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

Question

Solving Initial Value Problems

Solve the initial value problems in Exercises 15–20.

t dy/dt + 2y = t³, t > 0, y(2) = 1

Accepted Answer

Welcome back everyone. Solve the initial value problem. T multiplied by Dy divided by DT plus Y equals t2. T is greater than 0. Y at 1 equals 0. For this problem, let's recognize the pattern on the left hand side of the equation. In particular, what we can do is show that c multiplied by d y divided by dc plus now in front of y we don't have anything right. But we have to remember that the derivative oft is equal to 1, right? So we can just say 1 multiplied by y and 1 can be rewritten as the derivative of t, right? This is equal to t2. For simplicity, we can replace d y divided by dt with y, and I will notice that on the left hand side we basically have the product rule of differentiation, which essentially means that if we go in reverse, that would be the product between C and Y. So on the left hand side overall we have the derivative oft multiplied by y, and this is equal to t2. We can now rewrite the differential form d divided by dt oft multiplied by y is equal to t2. Let's multiply both sides by dc so that the d of cy equals c 2 multiplied by dt, which allows us to integrate both sides. On the left hand side, the integral of the differential is going to be cy itself. On the right hand side, we can apply the power rule to get c cubed divided by 3 plus a constant of integration c. And now let's go ahead and identify C, knowing that. If t is equal to 1, y is equal to 0. So substituting c equals 1 and y equals 0, we get 1 multiplied by 0 equals. One cub which is one. Divided by 3 plus C, right? So 0 equals 1/3 plus c, meaning c is equal to -1/3. Well then, so what we can do now is substitute the value of C. To gett multiplied by y equals. T cubed divided by 3. 1/3. We can also use the. Common denominator. Which is 3. And that'd be T cubed. -1 divided by 3 and finally we can solve for y. Y is going to be equal to. C cubed minus 1 divided by 3 t. In other words, we can divide both sides by t, right? Well then, we have our final answer and thank you for watching.

Solving Initial Value ProblemsSolve the initial value problems in Exercises 15–20.t dy/dt + 2y = t³, t > 0, y(2) = 1

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

t dy/dt + 2y = t³, t > 0, y(2) = 1