Consider the differential equation x u^''-(x+1) u^'+u=0. Suppose we know the solution to the ini

Consider the differential equation x u^''-(x+1) u^'+u=0....

Consider the differential equation $x u^{\prime \prime}-(x+1) u^{\prime}+u=0$. Suppose we know the solution to the initial value problem $u(1)=2, u^{\prime}(1)=1$ is $u(x)=x+1$, while the solution to the initial value problem $u(1)=1, u^{\prime}(1)=1$ is $u(x)=e^{x-1}$. (a) What is the solution to the initial value problem $u(1)=3, u^{\prime}(1)=-2$ ? (b) What is the general solution to the differential equation?

Key Concepts

Determination of Constants Using Initial Conditions

Once the general solution is expressed as a linear combination of independent solutions, the unknown constants can be calculated by substituting the given initial conditions (values of the function and its derivative) into the general solution and solving the resulting system of equations.

Initial Value Problems (IVP) in Differential Equations

An initial value problem specifies the value of the unknown function and its derivatives at a particular point. The given initial conditions are used together with the general solution of the differential equation to determine the specific constants that make the solution satisfy those conditions.

Second Order Linear Homogeneous Differential Equations

These are differential equations involving the second derivative of an unknown function, where all terms depend linearly on the function and its derivatives. The key property is that the set of solutions forms a vector space, meaning any linear combination of solutions is also a solution.

Superposition Principle

In the context of linear differential equations, the superposition principle states that if you have two independent solutions, then any linear combination of these solutions is also a solution. This allows the construction of the general solution as a sum of multiples of the independent solutions.

Transcript

00:01 So when we have a separable differential equation, we literally separate it, meaning we get y on one side and x on the other side, ignoring the derivative for now.

00:13 So basically what i'm saying is we're going to subtract y from both sides first, so we have everything in terms of y on one side and terms of x on the other side.

00:22 So we get when we subtract y, right, because this here is y.

00:28 We get x times y prime is equal to y squared minus y it's often helpful to rewrite y prime as d y by d x and on the right side we could might as well factor that y out and then from here we've separated it now we can essentially um we can essentially multiply both sides by d x um um but, and the order is up to you in terms of how you do this, but of course from now, we can't, you know, integrate y in terms of x in terms of y like that and the other way around here.

01:16 What you want to do first is we want to get everything that's in terms of y on the side with dy, and that's in terms of x on the side with dx.

01:25 So basically what we need to do is multiply both sides by x and multiply both sides by y times y minus 1.

01:31 And when we do that, we basically get that 1 over y times y minus 1.

01:39 I don't know what i said earlier.

01:41 I meant to say y times y minus 1, but yeah, is dy is equal to 1 over x, dx.

01:50 Now we can integrate both sides.

01:54 For the left side, we can do the partial fraction method, in which case you get integral of 1 over y minus 1, minus 1 over y over y.

02:11 And then on the right side, that's easy.

02:15 We know that's long of absolute value x, and then we can put an arbitrary constant of c.

02:20 On the left side, it's also easy to integrate...

Please give Ace some feedback