SATHEE: Differential Equations 2 Question 19

Differential Equations 2 Question 19

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

(a) 3

(b) 2

(d) 0

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Answer:

Correct Answer: 20. (a)

Solution:

Given, $x d y = y (d x + y d y), y > 0$

$\begin{aligned} \Rightarrow x d y - y d x = y^{2} d y \\ \Rightarrow \frac{x d y - y d x}{y^{2}} = d y \Rightarrow d \frac{x}{y} = - d y \end{aligned}$

On integrating both sides, we get

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

Since,

$y (1) = 1 \Rightarrow x = 1, y = 1$

$∴ c = 2$

Now, Eq. (i) becomes, $\frac{x}{y} + y = 2$

Again, for $x = - 3$

$\begin{array}{ll} \Rightarrow & - 3 + y^{2} = 2 y \\ \Rightarrow & y^{2} - 2 y - 3 = 0 \\ \Rightarrow & (y + 1) (y - 3) = 0 \end{array}$

As $y > 0$ , take $y = 3$ , neglecting $y = - 1$ .

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Differential Equations 2 Question 19

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

Answer:

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