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Verify that $y(x)=x /(x+1)$ is a solution to the differential equation$$y+\frac{d^{2} y}{d x^{2}}=\frac{d y}{d x}+\frac{x^{3}+2 x^{2}-3}{(1+x)^{3}}$$

$L H S=R H S$ $y(x)=\frac{x}{x+1}$ is a solution

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Boston College

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in this problem, we have a second order differential equation for which we want to determine if the falling function is a solution. Well, since the equation itself is of second order, the first thing we need to do is compute the first derivative D y T X, then the second derivative due to wide the X two. So first, that first derivative D Y DX can be done through the quotient rule. And with the quotient rule, we first take the derivative of the Numerator X, which will be one itself and multiply it by the denominator. So copy X plus one that'll be minus the same kind of process. But in reverse order this time we copy the numerator X and multiply it by the drift of the denominator X plus one, which is just a one itself. And this will be all over the denominator squared or X plus one to the second power. Since we're going have to do quite a bit of work with these derivatives, it's a great idea to simplify these as much as possible. We have X plus one minus X itself left in the numerator, so the first rib have simplifies down to one divide by X plus one squared. Let's go off to next. The second derivative, de Tu y DX two for the second derivative can be done in a number of ways. Let's express the first derivative as follows. It is X plus one to the power of negative, too. We want to take the derivative of that result expressed in Newton's prime notation. The result will be first, the drift of the outer function negative too Times X plus one to the power of negative three and through the chain rule, we take the derivative of the Inter Function X plus one. Luckily, in this case, that derivative is times one, and so we can simplify this entire results into negative to divide by X plus one to the power of three. Now we're ready for the next step of this process. In order to verify that this function is a solution, we need to go back to the original differential equation and work in multiple steps. Let's first try out the right hand side, all right, or a just for right hand side to see how that expression simplifies the right hand side of this differential equation is first Dwight D. X, but we have by substitution that D y t X is one over X plus one to the power of To there will be, plus the function X cubed plus two X squared minus three. We'll divide by X plus one to the power of three and where to do a quick check on this? Whether it's a solution to the difference, you'll equation. We should simplify this answer as much as possible. It's all come down to the first fraction. Multiply it by X plus one in the numerator and denominator, and the result will then be equal to X plus one plus X cubed plus two X squared minus three all over the common denominator of X plus one. Cute. So now this simplifies down to X cubed plus two x squared plus X minus two. All divide by X plus one to the power of three. And that's our simplified form of the right hand side. Now let's call this portion of the differential equation the left hand side. Our job is to do the following. We're going to substitute as before, into the left hand side, and our goal is to try to get to this format here. If we can get to this, resulting expression after simplifying than a work will be complete. That indeed why of X equals X divide by X plus one is a solution. If we cannot get to that step, then we would know that this is not a solution. So it's now start off again from the left hand side. The left hand side first involves the function. Why so by substitution? Why is X divide by X Plus one? They'll be plus the second order derivative, which we simplify down to negative to divide by X plus one cubed so B plus negative, too. Divide by X plus one to the power three Recall. Their goal is to simplify this as much as possible. So let's go to the denominator of the very first fraction and multiply it by X plus one the power of to and again X plus one to the power of to and the numerator of that fraction. Then the result will be X times X plus one squared minus two. All divide by X plus one to the power of three. So we're nearly there. Let's first work on expanding this expression. Here we get X times X squared plus two X plus one, then minus two. All divide by X plus one to the power of three. Now, once we distribute the variable X into the group, we obtain X cubed plus two X squared plus X then copied My is too all divide by X plus one to the power of three. Comparing both sides the right hand side with the left hand side we find that these expressions are equal So the left hand side is equal to the right hand side. And why of tea? Excuse me. Why of X, which is equal to X divide by X plus one is a solution, since these expressions are both equal.

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