y%3D1%2F1%2Bsinx.jpeg "(d^2+1)y=1/1+sinx")
Solving the Differential Equation (d^2+1)y=1/1+sinx
Introduction
In this article, we will discuss how to solve the differential equation (d^2+1)y=1/1+sinx. This equation is a type of second-order linear ordinary differential equation (ODE).
Method of Solution
To solve this equation, we will use the method of undetermined coefficients. This method involves assuming a particular solution of the form y_p = A + Bsinx + Ccosx, where A, B, and C are constants to be determined.
Step 1: Find the Homogeneous Solution
First, we need to find the homogeneous solution of the equation, which is obtained by setting the right-hand side of the equation to zero:
(d^2+1)y = 0
To solve this equation, we can use the characteristic equation method. The characteristic equation is:
r^2 + 1 = 0
Solving for r, we get:
r = ±i
Therefore, the homogeneous solution is:
y_h = c_1 cosx + c_2 sinx
where c_1 and c_2 are constants.
Step 2: Find the Particular Solution
Next, we need to find the particular solution of the equation. We assume a particular solution of the form:
y_p = A + Bsinx + Ccosx
Substituting this into the original equation, we get:
(d^2+1)y_p = (d^2+1)(A + Bsinx + Ccosx) = 1/1+sinx
Simplifying and equating coefficients, we get:
A = 1/2, B = -1/2, and C = 0
Therefore, the particular solution is:
y_p = 1/2 - 1/2sinx
Step 3: Find the General Solution
The general solution is the sum of the homogeneous and particular solutions:
y = y_h + y_p = c_1 cosx + c_2 sinx + 1/2 - 1/2sinx
Simplifying, we get:
y = c_1 cosx + (c_2 - 1/2)sinx + 1/2
This is the general solution of the differential equation (d^2+1)y=1/1+sinx.
Conclusion
In this article, we have shown how to solve the differential equation (d^2+1)y=1/1+sinx using the method of undetermined coefficients. We have obtained the general solution of the equation, which involves arbitrary constants c_1 and c_2.
