Solve the given initial-value problem. y^'+2 x^-1 y=4 x, y(1)=2

Name: Problem 16, Chapter 1: Differential Equations and Linear Algebra
Uploaded: 2020-06-10T00:42:35-08:00
Duration: 2 min 41 s
Channel: Kyle Findley
Description: Solve the given initial-value problem. y^'+2 x^-1 y=4 x, y(1)=2

Solve the given initial-value problem. y^'+2 x^-1 y=4 x, ...

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Key Concepts

Initial Value Problem

An initial value problem is a differential equation accompanied by one or more conditions specifying the value of the unknown function at a given point. This condition is used to determine a unique solution to the differential equation, ensuring that the particular solution fits the specific scenario or phenomenon being modeled.

First-Order Linear Differential Equations

A first-order linear differential equation involves the first derivative of the unknown function and is linear in the function itself. Such equations can be written in the standard form y' + p(x)y = q(x), where p(x) and q(x) are given functions. They are widely studied due to their applicability in modeling physical processes and their relatively straightforward solution methods.

Integrating Factor Method

The integrating factor method is a technique used to solve first-order linear differential equations. By multiplying the entire differential equation by an integrating factor—typically chosen as e^(?p(x) dx)—the left-hand side of the equation becomes the derivative of the product of the integrating factor and the unknown function. This allows for the equation to be integrated directly, leading to the general solution from which particular solutions can be obtained using initial conditions.

Transcript

00:01 Let's get to solving this problem, shall we? so to solve any first order linear differential equation, we need to put it in the standard form.

00:09 So the standard form would be y prime plus p of x.

00:14 Y is equal to q of x, where everything in front of the y is our p of x and everything on the right side is q of x.

00:20 So to solve this, we need to find integrating factor, integrating factor is always e to the integral of p of x.

00:28 So what's in front of y is our p of x.

00:30 So it's just going to be 2 over x.

00:32 So we need to do the integral of 2 over x, which if you just pull the 2 on the outside, the integral of 1 over x is going to be l and x.

00:41 So we end up with e to the 2 of lnx.

00:44 Now the log rules, we pull the 2 up top.

00:47 So end up with e to the ln of x squared.

00:50 Finally, when you do e to the ln, whatever is inside of the bracket of the ln gets popped out.

00:55 So our integrating factor is just x squared.

00:57 Now we just multiply both sides of the equation by the integrating factor.

01:00 So we end up with y prime plus 2 over x y x squared is equal to 4 x cubed because we just multiply x by x squared.

01:13 So now on the left hand side and the right hand side we take the integrals.

01:19 And the rule is that the left hand side is always y times their integrating factor...

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