Solving Initial Value Problems
Solve the initial value p...

Question

Solving Initial Value Problems

Solve the initial value problems in Exercises 15–20.

θ dy/dθ + y = sin θ, θ > 0, y(π/2) = 1

Accepted Answer

Welcome back everyone. Solve the initial value problem R multiplied by Dy divided by dr plus y equals cosine R, where r is greater than 0, y at pi is equal to 0. For this problem, let's go ahead and notice the pattern on the left hand side. We have r multiplied by the derivative of y with respect to r. Plus why. Or basically 1 multiplied by y. Additionally, we can rewrite one as the derivative of R, right? The derivative of R with respect to r. And this is equal to cosine r. Now what we're going to notice is that on the left hand side we have the product rule of differentiation r multiplied by the derivative of y plus y multiplied by the derivative of r. So this is basically the derivative with respect to r of r multiplied by y. And this is equal to cosine r. And from here we can multiply both sides by dr to show that the d of r y is equal to cosine of r dr, which allows us to integrate both sides of our equation. On the left hand side we get RY, the integral of the differential. On the right hand side we get the integral of cosine r, which is sin r plus a constant of integration c. And now let's go ahead and identify the constant C. C is R Y minus sign R. And we can use the initial condition if r is equal to pi, y is equal to 0. So we get r, which is pi multiplied by y, which is 0, minus sin of r, which is. Sign off pi. And this is equal to 0. Therefore, are why. is equal to sin r. And we can finally solve for y itself. Y is equal to sin r. Divided by R. That's our final answer and thank you for watching.

Solving Initial Value ProblemsSolve the initial value problems in Exercises 15–20.θ dy/dθ + y = sin θ, θ > 0, y(π/2) = 1

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Solving Initial Value Problems
Solve the initial value problems in Exercises 15–20.

θ dy/dθ + y = sin θ, θ > 0, y(π/2) = 1