Differential Equations Question 288
Question: The solution to the differential equation $ \frac{dy}{dx}=\frac{yf’(x)-y^{2}}{f(x)} $ Where $ f(x) $ is a given function is
Options:
A) $ f(x)=y(x+c) $
B) $ f(x)=cxy $
C) $ f(x)=c(x+y) $
D) $ yf(x)=cx $
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Answer:
Correct Answer: A
Solution:
[a] We have $ \frac{dy}{dx}=\frac{f’(x)}{f(x)}y-\frac{y^{2}}{f(x)} $
$ \Rightarrow \frac{dy}{dx}-\frac{f’(x)}{f(x)}y=-\frac{y^{2}}{f(x)} $ Divide by $ y^{2}:{y^{-2}}\frac{dy}{dx}-{y^{-1}}\frac{f’(x)}{f(x)}=-\frac{1}{f(x)} $ Put $ {y^{-1}}=z\Rightarrow -{y^{-2}}\frac{dy}{dx}=\frac{dz}{dx} $ $ -\frac{dz}{dx}-\frac{f’(x)}{f(x)}(z)=-\frac{1}{f(x)} $
$ \Rightarrow \frac{dz}{dx}+\frac{f’(x)}{f(x)}z=\frac{1}{f(x)} $ I.F. $ ={e^{\int{\frac{f’(x)}{f(x)}dx}}}={e^{\log f(x)}}=f(x) $
$ \therefore $ The solution is $ z(f(x))=\int{\frac{1}{f(x)}(f(x))dx+c} $
$ \Rightarrow {y^{-1}}(f(x))=x+c\Rightarrow f(x)=y(x+c) $
BETA
