Question

Use separation of variables to solve the IVPs y′ =2(t+1)y.

Verify your answer by substituting it into the equation.

Answer 1

To solve the IVP y' = 2(t+1)y, separation of variables is used leading to the general solution y = Ce^(t^2+2t), where C is an integration constant. Without initial conditions provided, we cannot find the specific solution or fully verify it.

Step-by-step explanation:

To solve the IVP y' = 2(t+1)y, we employ separation of variables technique. The differential equation can be rewritten to allow separation of the variables y and t:

1. Rewrite the equation as dy/y = 2(t+1) dt.
2. Integrate both sides to find the solutions for y and t separately. For the left side, integrate with respect to y, and for the right side, integrate with respect to t.
3. The integration yields ln|y| = t2 + 2t + C, where C is the constant of integration.
4. Exponentiate both sides to solve for y: y = Cet^2+2t.
5. Apply the initial conditions given in the problem to find the value of the constant C. In this case, no initial value for y or t is given, thus we cannot determine C.

Next, we verify the solution by substituting y into the original equation:

1. Compute the derivative y' from y = Cet^2+2t.
2. Substitute y and y' back into the original differential equation y' = 2(t+1)y.
3. Ensure that the substitution holds true for the equation to verify the solution.

Unfortunately, without an initial condition, we cannot verify the solution completely because the constant C remains undetermined.