Differential Equations initial value problem

In summary, To solve the given initial value problem, use the integrating factor method by finding u(t) such that d(uy)/dt = (1/2)t^3 and then solve for f(t) by multiplying both sides of the equation by u(t) and integrating. You can also use a clever substitution, such as t^αy, to simplify the equation.
  • #1
hocuspocus102
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Homework Statement



Let f(t) be the solution to the initial value problem 2t(dy/dt)+y=t^4
with f(0) = 0
find f(t).

Homework Equations





The Attempt at a Solution



I tried to do this by separating variables but that hasn't gotten me very far. I don't know if I can do it by doing that thing where you find an integrating factor and like raising e^(some integral)? It didn't really make any sense to me when I tried it, so could someone explain it please? Thanks!
 
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  • #2
Or a cleaver substitution? Something of the form [tex]t^\alpha y[/tex] with some handy value of [tex]\alpha[/tex]? Does not the left hand side of your equation look like a derivative of a something?
 
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  • #3
Because it is a linear equation, there is a formula for the "integrating factor". Write it as [itex]dy/dx+ y/2t= (1/2)t^3[/itex].

Now look for u(t) such that [itex]d(uy)/dt= u (dy/dt)+ (du/dt)y= u dy/dt+ (u/2t)y[/itex]
 
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FAQ: Differential Equations initial value problem

What is a differential equation initial value problem?

A differential equation initial value problem is a mathematical equation that describes the relationship between a function and its derivatives. It involves finding a function that satisfies both the differential equation and a set of initial conditions.

How do you solve a differential equation initial value problem?

To solve a differential equation initial value problem, you must first identify the type of differential equation and then use various techniques such as separation of variables, integrating factors, or power series to find a solution that satisfies both the equation and the initial conditions.

What are initial conditions in a differential equation initial value problem?

Initial conditions refer to the values of the function and its derivatives at a specific point in the domain. These conditions are used to determine a specific solution to the differential equation.

What is the difference between an ordinary and a partial differential equation initial value problem?

An ordinary differential equation involves a single independent variable, while a partial differential equation involves multiple independent variables. As a result, the initial conditions for an ordinary differential equation are specified at a single point, while for a partial differential equation they are specified on a portion of the domain.

Why are differential equation initial value problems important?

Differential equation initial value problems have important applications in various fields, including physics, engineering, economics, and biology. They are used to model real-world phenomena and make predictions about how these systems will behave over time.

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