Answer:
Explanation:
We can use the method of undetermined coefficients to solve this differential equation. First, we will need to find the solution to the homogeneous equation and the particular solution to the non-homogeneous equation.
For the homogeneous equation, we will use the form y"+ky=0, where k is a constant. We can find the solutions to this equation by letting y=e^mx,
y"=m^2e^mx -> (m^2)e^mx+k*e^mx=0, therefore (m^2+k)e^mx=0
(m^2+k) should equal 0 for the equation to have a non-trivial solution. Therefore, m=±i√(k), and the general solution to the homogenous equation is y=A*e^i√(k)x+Be^-i√(k)*x.
Now, we need to find the particular solution to the non-homogeneous equation. We can use the method of undetermined coefficients to find the particular solution. We will let yp=a0+a1x+a2x^2+.... As the derivative of a sum of functions is the sum of the derivatives, we get
yp″=a1+2a2x....yp‴=2a2+3a3x+....
Substituting the general solution into the non-homogeneous equation, we get
a0+a1x+a2x^2+...+2a2x+3a3x^2+...=Y(4)
So, the coefficient of each term in the expansion of the left hand side should equal the coefficient of each term in the expansion of the right hand side. Since there is only one term on the right hand side, we get the recurrence relation:
a(n+1)=Y(n-2)/n^2
From this relation, we can find all the coefficients in the expansion for the particular solution. Using this particular solution, we can find the total solution to the differential equation as the sum of the homogeneous solution and the particular solution.