Initial Value Problems
Solve the initial value problems ...

Question

Initial Value Problems

Solve the initial value problems in Exercises 71–90.

d³y/dx³ = 6; y″(0) = −8, y′(0) = 0, y(0) = 5

Accepted Answer

Welcome back everyone. Solve the initial value problem. The third derivative of y with respect to x is equal to -9. The second derivative at 0 is 4. The first derivative at 0 is 2, and y of 0 is equal to -7. So first of all, we know that the 3rd derivative of y of x is equal to -9. What we can do is integrate both sides with respect to x, and this is how we will get the second derivative, which is y x. So y x must be equal to the integral of 9 dx, which is equal to -9 x plus a constant of integration c1. We know that y at 0 is equal to 4, so we can use this initial condition to solve for c1, right? We can show that C1 is equal toy of x. Plus 9 x and we can substitute the given values. So c1 is equal toy at 0 + 9 multiplied by 0. We're going to use that x equals 0, and that's basically y of 0, which is 4. And this is how we got c1. So we can now show that y x is equal to -9 x. Plus 4 Well done. So from here we can identify the first derivative by integrating both sides. Let's go ahead and do that. Integrating the left hand side, we will get y x is equal to. The integral of -9 x + 4 d x. Integrating, we will get -9 multiplied by x2 divided by 2. Here we are applying the power rule plus 4 x plus a constant of integration c2. Simplifying, we can show that this is equal to -9 halves x2 + 4 x + c2. And now let's find C2. So C2 is going to be equal to Y of X. Plus 9 5 x 2 minus 4 x And again we can use the initial condition when X is 0, Y is 2. So y becomes 2, right? We can take 2 plus 9 halves multiplied by 0 minus 4 multiplied by 0, which is equal to 2, and that's our constant of integration C2, so we can rewritey of x. As -9 halves x2 + 4 x Followed by C 2, so we're adding 2. And finally we will find y of x so we can integrate both sides to show that y of x is equal to the integral of -9 halves x2 + 4 x + 2 d x. And now we can integrate, so that would be -9 halves multiplied by x cubed divided by 3 using the power rule plus 4 multiplied by x 2 divided by 2. Plus 2 x + c 3, which is going to be our final constant of integration. So now let's simplify. 9 divided by 3 gives us 3, right, so we can just write -3 halves. X cubed Plus 4 divided by 2 gives us 2 So we get 2 multiplied by x. 2 + 2 x + C 3 And again we know that y is equal to -7. When x is 0, so -7 equals 0 + 0 + 0 + c3. So C3 is just -7 using the same method as before. Now what does that mean? Well, we got our solution y of x is equal to -3 halves x cubed plus 2 x 2 plus. 2 x minus 7 And that's our final answer for this problem. Thank you for watching.

Initial Value ProblemsSolve the initial value problems in Exercises 71–90.d³y/dx³ = 6; y″(0) = −8, y′(0) = 0, y(0) = 5

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Initial Value Problems
Solve the initial value problems in Exercises 71–90.
d³y/dx³ = 6; y″(0) = −8, y′(0) = 0, y(0) = 5