{"id":260616,"date":"2025-07-19T15:41:35","date_gmt":"2025-07-19T15:41:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=260616"},"modified":"2025-07-19T15:41:37","modified_gmt":"2025-07-19T15:41:37","slug":"chegg-study-guided-solutions-lon-capa-c2-diffeq-31-2000-https","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/19\/chegg-study-guided-solutions-lon-capa-c2-diffeq-31-2000-https\/","title":{"rendered":"chegg Study Guided Solutions LON-CAPA c2-diffeq-31 2000 Https"},"content":{"rendered":"\n

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 11<\/em>exp(9) Submit Answer Incorrect. Tries 1\/8 Previous Tries Post Discussion<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To solve the second-order homogeneous differential equation given as:y\u2032\u2032+9y=0y” + 9y = 0y\u2032\u2032+9y=0<\/p>\n\n\n\n

with initial conditions:y(0)=2,y\u2032(0)=\u221214y(0) = 2, \\quad y'(0) = -14y(0)=2,y\u2032(0)=\u221214<\/p>\n\n\n\n

Step 1: Solve the characteristic equation.<\/strong><\/p>\n\n\n\n

For a second-order linear homogeneous differential equation of the form:y\u2032\u2032+ay\u2032+by=0y” + ay’ + by = 0y\u2032\u2032+ay\u2032+by=0<\/p>\n\n\n\n

The characteristic equation is:r2+9=0r^2 + 9 = 0r2+9=0<\/p>\n\n\n\n

Solving for rrr, we get:r=\u00b13ir = \\pm 3ir=\u00b13i<\/p>\n\n\n\n

Thus, the general solution of the differential equation is:y(t)=C1cos\u2061(3t)+C2sin\u2061(3t)y(t) = C_1 \\cos(3t) + C_2 \\sin(3t)y(t)=C1\u200bcos(3t)+C2\u200bsin(3t)<\/p>\n\n\n\n

where C1C_1C1\u200b and C2C_2C2\u200b are constants to be determined by the initial conditions.<\/p>\n\n\n\n

Step 2: Apply the initial conditions.<\/strong><\/p>\n\n\n\n

We are given that:y(0)=2andy\u2032(0)=\u221214y(0) = 2 \\quad \\text{and} \\quad y'(0) = -14y(0)=2andy\u2032(0)=\u221214<\/p>\n\n\n\n

First, evaluate y(0)y(0)y(0):y(0)=C1cos\u2061(0)+C2sin\u2061(0)=C1y(0) = C_1 \\cos(0) + C_2 \\sin(0) = C_1y(0)=C1\u200bcos(0)+C2\u200bsin(0)=C1\u200b<\/p>\n\n\n\n

So, C1=2C_1 = 2C1\u200b=2.<\/p>\n\n\n\n

Next, find y\u2032(t)y'(t)y\u2032(t):y\u2032(t)=\u22123C1sin\u2061(3t)+3C2cos\u2061(3t)y'(t) = -3C_1 \\sin(3t) + 3C_2 \\cos(3t)y\u2032(t)=\u22123C1\u200bsin(3t)+3C2\u200bcos(3t)<\/p>\n\n\n\n

Now, evaluate y\u2032(0)y'(0)y\u2032(0):y\u2032(0)=\u22123C1sin\u2061(0)+3C2cos\u2061(0)=3C2y'(0) = -3C_1 \\sin(0) + 3C_2 \\cos(0) = 3C_2y\u2032(0)=\u22123C1\u200bsin(0)+3C2\u200bcos(0)=3C2\u200b<\/p>\n\n\n\n

We are given that y\u2032(0)=\u221214y'(0) = -14y\u2032(0)=\u221214, so:3C2=\u2212143C_2 = -143C2\u200b=\u221214<\/p>\n\n\n\n

Thus, C2=\u2212143C_2 = -\\frac{14}{3}C2\u200b=\u2212314\u200b.<\/p>\n\n\n\n

Step 3: Write the solution.<\/strong><\/p>\n\n\n\n

Now that we have C1=2C_1 = 2C1\u200b=2 and C2=\u2212143C_2 = -\\frac{14}{3}C2\u200b=\u2212314\u200b, the solution to the differential equation is:y(t)=2cos\u2061(3t)\u2212143sin\u2061(3t)y(t) = 2 \\cos(3t) – \\frac{14}{3} \\sin(3t)y(t)=2cos(3t)\u2212314\u200bsin(3t)<\/p>\n\n\n\n

Step 4: Verify the solution.<\/strong><\/p>\n\n\n\n

Finally, verify the solution by checking the initial conditions:<\/p>\n\n\n\n