If x d y=y(d x+y d y) ; y(1)=1 and y(x) > 0 , then what is y(-3) equal to?

Correct answer: A. 3 aligned & x d y=y d x+y^2 d y & x · (d y)/(d x)=y+y^2 · (d y)/(d x) & (y-x · (d y)/(d x))/(y^2)=(-d y)/(d x) aligned Integrating both sides (x)/(y)=-y+c Given, x=1, y=1 aligned & (1)/(1)=-1+c & c=2 & x y +y=2 & x+y^2=2 y & -3+y^2-2 y=0 & y^2-2 y-3=0 & y=3,-1 & y=3 aligned

NDA 2017

Exam: NDA
Year: 2017
Subject: Mathematics
Chapter: Differential Equations

NDA 2017 · Mathematics · Differential Equations

If x d y=y(d x+y d y) ; y(1)=1 and y(x) > 0 , then what is y(-3)...

x d y=y(d x+y d y) ; y(1)=1

and

y(x) > 0

, then what is

y(-3)

equal to?

Options

Correct answer

A. 3

Step-by-step solution

\begin{aligned} & x d y=y d x+y^2 d y \\ & \Rightarrow x \cdot \frac{d y}{d x}=y+y^2 \cdot \frac{d y}{d x} \\ & \Rightarrow \frac{y-x \cdot \frac{d y}{d x}}{y^2}=\frac{-d y}{d x} \end{aligned}

Integrating both sides

\frac{x}{y}=-y+c

Given,

\mathrm{x}=1, \mathrm{y}=1

\begin{aligned} & \Rightarrow \frac{1}{1}=-1+\mathrm{c} \\ & \Rightarrow \mathrm{c}=2 \\ & \therefore \frac{\mathrm{x}}{\mathrm{y}}+\mathrm{y}=2 \\ & \Rightarrow \mathrm{x}+\mathrm{y}^2=2 \mathrm{y} \\ & \Rightarrow-3+\mathrm{y}^2-2 \mathrm{y}=0 \\ & \Rightarrow \mathrm{y}^2-2 \mathrm{y}-3=0 \\ & \mathrm{y}=3,-1 \\ & \therefore \mathrm{y}=3 \end{aligned}

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