Solve the initial value problems in Exercises 67–70 for x as a...

Question

Solve the initial value problems in Exercises 67–70 for x as a function of t.

(t + 1) (dx/dt) = x² + 1 (for t > -1), x(0) = 0

Accepted Answer

Welcome back everyone. Find the solution to the initial value problem T + 4 DZ divided by DT equals Z2 + 9 for C greater than -4 and Z at 0 equals 0. Now for this problem, let's use separation of variables. Let's rewrite the original equation that'd bec + 4. DC divided by DC equals. Z2 + 9 In order to separate the variables, let's notice that on the right hand side we have z2 + 9. On the left hand side we have dz term. So basically what we can do is divide both sides by z2 + 9, which is going to give us dz divided by. Z2 + 9, so that's how we group the z terms. And on the left hand side we have T + 4 divided by dT. So let's multiply both sides by dt, which gives us dt on the right hand side, and divided both sides by T plus 4. Well, then, so we end up with. A new form which is dz divided by z2 + 9 equals DC divided by Z + 4. And from here we are ready to integrate both sides. So let's go ahead and. Integrate the first expression, we have the integral of dz divided by z2 + 9. And the value of this integral is going to be 1 divided by square root of 9, right? Because our integral is in a form. Dz divided by z2 plus a 2. Remember that the integral of. D X Divided by x2 plus a 2 is equal to 1 divided by a inverse of tangent of x divided by a plus a constant of integration c1. So in this case, our A is 3, right? Because. We have 9, which is 3 squad. So the integral on the left is going to be 1/3 inverse of tangent of our variable which is z divided by 3. Because again, A is 3, right? And on the right hand side we have an elementary integral. Its value is ln. Of the absolute value oft + 4. And we're going to add the constant of integration c. Additionally, we can drop the absolute value because according to the problem c is greater than -4. So T + 4 is going to be greater than 0 as a result. And from this expression Using the initial condition, we can find the value of c. We know that when c is equal to 0, z is also zero. So let's set each value equal to 0. On the left hand side we have 1/3 inverse of tangent of 0 divided by 3, which is 0. Equals ln of 0 + 4, which is basically ln of 4 plus c. Now inverse of tangent of 0 is equal to 0, so on the left hand side. We're going to have 0 overall. And on the right hand side we have LN of 4 + C. Solving for C, C is equal to 0 minus 104, which is negative LN of 4. Well then, so now we can show that. 1/3 inverse of tangent of z divided by 3. Equals ln of c + 4 minus 104. In other words, we are replacing the constant c with negative 104. And using the properties of logarithms, we can rewrite it as ln of c + 4 divided by 4. Finally, we can solve for the value of z. In order to do that, let's start by multiplying both sides by 3. So that inverse of tangent of c divided by 3 is going to be equal to. We Alan of. C + 4 divided by 4. Now let's take tangent of both sides, so that z divided by 3 is going to be tangent of. 3 Ellen off. C + 4 divided by 4. And finally, if we multiply both sides by 3, we're going to have z equals. 3 multiplied by a tangent of 3 ln oft plus 4 divided by 4. This is going to be our final answer for this problem. Thank you for watching.

Solve the initial value problems in Exercises 67–70 for x as a function of t.(t + 1) (dx/dt) = x² + 1 (for t > -1), x(0) = 0

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Solve the initial value problems in Exercises 67–70 for x as a function of t.
(t + 1) (dx/dt) = x² + 1 (for t > -1), x(0) = 0