Find the solution to the initial-value problem. (d y)/(d x)=y^3 x e^x^2, y(0)=1

Name: Problem 135, Chapter 4: Calculus Volume 2
Uploaded: 2020-07-08T12:50:21-08:00
Duration: 6 min 9 s
Channel: Carl David Cepeda
Description: Find the solution to the initial-value problem. (d y)/(d x)=y^3 x e^x^2, y(0)=1

Find the solution to the initial-value problem. (d y)/(d...

Key Concepts

Separable Differential Equations

A separable differential equation is one where the variables can be separated on different sides of the equation. This property allows the equation to be rewritten in the form f(y) dy = g(x) dx, making it easier to integrate each side individually. This concept is central in solving many ordinary differential equations by reducing them to integrals.

Separation of Variables

This technique involves rearranging a differential equation so that all terms involving one variable are on one side and all terms involving the other variable are on the opposite side. Once separated, each side can be integrated independently, providing a path to the solution of the differential equation.

Integration Techniques

Once the differential equation is separated, the next key step is performing the integrations on both sides. This might involve using techniques such as substitution, integration of exponential functions, or power rules. Mastery of various integration techniques is essential for solving a wide range of differential equations.

Incorporating Initial Conditions

After finding the general solution through integration, the next step is to use the initial condition to determine the constant of integration. This process ensures that the solution is tailored to the specific scenario described by the initial-value problem, thereby yielding a unique solution.

Transcript

00:01 Alright, so we have an initial value problem, but we have to get the general solution first by variable separation.

00:11 Okay, so this given should have y -quit -k, okay? so this becomes d -y over y -q, because x, e -rays to x squared, d -x.

00:27 Okay, and then get the anti -derivative of both sides of the equation.

00:31 Let's evaluate them individually.

00:36 So we have an integral screen integral of d .y over y cubed.

00:46 This can be written as integral of y raised negative 3dy, which is equal to y raised to negative 2 over negative 2.

00:58 Or, okay, negative 1 over 2.

01:04 Square.

01:06 Okay.

01:07 So this will be our left -hand side.

01:09 Next for the right -hand side, we have integral of x -e -rease -2 -x squared, dx.

01:16 Okay, we'll integrate this by substitution.

01:22 So we'll let u be equal to x squared and then du be equal to 2x dx.

01:30 Therefore, x -d -x is cancel out this one x ds is equal to d u over two okay so by substitution this equation becomes erase to u d u and then don't forget over two okay and then we have one half integral of erase to u d u or one half erase t okay so substituting our u back to our x squared we have one half e raise 2 x squared okay so this will be our right hand side there go our differential equation or the solution of our differential equation is negative 1 over 2 y squared because 1 half e raised 2 x squared and don't forget the arbitrary constant plus c okay and then solving for y okay, okay, we'll have, okay, solving for y, we have, let's do this, okay, so we can solve for y by multiplying both sides by two first.

02:59 Okay, so we have negative one over the y squared, equals erase to x squared, okay, our constant plus c will remain plus c, okay? if we're going to write two here, it doesn't matter, okay, because it's still a constant.

03:16 Okay, so let's not.

03:22 Or initially you can write two here for our calculation of the initial value problem.

03:29 Okay...

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