Show that y(t)=3 /(6 t-11) is a solution of y^'=-2 y^2, y(2)=3. Sketch this solution and discuss

step 1 Hello and welcome to problem 16 of chapter 2 section 4. Here we're asked to solve the given init

Show that y(t)=3 /(6 t-11) is a solution of y^'=-2 y^2,...

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Key Concepts

Verification of Differential Equation Solutions

This involves substituting the proposed function into the differential equation. By differentiating the function and showing that the derivative satisfies the given ODE, one confirms that the function is indeed a solution. This process reinforces understanding of the relationship between functions and their derivatives.

Initial Value Problems (IVP)

An IVP pairs a differential equation with a specified initial condition. This condition, often given as the value of the solution at a particular point, is critical for determining a unique solution among all potential solutions of the differential equation.

Interval of Existence

This concept refers to the set of values of the independent variable over which the solution to a differential equation remains valid and free from singularities. It involves identifying points where the solution may become undefined, ensuring that only the appropriate segment of the solution is considered.

Qualitative Analysis and Sketching of Solutions

Sketching the solution graphically helps visualize its behavior, including key features such as the initial condition and any asymptotic behavior. This analysis aids in understanding the overall dynamics of the system described by the differential equation.

Transcript

00:01 Hello and welcome to problem 16 of chapter 2 section 4.

00:04 Here we're asked to solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value value.

00:13 Y prime is equal to t squared all over y times 1 plus t cubed and of course y value at zero is to why not.

00:21 So let's start off by solving this differential equation.

00:25 Since it's separable, we can get the y into the left side.

00:29 So we'll have y prime times.

00:31 Times y is equal to t squared over one plus t cubed here we can just take the integral of both sides and we'll have um one half y squared is equal to the integral of t squared over one plus t cubed let's just use u substitution so if u is equal to one plus t cubed for d u is equal to um three t squared so we're left with the integral of d t.

01:10 So we're left with the integral of u over d u times one third, i believe.

01:19 So it's yeah, like that.

01:24 And of course, this is just a natural log.

01:26 So we'll have one, back to the green, 1 half y squared is equal to 1 third, the natural log of 1 plus t cubed plus c on the outside.

01:46 Here from here, let's just solve for y.

01:49 So we can multiply by two and get two thirds.

01:55 And then we can just take the square root.

01:57 So y is equal to square root of two thirds...