Solve the initial value problems in Exercises 87 and 88.

Question

88. d²y/dx² = sec²x, y(0)=0 and y'(0)=1

Accepted Answer

Welcome back everyone. Solve the initial value problem. The second derivative of y with respect to x is equal to 2 e to the power of 2 x. The first derivative at 0 is equal to 3, and y at 0 is equal to 1. So what we're going to do is identify the first derivative by integrating the second derivative. Let's show that y x is equal to the integral of the second derivative with respect to x. So the integral of 2 e to the power of 2 x dx. We can factor out 2 and integrate e to the power of 2 x dx. We're going to get 2 multiplied by e to the power of 2 x, and according to the reverse chain rule, we want to divide by 2. Plus a constant of integration c 1 we can now cancel out the two and we're going to get each to the power of 2 x plus c1. From here we can use the initial condition. Such that when x is equal to 0, y is equal to 3. So we're going to set the left hand side equal to 3, and the right hand side is going to contain x equals 0. So that's e to the power of 0 plus c1. Simplifying c1 is equal to 3 minus e to the power of 0, which is 3 minus 1, and that's 2. Well then, so now we have our first derivative y x equals e to the power of 2 x + 2. And finally, we can identify Y of X. By taking the derivative y x and integrating it, so we're going to take the integral of e to the power of 2 x plus 2dx. Let's calculate the integral that would be e to the power of 2 x divided by 2. Plus 2 x plus a constant of integration c 2. Let's use the initial condition when x is 0, y is 1, so we're going to set 1 equals. 1/2 eats the power of 0 + 2 multiplied by 0 + c2. Simplifying, we got 1 equals 1/22 plus c2 and therefore the constant of integration c2 is going to be 1 minus 1/2, which is 1/22. And therefore our solution y of x is equal to 1/2 e to the power of 2 x plus 2 x plus 1/22. That's our final answer. Let's label it and thank you for watching.

Solve the initial value problems in Exercises 87 and 88.88. d²y/dx² = sec²x, y(0)=0 and y'(0)=1

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Solve the initial value problems in Exercises 87 and 88.
88. d²y/dx² = sec²x, y(0)=0 and y'(0)=1