(d^2-dd'-2d'^2)z=(y-1)e^x

3 min read Jul 03, 2024

Solving the Differential Equation: (d^2 - dd' - 2d'^2)z = (y-1)e^x

In this article, we will discuss the solution to the differential equation (d^2 - dd' - 2d'^2)z = (y-1)e^x. This is a second-order linear differential equation, which can be solved using various techniques.

Step 1: Simplify the Equation

To simplify the equation, let's first combine the terms with d'^2:

(d^2 - dd' - 2d'^2)z = (y-1)e^x

d^2z - dd'z - 2d'^2z = (y-1)e^x

Now, let's try to find the general solution to the homogeneous equation:

d^2z - dd'z - 2d'^2z = 0

Step 2: Find the Homogeneous Solution

To find the homogeneous solution, we can assume that z = e^(rx) is a solution, where r is a constant. Substituting this into the homogeneous equation, we get:

(r^2 - r^2 - 2r^2)e^(rx) = 0

Simplifying the equation, we get:

-2r^2e^(rx) = 0

This implies that r = 0 or r = ±√2. Therefore, the general solution to the homogeneous equation is:

z_h = c1e^(√2x) + c2e^(-√2x)

where c1 and c2 are arbitrary constants.

Step 3: Find the Particular Solution

To find the particular solution, we can use the method of undetermined coefficients. Assume that the particular solution has the form:

z_p = Ae^x + Bxe^x

Substituting this into the original equation, we get:

(A + Bx)e^x - (A + B)e^x - 2(B + Bx)e^x = (y-1)e^x

Simplifying the equation, we get:

(A - B - 2B)e^x + Bxe^x = (y-1)e^x

Comparing coefficients, we get:

A - B - 2B = y - 1

B = 0

Solving for A, we get:

A = y - 1

Therefore, the particular solution is:

z_p = (y-1)e^x

Step 4: Find the General Solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions:

z = z_h + z_p

z = c1e^(√2x) + c2e^(-√2x) + (y-1)e^x

This is the general solution to the differential equation (d^2 - dd' - 2d'^2)z = (y-1)e^x.