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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Differential Equations

BITSAT 2013 · Mathematics · Differential Equations

The solution to the differential equation (d y)/(d x)= y f^ (x)-y^...

The solution to the differential equation dydx=yf′(x)−y2f(x)\frac{d y}{d x}=\frac{y f^{\prime}(x)-y^{2}}{f(x)} where f(x)f(x) is a given function is

Options

  1. Af(x)=y(x+c)f(x)=y(x+c)
  2. Bf(x)=cxyf(x)=c x y
  3. Cf(x)=c(x+y)f(x)=c(x+y)
  4. Dyf(x)=cx\mathrm{yf}(\mathrm{x})=\mathrm{cx}

Correct answer

A. f(x)=y(x+c)f(x)=y(x+c)

Step-by-step solution

We have dydx=f′(x)f(x)y−y2f(x)⇒dydx−f′(x)f(x)y=−y2f(x)\frac{d y}{d x}=\frac{f^{\prime}(x)}{f(x)} y-\frac{y^{2}}{f(x)}\Rightarrow \frac{d y}{d x}-\frac{f^{\prime}(x)}{f(x)} y=-\frac{y^{2}}{f(x)} Divide by y2:y−2dydx−y−1f′(x)f(x)=−1f(x)y^{2}: y^{-2} \frac{d y}{d x}-y^{-1} \frac{f^{\prime}(x)}{f(x)}=-\frac{1}{f(x)} Put y−1=z⇒−y−2dydx=dzdx−dzdx−f′(x)f(x)(z)=−1f(x)⇒dzdx+f′(x)f(x)z=1f(x)IF=e∫f′(x)f(x)dx=elog⁡f(x)=f(x)∴y^{-1}=z \Rightarrow-y^{-2} \frac{d y}{d x}=\frac{d z}{d x}-\frac{d z}{d x}-\frac{f^{\prime}(x)}{f(x)}(z)=-\frac{1}{f(x)}\Rightarrow \frac{d z}{d x}+\frac{f^{\prime}(x)}{f(x)} z=\frac{1}{f(x)}I F=e^{\int \frac{f^{\prime}(x)}{f(x)} d x}=e^{\log f(x)}=f(x)\therefore The solution is z(f(x))=∫1f(x)(f(x))dx+c⇒y−1(f(x))=x+c⇒f(x)=y(x+c)z(f(x))=\int \frac{1}{f(x)}(f(x)) d x+c\Rightarrow y^{-1}(f(x))=x+c \Rightarrow f(x)=y(x+c)
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