Consider the following 2nd order ODE for y (where the independent variable is t):

2y″+3y′=0

1) Find the general solution to the above ODE.

2) Use the initial conditions y(0)=6, y'(0)=0 to find the particular solution to this differential equation.

3) Use your particular solution to find the value of y when t=6 and enter your answer below.

If your answer is not an integer, enter it to 1 decimal place, otherwise enter an integer value.

Question

Consider the following 2nd order ODE for y (where the independent variable is t):

2y″+3y′=0

1) Find the general solution to the above ODE.

2) Use the initial conditions y(0)=6, y'(0)=0 to find the particular solution to this differential equation.

3) Use your particular solution to find the value of y when t=6 and enter your answer below.

If your answer is not an integer, enter it to 1 decimal place, otherwise enter an integer value.

BlackTom AI · Accepted Answer

Characteristic equation 2r^2 + 3r = 0 gives r(2r + 3) = 0, so r = 0 or r = −3/2. Thus the general solution is a linear combination of the corresponding exponentials:
1) Relevant concepts:
- Linear constant-coefficient homogeneous ODEs lead to a characteristic polynomial whose roots give exponentials e^{rt}.
- Repeated/zero roots produce constant terms (r = 0 gives e^{0t} = 1).
2) Solve the characteristic equation step by step:
- 2r^2 + 3r = 0
- r(2r + 3) = 0 ⇒ r = 0 or r = −3/2.
- Therefore general solution: y(t) = C1 e^{0·t} + C2 e^{−(3/2)t} = C1 + C2 e^{−3t/2}.
3) Compute derivative and apply initial conditions:
- y'(t) = C2 · (−3/2) e^{−3t/2}.
- From y'(0) = 0: y'(0) = (−3/2) C2 = 0 ⇒ C2 = 0.
- From y(0) = 6: y(0) = C1 + C2 = C1 + 0 = 6 ⇒ C1 = 6.
4) Particular solution and evaluation:
- The particular solution is y(t) = 6 (a constant function).
- Therefore y(6) = 6.
5) Why alternatives are incorrect:
- Any nonzero C2 would introduce an exponential decay term C2 e^{−3t/2} and produce a nonzero initial derivative y'(0) = −(3/2)C2, which contradicts y'(0)=0. So the decaying exponential coefficient must be zero.
- A purely exponential (no constant) solution would not satisfy y(0)=6 unless adjusted; the correct form must include the constant term from r = 0.
6) Conclusion:
All steps confirm the particular solution is y(t) = 6, and thus y(6) = 6, matching the provided answer.

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