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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Differential Equations

BITSAT 2015 · Mathematics · Differential Equations

If φ(x) is a differential function, then the solution of the...

If ϕ(x)\phi(\mathrm{x}) is a differential function, then the solution of the differential equation dy+{yϕ′(x)−ϕ(x)ϕ′(x)}dx=0,\mathrm{dy}+\left\{\mathrm{y} \phi^{\prime}(\mathrm{x})-\phi(\mathrm{x})\right.\left.\phi^{\prime}(x)\right\} d x=0, is

Options

  1. Ay={ϕ(x)−1}+Ce−ϕ(x)y=\{\phi(x)-1\}+C e^{-\phi(x)}
  2. Byϕ(x)={ϕ(x)}2+Cy \phi(x)=\{\phi(x)\}^{2}+C
  3. Cyeφ(x)=ϕ(x)eϕ(x)+Cy e^{\varphi(x)}=\phi(x) e^{\phi(x)}+C
  4. Dy−ϕ(x)=ϕ(x)e−ϕ(x)y-\phi(x)=\phi(x) e^{-\phi(x)}

Correct answer

A. y={ϕ(x)−1}+Ce−ϕ(x)y=\{\phi(x)-1\}+C e^{-\phi(x)}

Step-by-step solution

Given differential equation is dy+{yϕ′(x)−ϕ(x)ϕ′(x)}dx=0⇒dydx+ϕ′(x)y=ϕ(x)ϕ′(x)\mathrm{dy}+\left\{y \phi^{\prime}(x)-\phi(x) \phi^{\prime}(x)\right\} d x=0\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\phi^{\prime}(\mathrm{x}) \mathrm{y}=\phi(\mathrm{x}) \phi^{\prime}(\mathrm{x}) which is a linear differential equation with P=ϕ′(x),Q=ϕ(x)⋅ϕ′(x)\quad P=\phi^{\prime}(x), Q=\phi(x) \cdot \phi^{\prime}(x) and I⋅F=e∫ϕ′(x)dx=eϕ(x)∴I \cdot F=e^{\int \phi^{\prime}(x) d x}=e^{\phi(x)}\therefore Solution is y⋅eϕ(x)=∫ϕ(x)⋅ϕ′(x)eϕ(x)dx+C⇒y⋅eϕ(x)=∫ϕ(x)⋅eϕ(x)ϕ′(x)dx+C⇒y⋅eϕ(x)=ϕ(x)eϕ(x)−∫ϕ′(x)eϕ(x)dx+C⇒y⋅eϕ(x)=ϕ(x)eϕ(x)−eϕ(x)+C⇒y=⌈ϕ(x)−1]+Ce−ϕ(x)y \cdot e^{\phi(x)}=\int \phi(x) \cdot \phi^{\prime}(x) e^{\phi(x)} d x+C\Rightarrow y \cdot e^{\phi(x)}=\int \phi(x) \cdot e^{\phi(x)} \phi^{\prime}(x) d x+C\Rightarrow y \cdot e^{\phi(x)}=\phi(x) e^{\phi(x)}-\int \phi^{\prime}(x) e^{\phi(x)} d x+C\Rightarrow y \cdot e^{\phi(x)}=\phi(x) e^{\phi(x)}-e^{\phi(x)}+C\Rightarrow y=\lceil\phi(x)-1]+C e^{-\phi(x)}
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