Solve the given initial-value problem. (d y)/(d x)=(y^2-1)/(x^2-1), y(2)=2

Name: Problem 24, Chapter 8: Calculus, Early Transcendentals
Uploaded: 2021-08-15T06:53:05-08:00
Duration: 4 min 1 s
Channel: Darshan Maheshwari
Description: Solve the given initial-value problem. (d y)/(d x)=(y^2-1)/(x^2-1), y(2)=2

step 1 This question we have to solve this initial value problem. So let's do a cross multiplication he

Solve the given initial-value problem. (d y)/(d...

Key Concepts

Separable Differential Equations

A separable differential equation is one in which the variables can be rearranged so that each side of the equation depends on only one variable. This typically involves rewriting the differential equation in the form N(y) dy = M(x) dx, which then allows the integration of both sides with respect to their respective variables.

Separation of Variables

Separation of variables is a technique used to solve differential equations by isolating the functions of each independent variable on opposite sides of the equation. After separating the variables, each side of the equation is integrated separately, leading to a solution that may be explicit or implicit.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic method used to break down a complex rational expression into a sum of simpler fractions. This is especially useful in integration when the integrand is a rational function that can be integrated term by term after decomposition.

Initial Value Problems

Initial value problems involve differential equations along with specified values of the unknown function at a given point. These conditions are used to determine the constant(s) of integration that arise during the process of solving the differential equation, and they ensure that the solution satisfies both the differential equation and the initial condition.

Transcript

00:01 This question we have to solve this initial value problem.

00:03 So let's do a cross multiplication here so that we if we can separate the variables.

00:09 So we have dy over y square minus 1 and this should be equal to dx over x square minus 1 integration.

00:18 All right.

00:19 So in order to integrate this, we can we can use a partial fraction.

00:27 So we'll do only partial fraction for any one because that's basically the same except the variable change.

00:35 So let's do it over the right.

00:37 So we have one over.

00:40 Before that, let's rewrite this so that it looks like we can do a partial fraction.

00:44 So this is nothing but a difference of perfect squares, y square minus one square.

00:48 So that can be written as using the identity.

00:50 That's y plus one times y minus one.

00:55 And likewise, this is also x plus one times.

00:58 X minus one and all the partial fraction can be done so let's do in for any one if we take y plus one times y minus one that can be written as a over y plus one plus b over y minus one so this can be written as one is equal to a times y minus one plus b times y plus one so if y is one then one is equal to this will become zero and this will become two b this means that the value of b is one over two and if y is minus one then this means that this will become zero and here we have minus two a which means that the value of a is minus one over minus one over two so we have the value of a and b so this can be rewritten as i'm just going to write the left side uh because the right side is going to be the replica so it's minus one over two over y plus one d y plus one over two over y minus one this is equal to minus 1 over 2 over x plus 1 d x...

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