Solve the initial value problems in Exercises 55–58.
55....

Question

Solve the initial value problems in Exercises 55–58.

55. dy/dt = e^t sin(e^t − 2),y(ln 2) = 0

Accepted Answer

Welcome back everyone. Solve the initial value problem Dy divided by DT equals 2e to the power of t, cosine of e to the power of t plus 3, and y at 0 is equal to 4. To solve this differential equation, we're going to use separation of variables, and what we can do is multiply both sides by DT so that DY Is going to be equal to 2e the power of t. Cosine of e to the power of t + 3. They say And now we can integrate both sides. On the left hand side, the integral of Dy is simply y. And on the right hand side we can first of all factor out 2 and integrate e to the power of t. Cosine of e to the power of t + 3 dt. We can introduce the u substitution. Let's suppose that u is equal to e to the power of t + 3. Now the derivative of u with respect to t is going to be the derivative of e to the power of t plus 3, which is e to the power of t. And now what we can do is simply notice that within the integrant, we have e to the power of TDT, right? So we can multiply both sides by dt to show that the DU is equal to e to the power of TDT. And now we, we can rewrite the integral as 2 integral of cosine of u multiplied by each to the power of TDT which is just D U. Let's integrate cosine. The integral of cosine is sine, so we get 2 multiplied by sine of u plus a constant of integration c because that's an indefinite integral. And finally, because u is equal to e to the power of t + 3, we get two sins. E to the power of T + 3 + c. So this is how we get our integral. The next step is to find the constant c. Knowing that if. T is equal to 0. Y is equal to 4. This is the initial condition. We're going to set the left hand side equal to 4. And on the right hand side, t is going to become 0, so we get 4 equals. 2 sin of e to the power of 0 + 3 + c. Each to the power of 0 is 1, and 1 + 3 gives us 4, so we get to sin of 4 plus c. And out to sulfur sea we can subtract 2 sin 4 from both sides. Thereforec is equal to 4 minus. 2 sin of 4 and we can substitute this constant into our solution to get y of t equals. 2 sin of e to the power of t + 3 minus. To sign off 4 + 4. That's our final answer. Let's label it and thank you for watching.

Solve the initial value problems in Exercises 55–58.55. dy/dt = e^t sin(e^t − 2),y(ln 2) = 0

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Solve the initial value problems in Exercises 55–58.
55. dy/dt = e^t sin(e^t − 2),y(ln 2) = 0