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

Question

If x d y=y(d x+y d y), y(1)=1 and y(x)>0, then y(-3) is equal to (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Given, (x d y-y d x/y2)=d y ⇒ d((x/y))=-d y ⇒ (x/y)=-y+C As, y (1)=1 ⇒ C =2 ⇒ ( x / y )+ y =2 Put x=-3, we get -3+y2=2 y ⇒(y+1)(y-3)=0 text As y >0 text So, y(x=-3)=3 .

Q. If $x d y = y (d x + y d y), y (1) = 1$ and $y (x) > 0$ , then $y (- 3)$ is equal to

1

2

3

4

Given,
$\frac{x d y - y d x}{y ^{2}} = d y \Rightarrow d (\frac{x}{y}) = - d y \Rightarrow \frac{x}{y} = - y + C$
As, $y (1) = 1 \Rightarrow C = 2 \Rightarrow \frac{x}{y} + y = 2$
Put $x = - 3$ , we get
$- 3 + y^{2} = 2 y \Rightarrow (y + 1) (y - 3) = 0$
$As y > 0$
$So, y (x = - 3) = 3.$

Q. If xdy=y(dx+ydy),y(1)=1 and y(x)>0, then y(−3) is equal to

1

2

3

4

Given, y2xdy−ydx​=dy⇒d(yx​)=−dy⇒yx​=−y+C As, y(1)=1⇒C=2⇒yx​+y=2 Put x=−3, we get −3+y2=2y⇒(y+1)(y−3)=0 As y>0 So, y(x=−3)=3.

Q. If $x d y = y (d x + y d y), y (1) = 1$ and $y (x) > 0$ , then $y (- 3)$ is equal to

Given,
$\frac{x d y - y d x}{y ^{2}} = d y \Rightarrow d (\frac{x}{y}) = - d y \Rightarrow \frac{x}{y} = - y + C$
As, $y (1) = 1 \Rightarrow C = 2 \Rightarrow \frac{x}{y} + y = 2$
Put $x = - 3$ , we get
$- 3 + y^{2} = 2 y \Rightarrow (y + 1) (y - 3) = 0$
$As y > 0$
$So, y (x = - 3) = 3.$