Differential Equations Question 310
Question: The gradient of the curve passing through (4, 0) is given by $ \frac{dy}{dx}-\frac{y}{x}+\frac{5x}{(x+2)(x-3)}=0 $ if the point (5, a) lies on the curve, then the value of a is
Options:
A) $ \frac{67}{12} $
B) $ 5\sin \frac{7}{12} $
C) $ 5\log \frac{7}{12} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] The differential equation is $ \frac{dy}{dx}-\frac{y}{x}=-\frac{5x}{(x+2)(x-3)} $
I.F = $ {e^{\int{( \frac{1}{x} )dx}}}={e^{-lnx}}=\frac{1}{x} $ Solution is $ y( \frac{1}{x} )=\int{( \frac{1}{x} )\times \frac{5x}{(x+2)(x-3)}dx=ln( \frac{x+2}{x-3} )+C} $ It passes through (4, 0), so $ C=-ln6 $
$ \therefore y=xln{ \frac{(x+2)}{6(x-3)} } $ Putting (5, a), we get a = 5 $ \ln ( \frac{7}{12} ) $
BETA
