Yes, in this case p(x)=-1. However, you're answer isn't correct as it doesn't satisfy the initial value problem. You have made a little slit in the second to last line of your working.mad_monkey_j said:Homework Statement
Find the solution to the initial value problem
dy/dx - y = e^3x
y(0) = 3
Homework Equations
e^∫p(x)
The Attempt at a Solution
Do I treat p(x) = -1?
I(x) = e^∫-1 = e^-x
e^-x(dy/dx) - ye^-x = e^3x . e^-x
e^-x(dy/dx) - e^-x . y = e^2x
e^-x . y = ∫e^2x
y = (2e^2x + c)/(e^-x)
y = C+2e^3x?
An initial value problem is a type of differential equation that involves finding a function that satisfies both a given differential equation and a set of initial conditions. These initial conditions specify the value of the function at a certain point or points.
An initial value problem can be solved by using various methods such as separation of variables, substitution, and integrating factors. These methods involve manipulating the given differential equation to isolate the dependent and independent variables, and then using the initial conditions to determine any constants of integration.
An initial value problem is unique because it requires finding a specific function that satisfies both the differential equation and the given initial conditions. This means that there is only one solution to the problem, rather than a general solution that can be found for other types of differential equations.
No, an initial value problem can only have one unique solution. This is because the initial conditions are used to determine any constants of integration, leaving no room for multiple solutions.
To check if a solution to an initial value problem is correct, you can plug the solution into the given differential equation and see if it satisfies both the equation and the initial conditions. It is also helpful to graph the solution to visualize if it aligns with the initial conditions.