Solving the Differential Equation (d^2-5d+6)y=e^4x
In this article, we will solve the differential equation (d^2-5d+6)y=e^4x
. This is a second-order linear differential equation, and we will use the method of undetermined coefficients to find the particular solution.
Step 1: Find the Homogeneous Solution
First, we need to find the homogeneous solution of the differential equation, i.e., the solution to the equation (d^2-5d+6)y=0
. To do this, we can use the characteristic equation, which is given by:
r^2 - 5r + 6 = 0
Factoring the characteristic equation, we get:
(r - 2)(r - 3) = 0
This gives us two roots, r = 2
and r = 3
. Therefore, the homogeneous solution is given by:
y_h = c1e^(2x) + c2e^(3x)
where c1
and c2
are arbitrary constants.
Step 2: Find the Particular Solution
Next, we need to find the particular solution of the differential equation, i.e., the solution to the equation (d^2-5d+6)y=e^4x
. To do this, we will use the method of undetermined coefficients.
Let's assume that the particular solution has the form:
y_p = Ae^(4x)
where A
is an undetermined coefficient.
Substituting this into the original differential equation, we get:
(d^2-5d+6)(Ae^(4x)) = e^4x
Simplifying the equation, we get:
16Ae^(4x) - 20Ae^(4x) + 6Ae^(4x) = e^4x
Collecting like terms, we get:
2Ae^(4x) = e^4x
Dividing both sides by e^4x
, we get:
2A = 1
A = 1/2
Therefore, the particular solution is given by:
y_p = (1/2)e^(4x)
Step 3: Find the General Solution
Finally, we can find the general solution by combining the homogeneous solution and the particular solution:
y = y_h + y_p
y = c1e^(2x) + c2e^(3x) + (1/2)e^(4x)
This is the general solution of the differential equation (d^2-5d+6)y=e^4x
.
Conclusion
In this article, we have solved the differential equation (d^2-5d+6)y=e^4x
using the method of undetermined coefficients. We have found the general solution, which is a combination of the homogeneous solution and the particular solution.