Differential Equations Question 243
Question: The solution of the equation $ x\int\limits_0^{x}{y(t)dt=(x+1)\int_0^{x}{ty(t)dt,x>0}} $ is
Options:
A) $ y=\frac{c}{x^{3}}{e^{x^{3}}} $
B) $ y=cx^{3}{e^{-x^{3}}} $
C) $ \frac{c}{x^{3}}{e^{-x}} $
D) None of these
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Answer:
Correct Answer: D
Solution:
[d] The equation is $ x\int_0^{x}{y(t)dt=(x+1)\int_0^{x}{ty(t)dt,}} $
- (i) Differentiating both the sides with respect to x, we get $ xy(x)+\int_0^{x}{y(t)dt=(x+1)xy(x)+\int_0^{x}{ty(t)dt}} $
$ \Rightarrow \int_0^{x}{y(t)dt=x^{2}y(x)+\int_0^{x}{ty(t)dt}} $
- (ii) Differentiating again with respect to x, we get $ y(x)=2xy(x)+x^{2}y’(x)+xy(x) $
$ \Rightarrow x^{2}\frac{dy}{dx}=(1-3x)y $
$ [writingy(x)=y] $
$ \Rightarrow \frac{dy}{y}=( \frac{1-3x}{x^{2}} )dx=( \frac{1}{x^{2}}-\frac{3}{x} )dx $ Integrating we get, $ \log y=-\frac{1}{x}-3\log x+a,a $ is constant
$ \Rightarrow \log y+3\log x=a-\frac{1}{x} $
$ \Rightarrow \log (yx^{3})=a-\frac{1}{x}\Rightarrow yx^{3}={e^{a-\frac{1}{x}}}=c.{e^{-\frac{1}{x}}} $ where $ c=e^{a}\therefore y=\frac{c}{x^{3}}{e^{-\frac{1}{x}}} $
BETA
