In Exercises 125–128 solve the differential equation.
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Question

In Exercises 125–128 solve the differential equation.

127. yy' = sec(y²)sec²(x)

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Solve the differential equation y multiplied by y is equal to sin of y to the 2 multiplied by cosine square of x. OK. So it appears for this particular prompt, we're asked to solve for the differential equation that is provided to us by the problem itself. So now that we know that we're ultimately trying to solve for this particular differential equation, we need to recall and note that in order to solve for this specific differential equation, we will need to recall and use the separation of variables method. By grouping all the terms involving y on one side of our expression and x on the other side of our expression, then we will need to integrate both sides by recalling and applying a u substitution method for the y variable specifically and a trigometric identity for the x variable. And then after that, we will need to algebraically isolate Y in order to obtain the final explicit solution. So with that in mind, that's a brief overview of what we're going to do. So our first step now is we need to separate the variables and set up the integration. So we need to recognize that Y is equal. To Dy by dx, and we will need to multiply by dx and divide by sin of y to the power of 2 in order to group our variables, so we'll be able to write y divided by sin of y to the power of 2. Dy will be equal to cosine squared of x dx. Fantastic. So, moving right along here, what we need to do now is we need to integrate. So we need to take the integral of Y multiplied by cosecant of Y to the power of 2, Dy, which will be equal to the integral of cosine squared of XDX. Fantastic. So with that said. Our second step now, so this is step 2, is we need to evaluate the Y integral by recalling and using our substitution method. So let us state that you for this problem will be equal to y to the 2, then that will mean that DU will be equal to 2YDY. Which will mean, so this will mean that Y multiplied by DY will be equal to 1/2 DU. Fantastic. So that means we can go ahead and write 1/2 multiplied by the integral of cosiant of u duu, which will be equal to 1/2 multiplied by the natural log of the absolute value of tangent of u divided by 2. Fantastic. So now when we substitute back our value of u is equal to y to the 2, we will be able to write that the left hand side is equal to 1/2 multiplied by the natural log of the absolute value of tangent of y to the 2 divided by 2. Fantastic. So moving right along here. Our 3rd step now is we need to evaluate our X integral by recalling and using a power reduction identity. So with that in mind, we need to recall the identity that cosine squared of x is equal to 1 plus cosine of 2 x divided by 2. Fantastic. So with that in mind, What we need to do now is we need to apply this identity. So when we apply this identity, we will be able to safely write that the integral of 1 plus cosine of 2 x divided by 2. D X will be equal to 1/2 multiplied by the integral of parentheses of 1 plus cosine of 2 XDX, which will be equal to. 1/2 multiplied by parentheses of x plus 1/2 multiplied by sin of 2 x plus c subscript 1 where c denotes some constant variable, so that will mean that our right hand side will be equal to x divided by 2 plus. Sign of 2 x divided by 4 plus C subscript 1. For simplicity's sake, I'll just call it C1. OK. So now, for our 4th step, so this is step 4 now, we need to equate our expression and isolate Y, meaning we need to combine the results that we have found so far, and we will need to multiply the entire equation expression. By 2 in order to simplify, so we'll be able to write that the natural log of the absolute value of tangent. Of, so this will be tangent of, as we should recall, Y to the power of 2 divided by, so it'll be y to the 2 divided by 2. And don't forget your absolute value symbols, which will be equal to x plus 1/2 multiplied by sin of 2 X. Plus 2 multiplied by C1. Fantastic. So, let us note, let's make a note here, off to the side, that C subscript 2 will be equal to 2 multiplied by C1. So C2 will be equal to 2 multiplied by C1. So now we need to exponentiate both sides. So when we exponentiate both sides, we'll be able to write that the absolute value of tangent of y to the 2 divided by 2 will be equal to e to the power of x plus. 1/2 Multiplied by sin of 2 x. Plus C2. So, we can remove the absolute value signs because we know we're dealing with a with anything inside of the absolute value sign will become positive, so we could just simply write. tangent of y to the 2 divided by 2 will be equal to plus or minus e to the power of x plus 1/2 multiplied by sin of 2 x plus c2. So now what we need to do is we need to simplify one last time, so we'll be able to write that tangent of Y to the power 2 divided by 2 will be equal to plus or minus E to the power of C2 multiplied by. E to the power of x plus 1/2 multiplied by sin of 2 X. Fantastic. So, moving right along here, and let's make a note that C, or which once again C denotes some constant variable, will be equal to plus or minus E to the power of C2. OK. So that means that we can go ahead and write that tangent of. Y to the power of 2 divided by 2. Will be Equal to c multiplied by e to the power of x. Plus 1/2 multiplied by sin of 2 x. OK, so we are on a roll here. So with that in mind. Our next step now is we need to, this is our 5th and final step. So for our final step, step number 5, we need to determine the final explicit form. So in order to do that, we will need to take inverse tangent of both sides, and then we will need to multiply by 2, and then we will need to take the square root in order to solve for y. So we can write this as y to the power 2 divided by 2 will be equal to inverse tangent of parentheses c multiplied by e to the power of x plus 1/2 multiplied by sin of 2 x. Fantastic. And once we take the square root and we are isolating and solve for Y, skipping a few steps, you'll just need to use some algebra, we will find that Y is simply equal to plus or minus the square root of 2 multiplied by inverse tangent of. C multiplied by e to the power of x plus 1/2 multiplied by sin of 2 x. And that's it. That is our final answer. Hooray, we did it. We solve for our final answer. So going back to look at the problem itself, we could safely say once again that our final answer for this specific differential equation is once again. Y is equal to plus or minus the square root of 2 multiplied by inverse tangent of c multiplied by e to the power of x plus 1/2 multiplied by sin of 2 x, and that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 125–128 solve the differential equation.127. yy' = sec(y²)sec²(x)

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In Exercises 125–128 solve the differential equation.
127. yy' = sec(y²)sec²(x)