If x d y=y d x+y^2 d y, y > 0 and y(1)=1 , then what is y(-3) equal to?

Correct answer: A. 3 only aligned & Given, x d y=y d x+y^2 d y & 1=(y)/(x) (d x)/(d y)+(y^2)/(x) & (d x)/(d y)+y=(x)/(y) aligned (d x)/(d y)-(x)/(y)=-y ...(1) aligned & P=-(1)/(y), Q=-y & IF=e^ ∫ P d y =e^ ∫-(1)/(y) d y =e^ - y =(1)/(y) aligned Now, Solution of d.E. aligned & x(I . F)=∫(Q . I . F .) dy & (x)/(y)=∫ (1)/(y)(-y) d y+C & (x)/(y)=∫-1 d y+C & (x)/(y)=-y+C & y(1)=1 aligned aligned & (1)/(1)=-1+C C=2 & (x)/(y)=-y+2 x=-y^2+2 y & y(-3) -3=-y^2+2 y & y^2-2 y-3=0 & y= +2 ± 4+12 2 =(2 ± 4)/(2) & y=3,-1 & Since y > 0 so y=3 aligned

NDA 2016

Exam: NDA
Year: 2016
Subject: Mathematics
Chapter: Differential Equations

NDA 2016 · Mathematics · Differential Equations

If x d y=y d x+y^2 d y, y > 0 and y(1)=1 , then what is y(-3) equal...

x d y=y d x+y^2 d y, y > 0

and

y(1)=1

, then what is

y(-3)

equal to?

Options

A3 only
B-1 only
CBoth -1 and 3
DNeither-1 nor 3

Correct answer

A. 3 only

Step-by-step solution

\begin{aligned} & \text{ Given, } x d y=y d x+y^2 d y \\ & \Rightarrow 1=\frac{y}{x} \frac{d x}{d y}+\frac{y^2}{x} \\ & \Rightarrow \frac{d x}{d y}+y=\frac{x}{y}\end{aligned}

\Rightarrow \frac{d x}{d y}-\frac{x}{y}=-y

...(1)

\begin{aligned} & P=-\frac{1}{y}, Q=-y \\ & \mathrm{IF}=e^{\int P d y}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=\frac{1}{y} \end{aligned}

Now, Solution of d.E.

\begin{aligned} & \mathrm{x}(\mathrm{I} . \mathrm{F})=\int(\mathrm{Q} . \mathrm{I} . \mathrm{F} .) \mathrm{dy} \\ & \frac{x}{y}=\int \frac{1}{y}(-y) d y+C \\ & \Rightarrow \frac{x}{y}=\int-1 d y+C \\ & \Rightarrow \frac{x}{y}=-y+C \\ & y(1)=1 \end{aligned}

$\begin{aligned} & \frac{1}{1}=-1+C \Rightarrow C=2 \\ & \Rightarrow \frac{x}{y}=-y+2 \Rightarrow x=-y^2+2 y \\ & \Rightarrow y(-3) \Rightarrow-3=-y^2+2 y \\ & \Rightarrow y^2-2 y-3=0 \\ & \Rightarrow y=\frac{+2 \pm \sqrt{4+12}}{2}=\frac{2 \pm 4}{2} \\ & \Rightarrow y=3,-1 \\ & \text{Since } y > 0 \text{ so } y=3\end{aligned}

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