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Welcome to a lesson on solutions to differential equations. In this video, we will confirm a given function is a solution to a differential equation and determine if a function is a solution to a differential equation. A solution to a differential equation is any function that satisfies the differential equation on the given interval. For our first example we want to verify that y equals one-sixteenth x to the the fourth is a solution to the differential equation dy/dx equals x times y to the one-half on the interval from negative infinity to positive infinity. So you're gonna want to verify that this function satisfies this differential equation. So we need to first find the derivative of y with respect to x and then perform substitution into this equation. So if y is equal to one-sixteenth x to the fourth, we apply the power rule here to find dy/dx. So dy/dx is gonna be equal to one-sixteenth times the derivative of x to the fourth which should be four x to the third. So the same as four over one. So this here's gonna simplify. There's one factor of four and four. And four factors of four in sixteen. So dy/dx is equal to one fourth x to the third. So now we have enough information to verify that this function is a solution to this differential equation. To verify that the dy/dx equals x times y the one-half. Which I'm gonna write as the square root of y. Now I'll perform substitution here for dy/dx and perform substitution here for y. So we're gonna have one fourth x to the third equals x times the square root of one-sixteenth x to the fourth. One-sixteenth x to the fourth is a perfect square so we have one-fourth x to the third equals x times the square root of one-sixteenth is one-fourth. The square root of x to the fourth is x to the second. Looks like we have it. One- fourth x to the third is equal to this will be one fourth x to the third therefore this verifies the given function is a solution to this differential equation. Now we want to verify that y equals four cosine two x plus six sine two x is a solution to y double prime plus four y equals zero. Notice to verify this we'll have to find y double prime but to find y double prime we'll first have to find y prime. So given y, let's find y prime. Y prime is going to be equal to four times the derivative of cosine two x which will require the chain rule because we have a composite function here with the inner function of two x. So the derivative of cosine two x is going to be negative sine two x times two. Let's put the times two here. The negative here and the sine two x here. Plus six times the derivative of sine two x, again applying the chain rule. This is gonna be cosine two x times two. Put the times two here and then cosine two x. So y prime is equal to negative eight sine two x plus twelve cosine two x. And now we need to find y double prime. So y double prime is gonna be equal to the derivative of negative eight sine two x. So we're gonna have negative eight times the derivative of sine two x which is going to be cosine two x times two because of the chain rule. So it's times two cosine two x plus 12 times the derivative of cosine two x. So again applying the chain rule the derivative is going to be negative sine two x times two. So there's the negative, there's two, and sine two x. Let's clean this up we have y double prime equals negative 16 cosine two x. This will be minus 24 sine two x. Now we have the information we need to verify the solution. We're trying to verify that y double prime plus four y equals zero. So we'll substitute all this for y double prime and we'll substitute the original function for y. We'll have negative 16 cosine two x minus 24 sine two x All of this is y double prime plus four times y equals zero. Or y is four cosine two x plus six sine two x. So now we'll distribute and see if it's equal to zero. So we'll distribute the four here and here. Things are looking good. This first part stays the same. So we'll have plus 16 cosine two x. and plus 24 sine two x. Again this is good news because notice how this term and this term are opposites. And so are these two, therefore we have zero equals zero. Verifying the given function is a solution to the differential equation. Okay I think we have time for one more. Is y equals e to the negative x plus two x e to the negative x a solution to this given differential equation. Notice how we'll have to find the first derivative here and the second derivative here. So we'll first find dy/dx. The derivative of e to the negative X will require the chain rule. This is going to be e to the negative x times the derivative of negative x. This is gonna be negative e to the negative x plus. To find the derivative of this term here we'll have to apply the product rule and the chain rule. So we'll have the first function two x times the derivative of the second. Which again it's gonna be e to the negative x times the derivative of negative x. This will be negative e to the negative x plus the second function e to the negative x times the derivative of the first function which is two x. So that'll be times two. Let's go ahead and clean this up. So we have dy/dx equals this is negative e to the negative x and this is plus two e to the negative x. So we'll have a positive one e to the negative x. This will be minus two x e to the negative x. Now to find the second derivative, we'll find the derivative of this. So second derivative of y with respect to x will equal the derivative of e to the negative x which we saw above is e to the negative x times negative one or just negative e to the negative x. Minus, now we have to apply the product rule with the chain rule to find the derivative of this term here. So we'll have the first function two x times the derivative of the second function. The derivative of e to the negative x is negative e to the negative x. Plus the second function e to the negative x times the derivative of the first which should be two. And it is important we have this in paranthesis so we're subtracting this entire quantity. Let's go ahead and clean this up. And notice here if we distributed negative one here we're going to have negative two e to the negative x. Here's a negative one e to the negative x. So we have negative three e to the negative x. And then this product here is gonna be a positive so plus two x e to the negative x. So now using the first derivative here we'll perform substitution into our equation. Now we'll use our second derivative here to perform substitution here. So we're going to have negative three e to the negative x plus two x e to the negative x plus two times dy/dx. So that's e to the negative x minus two x e to the negative x. Plus y or y as the original function. So we have e to the negative x plus two x e to the negative x equals zero. If this is true then our function is a solution, if it's not true then it's not. Let's go ahead and clear our parenthesis. Let's see if this is equal to zero. So then plus two e to the negative x minus four x e to the negative x. Now let's identify the like terms. Here's an e to the negative x as well as here and here. Looking at the coefficients we have negative three plus two and then plus one. Therefore these terms sum to zero. Notice the three reaming terms are also like terms. They each contain a factor of x e to the negative x. Here, here, and here. Notive if we add this first term with this third time here would have four x e to the x minus four x e to the x. Again would be zero Which verifies that the given function is a solution since we've just verified that zero equals zero. Which is what it takes to show the given function is a solution to the differential equation. Okay hope you found this helpful.