(Acos(2x) Bsin(2x)) - 4Ax(2cos(2x) - 2sin(2x)){% endmath %}

Step 4: Substitute into the Original Equation

Substitute (y_p), (y_p'), and (y_p'') back into the original equation: {% math %}y_p' - 4y_p 3sin(2x){% endmath %}

After simplifying, we get:

{% math %}-4Asin(2x) - 4Bcos(2x) - 4Axcos(2x) 4Bxsin(2x) - 4Axcos(2x) - 4Bxsin(2x) 3sin(2x){% endmath %}

Simplifying further, the (x) terms cancel out:

{% math %}-4Asin(2x) - 4Bcos(2x) 3sin(2x){% endmath %}

Step 5: Set Coefficients Equal

We set the coefficients equal to match the right-hand side:

{% math %}-4A 3 quad text{and} quad -4B 0{% endmath %}

This yields:

{% math %}A -frac{3}{4} quad text{and} quad B 0{% endmath %}

Step 6: Write the Particular Solution

Therefore, the particular solution is:

{% math %}y_p -frac{3}{4} xcos(2x){% endmath %}

Step 7: General Solution

The complete solution is the sum of the homogeneous and particular solutions:

{% math %}y y_h y_p C_1 e^{4x} - frac{3}{4} xcos(2x){% endmath %}

Conclusion

The final solution to the differential equation, incorporating the homogeneous part and particular solution, is:

{% math %}y C_1 cos(2x) C_2 sin(2x) - frac{3}{4} xcos(2x){% endmath %}

Here, (C_1) and (C_2) are arbitrary constants determined by initial conditions.