chegg Study Guided Solutions LON-CAPA c2-diffeq-31 2000 Https://homeworkz Diffeqr dhttps| Hom 0 https://homeworkzmath.pittedu/res/pitt/athanas/c2-diffeq-31 problem?symb-uploaded?2fpitt382 Connor Wilson (Student section: 18259) Iain Menu Contents Grades 2187 18259 MATH 0230 ANALYTIC GEOMETRY Course Contents PROBLEM SET 4 c2-diffeq-31 Determine the solution to the second-order homogeneous initial value differential equation 293′ 901″ J(0) = 2 Y(0) = 0) = -14 11exp(iOx)-36 11exp(9) Submit Answer Incorrect. Tries 1/8 Previous Tries Post Discussion
The Correct Answer and Explanation is:
To solve the second-order homogeneous differential equation given as:y′′+9y=0y” + 9y = 0y′′+9y=0
with initial conditions:y(0)=2,y′(0)=−14y(0) = 2, \quad y'(0) = -14y(0)=2,y′(0)=−14
Step 1: Solve the characteristic equation.
For a second-order linear homogeneous differential equation of the form:y′′+ay′+by=0y” + ay’ + by = 0y′′+ay′+by=0
The characteristic equation is:r2+9=0r^2 + 9 = 0r2+9=0
Solving for rrr, we get:r=±3ir = \pm 3ir=±3i
Thus, the general solution of the differential equation is:y(t)=C1cos(3t)+C2sin(3t)y(t) = C_1 \cos(3t) + C_2 \sin(3t)y(t)=C1cos(3t)+C2sin(3t)
where C1C_1C1 and C2C_2C2 are constants to be determined by the initial conditions.
Step 2: Apply the initial conditions.
We are given that:y(0)=2andy′(0)=−14y(0) = 2 \quad \text{and} \quad y'(0) = -14y(0)=2andy′(0)=−14
First, evaluate y(0)y(0)y(0):y(0)=C1cos(0)+C2sin(0)=C1y(0) = C_1 \cos(0) + C_2 \sin(0) = C_1y(0)=C1cos(0)+C2sin(0)=C1
So, C1=2C_1 = 2C1=2.
Next, find y′(t)y'(t)y′(t):y′(t)=−3C1sin(3t)+3C2cos(3t)y'(t) = -3C_1 \sin(3t) + 3C_2 \cos(3t)y′(t)=−3C1sin(3t)+3C2cos(3t)
Now, evaluate y′(0)y'(0)y′(0):y′(0)=−3C1sin(0)+3C2cos(0)=3C2y'(0) = -3C_1 \sin(0) + 3C_2 \cos(0) = 3C_2y′(0)=−3C1sin(0)+3C2cos(0)=3C2
We are given that y′(0)=−14y'(0) = -14y′(0)=−14, so:3C2=−143C_2 = -143C2=−14
Thus, C2=−143C_2 = -\frac{14}{3}C2=−314.
Step 3: Write the solution.
Now that we have C1=2C_1 = 2C1=2 and C2=−143C_2 = -\frac{14}{3}C2=−314, the solution to the differential equation is:y(t)=2cos(3t)−143sin(3t)y(t) = 2 \cos(3t) – \frac{14}{3} \sin(3t)y(t)=2cos(3t)−314sin(3t)
Step 4: Verify the solution.
Finally, verify the solution by checking the initial conditions:
- y(0)=2y(0) = 2y(0)=2, which matches the given initial condition.
- y′(0)=−14y'(0) = -14y′(0)=−14, which also matches the given initial condition.
Thus, the solution is correct. The correct answer is:y(t)=2cos(3t)−143sin(3t)y(t) = 2 \cos(3t) – \frac{14}{3} \sin(3t)y(t)=2cos(3t)−314sin(3t)
