If x³dy + xydx = x²dy + 2ydx; y(2) = e and x > 1, then y(4) is equal to:
If x³dy + xydx = x²dy + 2ydx; y(2) = e and x > 1, then y(4) is equal to:
Option 1 -
√e/2
Option 2 -
1/2 + √e
Option 3 -
√e
Option 4 -
1/2 + √e
-
1 Answer
-
Correct Option - 3
Detailed Solution:x³dy + xydx = 2ydx + x²dy
=> (x³-x²)dy = (2-x)ydx
=> dy/y = (2-x)/ (x² (x-1) dx
=> ∫ (dy/y) = ∫ (2-x)/ (x² (x-1)dx
Let (2-x)/ (x² (x-1) = A/x + B/x² + C/ (x-1)
=> 2-x = A (x-1) + B (x-1) + Cx²
=> C=1, B=-2 and A=-1
=> ∫ (dy/y) = ∫ (-1/x - 2/x² + 1/ (x-1)dx
=> lny = -lnx + 2/x + ln|x-1| + C
∴ y (2) = e
=> 1 = -ln2 + 1 + 0 + C
=> C = ln2
=> lny = -lnx + 2/x + ln|x-1| + ln2
at x = 4
=> lny (4) = -ln4 + 1/2 + ln3 + ln2
=> lny (4) = ln (3/4) + 1/2 = ln (3/2)e^ (1/2)
=> y (4) = (3/2)e^ (1/2)
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 686k Reviews
- 1800k Answers