Solving Initial Value Problems
Solve the initial value p...

Question

Solving Initial Value Problems

Solve the initial value problems in Exercises 15–20.

(x+1) dy/dx - 2 (x² + x)y = exp(x²) / (x+1), x > -1, y(0) = 5

Accepted Answer

Hello. In this video, we are going to be solving the given initial value problem. So we are given a differential equation that is the product of X + 2 multiplied by DYDX minus 2 multiplied by X2 + 2 X multiplied by Y, and that is equal to the fraction E raises the power of X2 divided by X + 2, and we are told that the initial condition is that Y 0 is equal to 3. In order to solve this initial value problem, the first thing we need to do is we need to solve for a general solution. Now, in order to solve for the general solution, we need to go ahead and solve the differential equation. So the first thing we want to check and make sure to do is that our differential symbol DYDX has a coefficient of 1. But what we want to do in this case is you want to simplify this equation by dividing the entirety of the equation by X + 2. Now, before we do that, let's simplify the second term. If we take a look at the second term, we can factor out an X term. So we can rewrite the differential equation as X + 2 multiplied by D Y D X minus 2 multiplied by the quantity X multiplied by X + 2 multiplied by Y. And that will equal toe raised the power of X 2 divided by X + 2. Now if we were to divide the entirety of this equation by x + 2, we can further simplify the differential equation to be D Y D X minus 2x multiplied by y, and that will equal toe raise the power of x 2 divided by the quantity x + 2 quantity squared. So if we take a look at this differential equation, we have a first order linear differential equation. The general form is DYDX plus a function of X, which is P of X multiplied by Y. And that will equal to another function of X Q X. So here if we compare this general form of the linear differential equation to ours. -2 X is going to be our PX function, and the right-hand side of this equation will be our QX function. So in order to simplify this differential equation, we need to identify our integrating factor. Now the integrating factor is defined as new, and that is equal to e raising the power of the integral of p of X. Now in this case here, we know that P X is equal to -2x, so our integrating factor will be defined as E raised to the integral of 2 X. This is going to simplify to be ease the power of negative x. Of g x 2 and this is going to further simplify, well, this will be the integrating factor which is e raised the power of x squared. So, the integrating factor is going to help us simplify the overall differential equation. What we want to do in this case is we want to take the integrating factor, which is e raised the power of x 2, and we want to multiply this to the entirety of our differential equation. So, once we multiply our integrating factor to the entirety of the equation, we can rewrote our differential equation to be E to the power of X2, DYDX minus 2X multiplied by E, raise the power of X2 multiplied by Y. And that will be equal to 1 divided by x + 2 quantity squared. Now, in order to continue simplifying, this is a little tricky to see at first, but notice that the left hand side of this differential equation is equivalent to a derivative. It is the derivative of the product of e to the negative x 2 power multiplied by y. So we can further simplify this differential equation by rewriting the left hand side as a derivative. So we will have the derivative with respect to x of e raised to the power of x2 multiplied by y, and that will equal to 1 divided by x + 2 quantity squared. Now in this case here, we need to go ahead and simplify by removing the differential symbol. In order to do this, we will integrate both sides of the equation. Now the left hand side will simplify to bee raised to the power of X 2 Y, and the right hand side will integrate to be 1 divided by X + 2 plus the generic constant C. Now, we want to make sure that Y is isolated on the left-hand side of the equation, which means that we need to remove this E to the power of X2. In order to do that, we will multiply the entirety of the equation bye raise the power of x2. So that will allow us to rewrite the general, the general solution as Y equal. To e raised the power of x2 multiplied by c minus 1 divided by x + 2. So what we have here is the general solution. But now we can use the general solution to find the specific initial value. Now recall that the problem tells us to use the initial condition of Y 0 equal to 3. So, if we use y of 0 equal to 3, we can solve for a generic constant C. This statement is telling us that the value of x is equal to 0 and the output y is going to equal to 3. So by applying this initial condition, we will get 3 is equal to E raised the power of 02 multiplied by C minus 1 divided by 0 + 2. Now anything raised to the zero power will simplify to be 1. So we are left with 3. is equal to 1 multiplied by c minus 1/22 and if we were to solve for c we get c is equal. To 7 divided by 2. So now that we have the value of the constant function or of the general constant, plugging this back into our general solution will give us Y is equal to E raise the power of X2 multiplied by 7 halves minus 1 divided by X + 2. And if we were to algebraically combine the terms in the parentheses, the final solution is going to be y is equal to e erase the power of x2 multiplied by 7 x + 12 divided by 2 multiplied by x + 2. This is going to be the solution to the given initial value problem. And with that being said, the overall solution to this problem is going to be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Solving Initial Value ProblemsSolve the initial value problems in Exercises 15–20.(x+1) dy/dx - 2 (x² + x)y = exp(x²) / (x+1), x > -1, y(0) = 5

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Solving Initial Value Problems
Solve the initial value problems in Exercises 15–20.

(x+1) dy/dx - 2 (x² + x)y = exp(x²) / (x+1), x > -1, y(0) = 5