Initial Value Problems

Solve the initial valu...

Question

Initial Value Problems

Solve the initial value problems in Exercises 71–90.

d²y/dx² = 2 − 6x; y′(0) = 4, y(0) = 1

Accepted Answer

Welcome back everyone. Findy given that the second derivative of y with respect to x is 3 minus 4 x. The derivative of y at 0 is equal to 5, and y at 0 is equal to 2. So first of all, what we can do is identify the first derivative. If we have the second derivative, we can just integrate the second derivative to get the first derivative. So we're going to take the integral of 3 minus 4 x dx, which is 3 x minus 4 multiplied by the integral of x, which is x 2 divided by 2, plus a constant of integration c1. And that's basically 3 x minus 2x2 plus C1. What we can do is identify C1. We know that Y1X is equal to 0 is equal to 5. So we can set the left-hand side equal to 5 and the right-hand side. Well, for x we're going to put 0, so we get 0 minus 0 plus c1, meaning C1 is equal to 5, right? 3 multiplied by 0 minus 2 multiplied by 02. This gives us 0 overall. So we just have C1 equals 5, and therefore the derivative of y is. 3 x minus 2 x 2 + 5 and from here we can find y which is the integral of the first derivative. So now we are going to integrate 3 x minus 2 x 2 + 5 d x. Let's go ahead and integrate. We get 3. The integral of x is x 2 divided by 2 minus 2. The integral of x squared is x cubed divided by 3. Plus the integral of 5 d x is 5 x because the integral of the x is x. And we're going to add a constant of integration C2. Or basically we can just call it C, right, because this is going to be our final function. So now let's simplify that. 3 halves x 2 minus 2/3 x cubed. + 5 x + c And we're going to find C knowing that. Y at 0 is equal to 2. So on the left hand side we have 2, on the right hand side x becomes equal to 0. So we get 0 minus 0 + 0 + C, meaning C equals 2. And therefore we can rewrite Y as let's arrange our terms. From the highest exponent on x to the lowest, so we get y equals -2/3 x cubed plus 3 halves x 2. Plus 5 x plus the final constant is 2. Well then, so that's our final answer for this problem. Thank you for watching.

Initial Value ProblemsSolve the initial value problems in Exercises 71–90.d²y/dx² = 2 − 6x; y′(0) = 4, y(0) = 1

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Initial Value Problems

Solve the initial value problems in Exercises 71–90.

d²y/dx² = 2 − 6x; y′(0) = 4, y(0) = 1