Find a second solution y2(x), using reduction of order: 1. x^2y'' − xy' + 2y = 0; y1 = x sin(ln x) Answer: y2 = x cos(ln x) 2. (1 − 2x − x^2)y'' + 2(1 + x)y' − 2y = 0; y1 = x + 1 Answer: y2 = x^2 + x + 2
Added by Nicholas P.
Step 1
For the first equation, we first find the complementary solution by assuming y = e^(rx). Substituting this into the equation, we get the characteristic equation r(r-1) + 2 = 0, which has roots r = 1, -2. Therefore, the complementary solution is y_c = c_1 x e^x + Show more…
Show all steps
Close
Your feedback will help us improve your experience
Madhur L and 94 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
x^2y'' - xy' + 2y = 0; y_1 = x sin(ln x) Answer: y_2 = x cos(ln x)
Disha S.
Consider the second-order homogeneous initial value problem, y''(x) − 2y'(x) + 2y(x) = 0. Given that y1(x) = e^x cos(x) and y2(x) = e^x sin(x) are solutions, find the solution that satisfies y(0) = 1 and y'(0) = 7. y(x) =
Likhit G.
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD