Initial Value Problems Solve the initial value problems in Exercises 71–90. dy/dx = 1/x² + x, x > 0; y(2) = 1

Welcome back everyone. Solve the initial value problem Dy divided by dx equals 3 x2 minus 4 divided by x2 square x greater than 0 with y of 1 equals 2. So for this problem, we're going to use separation of variables. Let's multiply both sides of this equation by DX. Then DY is going to be equal to 3 x 2 minus 4 divided by x 2 dx. And what we can do is simply integrate both sides, right? So on the left hand side, the integral of Dy is simply y. And on the right hand side we can apply the power rule. To apply the power rule, we're going to rewrite the second term as minus 4 x to the power of -2. So the integral of 3 x squared is going to be 3 multiplied by x cubed divided by 3, right? And for the second term, we're going to write -4, our constant multiplied by x, the power of -2 + 1 gives us -1 divided by -1. Plus a constant of integration c because we have indefinite integrals and now let's go ahead and simplify. Soy is going to be equal to x cubed minus 4 divided by -1 gives us 4 x to the power of -1, which can be written as 4 divided by x. Plus a constant c So now our final step is to find c, knowing that when x is equal to 1, y is equal to 2. So we're going to set 2 equals 1 cubed plus 4 divided by 1 plus c, which gives us 2 equals 1 + 4. That's 5 plus c. And therefore c is going to be equal to. 2 minus 5, which is -3. So from here we can now show that the solution is y equals x cubed plus 4 divided by x minus 3. That's our final answer, and thank you for watching.

Initial Value ProblemsSolve the initial value problems in Exercises 71–90.dy/dx = 1/x² + x, x > 0; y(2) = 1

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Initial Value Problems

Solve the initial value problems in Exercises 71–90.

dy/dx = 1/x² + x, x > 0; y(2) = 1