uber_kim said:Homework Statement
Solve the initial value problem:
t(dy/dt)+8y=t^3 where t>0 and y(1)=0
Homework Equations
None?
The Attempt at a Solution
It's a linear equation, so rearranged to dy/dt+8y/t=t^2.
Took the integrating factor e^(∫8/tdt)=t^8 and multiplied through.
(t^8)dy/dt+8t^7y=t^10.
∫(t^8y)'dt=∫t^10dt
t^8y=(t^11)/11 + C
Solve for C, I got -1/11.
Final solution is y=(t^3)/11 - 1/(11t^8)
This isn't right, though. Does anyone see where I made the mistake?
A first order differential equation is an equation that involves a function and its first derivative. It is used to model various physical and mathematical phenomena in science and engineering.
Initial conditions refer to the values of the function and its derivative at a specific point. These conditions are used to uniquely determine the solution to the differential equation.
To solve a first order differential equation with initial conditions, you can use various methods such as separation of variables, integrating factors, and substitution. The specific method used depends on the form of the equation.
The general solution to a first order differential equation with initial conditions is a family of solutions that satisfy the equation. It includes a constant of integration that is determined by the initial conditions.
Sure, let's consider the equation dy/dx = x + 2 with the initial condition y(0) = 1. Using separation of variables, we can rewrite the equation as dy = (x + 2)dx. Integrating both sides gives us y = (x^2/2) + 2x + C. Plugging in the initial condition, we get C = -1. Therefore, the solution to the differential equation with the given initial condition is y = (x^2/2) + 2x - 1.