Find the differential equation from the
function y=Ae^3x +Be^2x +Ce^x. Using the
method of determinant.
This might help:
http://tutorial.math.lamar.edu/Classes/DE/HOVariationOfParam.aspx
To find the differential equation from the given function y = Ae^(3x) + Be^(2x) + Ce^(x), we can use the method of determinants. The differential equation will involve the derivatives of y with respect to x.
1. Begin by finding the first derivative of y with respect to x.
dy/dx = 3Ae^(3x) + 2Be^(2x) + Ce^(x)
2. Now, find the second derivative of y with respect to x.
d²y/dx² = d/dx(3Ae^(3x) + 2Be^(2x) + Ce^(x))
= 9Ae^(3x) + 4Be^(2x) + Ce^(x)
3. Next, substitute the second derivative, d²y/dx², into the differential equation.
d²y/dx² = 9Ae^(3x) + 4Be^(2x) + Ce^(x)
4. The differential equation is obtained by equating the second derivative to zero, as follows:
9Ae^(3x) + 4Be^(2x) + Ce^(x) = 0
Therefore, the differential equation corresponding to the given function is:
9Ae^(3x) + 4Be^(2x) + Ce^(x) = 0
This equation represents the relationship between the function y = Ae^(3x) + Be^(2x) + Ce^(x) and its derivatives.